Instructor Solutions Manual for Introduction to Mathematical Statistics and Its Applications [5 ed.] 9780321694010, 0321694015

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INSTRUCTOR’S SOLUTIONS MANUAL

AN I NTRODUCTION TO MATHEMATICAL S TATISTICS AND ITS

APPLICATIONS

F IFTH E DITION

Richard J. Larsen Vanderbilt University

Morris L. Marx University of West Florida

This should be only distributed free of cost. If you have paid for this from an online solution manual vendor, you have been cheated. Copyright © 2012, 2006, 2001 Pearson Education, Inc. Publishing as Prentice Hall, 75 Arlington Street, Boston, MA 02116. All rights reserved. This manual may be reproduced for classroom use only. ISBN-13: 978-0-321-69401-0 ISBN-10: 0-321-69401-5

Contents

Chapter 2: Probability..................................................................................................................................................1 2.2 2.3 2.4 2.5 2.6 2.7

Samples Spaces and the Algebra of Sets ..........................................................................................................1 The Probability Function ..................................................................................................................................5 Conditional Probability.....................................................................................................................................7 Independence..................................................................................................................................................13 Combinatorics.................................................................................................................................................17 Combinatorial Probability ..............................................................................................................................23

Chapter 3: Random Variables....................................................................................................................................27 3.2 Binomial and Hypergeometric Probabilities ..................................................................................................27 3.3 Discrete Random Variables ............................................................................................................................34 3.4 Continuous Random Variables.......................................................................................................................37 3.5 Expected Values .............................................................................................................................................39 3.6 The Variance...................................................................................................................................................45 3.7 Joint Densities.................................................................................................................................................49 3.8 Transforming and Combining Random Variables..........................................................................................58 3.9 Further Properties of the Mean and Variance.................................................................................................60 3.10 Order Statistics ............................................................................................................................................64 3.11 Conditional Densities ..................................................................................................................................67 3.12 Moment-Generating Functions....................................................................................................................71

Chapter 4: Special Distributions ................................................................................................................................75 4.2 4.3 4.4 4.5 4.6

The Poisson Distribution ................................................................................................................................75 The Normal Distribution ................................................................................................................................80 The Geometric Distribution............................................................................................................................87 The Negative Binomial Distribution ..............................................................................................................89 The Gamma Distribution ................................................................................................................................91

Chapter 5: Estimation ................................................................................................................................................93 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Estimating Parameters: The Method of Maximum Likelihood and Method of Moments .............................93 Interval Estimation .........................................................................................................................................98 Properties of Estimators................................................................................................................................102 Minimum-Variance Estimators: The Cramér-Rao Lower Bound ................................................................105 Sufficient Estimators ....................................................................................................................................107 Consistency...................................................................................................................................................109 Bayesian Estimation .....................................................................................................................................111

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Contents

Chapter 6: Hypothesis Testing.................................................................................................................................113 6.2 6.3 6.4 6.5

The Decision Rule ........................................................................................................................................113 Testing Binomial Data - H0: p = po.......................................................................................................................................................................... 114 Type I and Type II Errors .............................................................................................................................115 A Notion of Optimality: The Generalized Likelihood Ratio........................................................................119

Chapter 7: Inferences Based on the Normal Distribution ........................................................................................121 7.3 Deriving the Distribution of

Y −µ

...............................................................................................................121

S/ n

7.4 Drawing Inferences about µ .........................................................................................................................123 2 7.5 Drawing Inferences about σ ........................................................................................................................127

Chapter 8: Types of Data: A Brief Overview ..........................................................................................................131 8.2 Classifying Data ...........................................................................................................................................131

Chapter 9: Two-Sample Inference ...........................................................................................................................133 9.2 9.3 9.4 9.5

Testing H 0 : µ X = µY .......................................................................................................................................133 Testing H 0 : σ X2 = σ Y2 —The F Test .................................................................................................................136 Binomial Data: Testing H 0 : p X = pY ...........................................................................................................138 Confidence Intervals for the Two-Sample Problem .....................................................................................140

Chapter 10: Goodness-of-Fit Tests ..........................................................................................................................143 10.2 10.3 10.4 10.5

The Multinomial Distribution ...................................................................................................................143 Goodness-of-Fit Tests: All Parameters Known.........................................................................................145 Goodness-of-Fit Tests: Parameters Unknown...........................................................................................148 Contingency Tables...................................................................................................................................154

Chapter 11: Regression............................................................................................................................................159 11.2 11.3 11.4 11.5

The Method of Least Squares....................................................................................................................159 The Linear Model......................................................................................................................................169 Covariance and Correlation.......................................................................................................................174 The Bivariate Normal Distribution............................................................................................................178

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Contents

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Chapter 12: The Analysis of Variance.....................................................................................................................181 12.2 The F test...................................................................................................................................................181 12.3 Multiple Comparisons: Tukey’s Method ..................................................................................................184 12.4 Testing Subhypotheses with Constrasts ....................................................................................................186 12.5 Data Transformations ................................................................................................................................188 Appendix 12.A.3 The Distribution of SSTR / (k − 1) When H1 Is True................................................................188 SSE / (n − k)

Chapter 13: Randomized Block Designs .................................................................................................................191 13.2 The F Test for a Randomized Block Design .............................................................................................191 13.3 The Paired t Test .......................................................................................................................................195

Chapter 14: Nonparametric Statistics ......................................................................................................................199 14.2 14.3 14.4 14.5 14.6

The Sign Test ............................................................................................................................................199 Wilcoxon Tests..........................................................................................................................................202 The Kruskal-Wallis Test ...........................................................................................................................206 The Friedman Test ....................................................................................................................................210 Testing for Randomness............................................................................................................................212

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Chapter 2: Probability Section 2.2: Sample Spaces and the Algebra of Sets 2.2.1

S = {( s, s, s ), ( s, s, f ), ( s, f , s ), ( f , s, s ), ( s, f , f ), ( f , s, f ), ( f , f , s ), ( f , f , f )} A=

{ (s, f , s), ( f , s, s)} ; B = {( f , f , f )}

2.2.2

Let (x, y, z) denote a red x, a blue y, and a green z. Then A = {(2,2,1), (2,1,2), (1, 2,2), (1,1,3), (1,3,1), (3,1,1)}

2.2.3

(1,3,4), (1,3,5), (1,3,6), (2,3,4), (2,3,5), (2,3,6)

2.2.4

There are 16 ways to get an ace and a 7, 16 ways to get a 2 and a 6, 16 ways to get a 3 and a 5, and 6 ways to get two 4’s, giving 54 total.

2.2.5

The outcome sought is (4, 4). It is “harder” to obtain than the set {(5, 3), (3, 5), (6, 2), (2, 6)} of other outcomes making a total of 8.

2.2.6

The set N of five card hands in hearts that are not flushes are called straight flushes. These are five cards whose denominations are consecutive. Each one is characterized by the lowest value in the hand. The choices for the lowest value are A, 2, 3, …, 10. (Notice that an ace can be high or low). Thus, N has 10 elements.

2.2.7

P = {right triangles with sides (5, a, b): a2 + b2 = 25}

2.2.8

A = {SSBBBB, SBSBBB, SBBSBB, SBBBSB, BSSBBB, BSBSBB, BSBBSB, BBSSBB, BBSBSB, BBBSSB}

2.2.9

(a) S = {(0, 0, 0, 0) (0, 0, 0, 1), (0, 0, 1, 0), (0, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1, ), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1, )} (b) A = {(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0, )} (c) 1 + k

2.2.10 (a) S = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)} (b) {2, 3, 4, 5, 6, 8} 2.2.11 Let p1 and p2 denote the two perpetrators and i1, i2, and i3, the three in the lineup who are innocent. Then S = {( p1 , i1 ), ( p1 , i2 ), ( p1 , i3 ), ( p2 , i1 ), ( p2 , i2 ), ( p2 , i3 ), ( p1 , p2 ), (i1 , i2 ), (i1 , i3 ), (i2 , i3 )} . The event A contains every outcome in S except (p1, p2). 2.2.12 The quadratic equation will have complex roots—that is, the event A will occur—if b2 − 4ac < 0.

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Chapter 2: Probability

2.2.13 In order for the shooter to win with a point of 9, one of the following (countably infinite) sequences of sums must be rolled: (9,9), (9, no 7 or no 9,9), (9, no 7 or no 9, no 7 or no 9,9), … 2.2.14 Let (x, y) denote the strategy of putting x white chips and y black chips in the first urn (which results in 10 − x white chips and 10 − y black chips being in the second urn). Then S = {( x, y ) : x = 0,1,...,10, y = 0,1,...,10, and 1 ≤ x + y ≤ 19} . Intuitively, the optimal strategies are (1, 0) and (9, 10). 2.2.15 Let Ak be the set of chips put in the urn at 1/2k minute until midnight. For example, A1 = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. Then the set of chips in the urn at midnight is ∞

∪(A

k

− {k + 1}) = ∅ .

k =1

2.2.16

2.2.17 If x2 + 2x ≤ 8, then (x + 4)(x − 2) ≤ 0 and A = {x: −4 ≤ x ≤ 2}. Similarly, if x2 + x ≤ 6, then (x + 3)(x − 2) ≤ 0 and B = {x: −3 ≤ x ≤ 2). Therefore, A ∩ B = {x: −3 ≤ x ≤ 2} and A ∪ B = {x: −4 ≤ x ≤ 2}. 2.2.18 A ∩ B ∩ C = {x: x = 2, 3, 4} 2.2.19 The system fails if either the first pair fails or the second pair fails (or both pairs fail). For either pair to fail, though, both of its components must fail. Therefore, A = (A11 ∩ A21) ∪ (A12 ∩ A22). 2.2.20 (a)

(c)

(b)

empty set

_____________________ −∞ ∞

(d)

2.2.21 40 2.2.22 (a) {E1, E2}

(b) {S1, S2, T1, T2}

(c) {A, I}

2.2.23 (a) If s is a member of A ∪ (B ∩ C) then s belongs to A or to B ∩ C. If it is a member of A or of B ∩ C, then it belongs to A ∪ B and to A ∪ C. Thus, it is a member of (A ∪ B) ∩ (A ∪ C). Conversely, choose s in (A ∪ B) ∩ (A ∪ C). If it belongs to A, then it belongs to A ∪ (B ∩ C). If it does not belong to A, then it must be a member of B ∩ C. In that case it also is a member of A ∪ (B ∩ C).

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Section 2.2: Sample Spaces and the Algebra of Sets

3

(b) If s is a member of A ∩ (B ∪ C) then s belongs to A and to B ∪ C. If it is a member of B, then it belongs to A ∩ B and, hence, (A ∩ B) ∪ (A ∩ C). Similarly, if it belongs to C, it is a member of (A ∩ B) ∪ (A ∩ C). Conversely, choose s in (A ∩ B) ∪ (A ∩ C). Then it belongs to A. If it is a member of A ∩ B then it belongs to A ∩ (B ∪ C). Similarly, if it belongs to A ∩ C, then it must be a member of A ∩ (B ∪ C). 2.2.24 Let B = A1 ∪ A2 ∪ … ∪ Ak. Then A1C ∩ A2C ∩ ... ∩ AkC = (A1 ∪ A2 ∪ …∪ Ak)C = BC. Then the expression is simply B ∪ BC = S. 2.2.25 (a) Let s be a member of A ∪ (B ∪ C). Then s belongs to either A or B ∪ C (or both). If s belongs to A, it necessarily belongs to (A ∪ B) ∪ C. If s belongs to B ∪ C, it belongs to B or C or both, so it must belong to (A ∪ B) ∪ C. Now, suppose s belongs to (A ∪ B) ∪ C. Then it belongs to either A ∪ B or C or both. If it belongs to C, it must belong to A ∪ (B ∪ C). If it belongs to A ∪ B, it must belong to either A or B or both, so it must belong to A ∪ (B ∪ C). (b) Suppose s belongs to A ∩ (B ∩ C), so it is a member of A and also B ∩ C. Then it is a member of A and of B and C. That makes it a member of (A ∩ B) ∩ C. Conversely, if s is a member of (A ∩ B) ∩ C, a similar argument shows it belongs to A ∩ (B ∩ C). 2.2.26 (a) (b) (c) (d) (e)

AC ∩ BC ∩ CC A∩B∩C A ∩ BC ∩ CC (A ∩ BC ∩ CC) ∪ (AC ∩ B ∩ CC) ∪ (AC ∩ BC ∩ C) (A ∩ B ∩ CC) ∪ (A ∩ BC ∩ C) ∪ (AC ∩ B ∩ C)

2.2.27 A is a subset of B. 2.2.28 (a) {0} ∪ {x: 5 ≤ x ≤ 10} (d) {x: 0 < x < 3} 2.2.29 (a) B and C

(b) {x: 3 ≤ x < 5} (e) {0} ∪ {x : 3 ≤ x ≤ 10}

(c) {x: 0 < x ≤ 7} (f) {0} ∪ {x : 7 < x ≤ 10}

(b) B is a subset of A.

2.2.30 (a) A1 ∩ A2 ∩ A3 (b) A1 ∪ A2 ∪ A3 The second protocol would be better if speed of approval matters. For very important issues, the first protocol is superior. 2.2.31 Let A and B denote the students who saw the movie the first time and the second time, respectively. Then N(a) = 850, N(b) = 690, and N [( A ∪ B)C ] = 4700 (implying that N(A ∪ B) = 1300). Therefore, N(A ∩ B) = number who saw movie twice = 850 + 690 − 1300 = 240. 2.2.32 (a)

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4

Chapter 2: Probability

(b)

2.2.33 (a)

(b)

2.2.34 (a)

A ∪ (B ∪ C)

(A ∪ B) ∪ C

(b)

A ∩ (B ∩ C)

(A ∩ B) ∩ C

2.2.35 A and B are subsets of A ∪ B. 2.2.36 (a)

( A ∩ B C )C = AC ∪ B

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Section 2.3: The Probability Function

5

(b)

B ∪ ( A ∪ B )C = AC ∪ B

(c)

A ∩ ( A ∩ B )C = A ∩ B C

2.2.37 Let A be the set of those with MCAT scores ≥ 27 and B be the set of those with GPAs ≥ 3.5. We are given that N(a) = 1000, N(b) = 400, and N(A ∩ B) = 300. Then N ( AC ∩ B C ) = N [( A ∪ B)C ] = 1200 − N(A ∪ B) = 1200 − [(N(a) + N(b) − N(A ∩ B)] = 1200 − [(1000 + 400 − 300] = 100. The requested proportion is 100/1200. 2.2.38

N(A ∪ B ∪ C) = N(a) + N(b) + N(c) − N(A ∩ B) − N(A ∩ C) − N(B ∩ C) + N(A ∩ B ∩ C) 2.2.39 Let A be the set of those saying “yes” to the first question and B be the set of those saying “yes” to the second question. We are given that N(a) = 600, N(b) = 400, and N(AC ∩ B) = 300. Then N(A ∩ B) = N(b) − N ( AC ∩ B) = 400 − 300 = 100. N ( A ∩ B C ) = N(a) − N(A ∩ B) = 600 − 100 = 500. 2.2.40 N [( A ∩ B)C ] = 120 − N(A ∪ B) = 120 − [N( AC ∩ B) + N(A ∩ B C ) + N(A ∩ B)] = 120 − [50 + 15 + 2] = 53

Section 2.3: The Probability Function 2.3.1

Let L and V denote the sets of programs with offensive language and too much violence, respectively. Then P(L) = 0.42, P(V) = 0.27, and P(L ∩ V) = 0.10. Therefore, P(program complies) = P((L ∪ V)C) = 1 − [P(L) + P(V) − P(L ∩ V)] = 0.41.

2.3.2

P(A or B but not both) = P(A ∪ B) − P(A ∩ B) = P(a) + P(b) − P (A ∩ B) − P(A ∩ B) = 0.4 + 0.5 − 0.1 − 0.1 = 0.7

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6

Chapter 2: Probability

2.3.3

(a) 1 − P(A ∩ B)

(b) P(b) − P(A ∩ B)

2.3.4

P(A ∪ B) = P(a) + P(b) − P(A ∩ B) = 0.3; P(a) − P(A ∩ B) = 0.1. Therefore, P(b) = 0.2. 3

2.3.5

1 ⎛5⎞ No. P(A1 ∪ A2 ∪ A3) = P(at least one “6” appears) = 1 − P(no 6’s appear) = 1 − ⎜ ⎟ ≠ . ⎝6⎠ 2 The Ai’s are not mutually exclusive, so P(A1 ∪ A2 ∪ A3) ≠ P(A1) + P(A2) + P(A3).

2.3.6

P(A or B but not both) = 0.5 – 0.2 = 0.3

2.3.7

By inspection, B = (B ∩ A1) ∪ (B ∩ A2) ∪ … ∪ (B ∩ An). 2.3.8

(a)

(b)

(b)

2.3.9

P(odd man out) = 1 − P(no odd man out) = 1 − P(HHH or TTT) = 1 −

2 3 = 8 4

2.3.10 A = {2, 4, 6, …, 24}; B = {3, 6, 9, …, 24); A ∩ B = {6, 12, 18, 24}. 12 8 4 16 + − = . Therefore, P(A ∪ B) = P(a) + P(b) − P(A ∩ B) = 24 24 24 24 2.3.11 Let A: State wins Saturday and B: State wins next Saturday. Then P(a) = 0.10, P(b) = 0.30, and P(lose both) = 0.65 = 1 − P(A ∪ B), which implies that P(A ∪ B) = 0.35. Therefore, P(A ∩ B) = 0.10 + 0.30 − 0.35 = 0.05, so P(State wins exactly once) = P(A ∪ B) − P(A ∩ B) = 0.35 − 0.05 = 0.30.

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Section 2.4: Conditional Probability

7

2.3.12 Since A1 and A2 are mutually exclusive and cover the entire sample space, p1 + p2 = 1. 1 5 But 3p1 − p2 = , so p2 = . 2 8 2.3.13 Let F: female is hired and T: minority is hired. Then P(f) = 0.60, P(T) = 0.30, and P(FC ∩ TC) = 0.25 = 1 − P(F ∪ T). Since P(F ∪ T) = 0.75, P(F ∩ T) = 0.60 + 0.30 − 0.75 = 0.15. 2.3.14 The smallest value of P[(A ∪ B∪ C)C] occurs when P(A ∪ B ∪ C) is as large as possible. This, in turn, occurs when A, B, and C are mutually disjoint. The largest value for P(A ∪ B ∪ C) is P(a) + P(b) + P(c) = 0.2 + 0.1 + 0.3 = 0.6. Thus, the smallest value for P[(A ∪ B ∪ C)C] is 0.4. 2.3.15 (a) XC ∩ Y = {(H, T, T, H), (T, H, H, T)}, so P(XC ∩ Y) = 2/16 (b) X ∩ YC = {(H, T, T, T), (T, T, T, H), (T, H, H, H), (H, H, H, T)} so P(X ∩ YC) = 4/16 2.3.16 A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} A ∩ BC = {(1, 5), (3, 3), (5, 1)}, so P(A ∩ BC) = 3/36 = 1/12. 2.3.17 A ∩ B, (A ∩ B) ∪ (A ∩ C), A, A ∪ B, S 2.3.18 Let A be the event of getting arrested for the first scam; B, for the second. We are given P(a) = 1/10, P(b) = 1/30, and P(A ∩ B) = 0.0025. Her chances of not getting arrested are P[(A ∪ B)C] = 1 − P(A ∪ B) = 1 − [P(a) + P(b) − P(A ∩ B)] = 1 − [1/10 + 1/30 − 0.0025] = 0.869

Section 2.4: Conditional Probability P (sum = 10 and sum exceeds 8) P (sum exceeds 8) P(sum = 10) 3 / 36 3 = = = . P (sum = 9,10,11, or 12) 4 / 36 + 3 / 36 + 2 / 36 + 1/ 36 10

2.4.1

P(sum = 10|sum exceeds 8) =

2.4.2

P(A|B) + P(B|A) = 0.75 =

P ( A ∩ B ) P( A ∩ B) 10 P ( A ∩ B ) + = + 5 P ( A ∩ B ) , which implies that P( B ) P ( A) 4

P(A ∩ B) = 0.1. P( A ∩ B) < P ( A) , then P(A ∩ B) < P(a) ⋅ P(b). It follows that P( B ) P ( A ∩ B ) P( A) ⋅ P ( B ) < P(B|A) = = P(b). P( A) P ( A)

2.4.3

If P(A|B) =

2.4.4

P(E|A ∪ B) =

2.4.5

The answer would remain the same. Distinguishing only three family types does not make them equally likely; (girl, boy) families will occur twice as often as either (boy, boy) or (girl, girl) families.

P ( E ∩ ( A ∪ B )) P( E ) P ( A ∪ B ) − P ( A ∩ B) 0.4 − 0.1 3 = = = = . P( A ∪ B ) P( A ∪ B) P( A ∪ B) 0.4 4

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8

Chapter 2: Probability

2.4.6

P(A ∪ B) = 0.8 and P(A ∪ B) − P(A ∩ B) = 0.6, so P(A ∩ B) = 0.2. Also, P(A|B) = 0.6 = P( A ∩ B) 0.2 1 1 2 = and P(a) = 0.8 + 0.2 − = . , so P(b) = P( B ) 0.6 3 3 3

2.4.7

Let Ri be the event that a red chip is selected on the ith draw, i = 1, 2. 3 1 3 Then P(both are red) = P(R1 ∩ R2) = P(R2 | R1)P(R1) = ⋅ = . 4 2 8

2.4.8

P(A|B) =

2.4.9

Let Wi be the event that a white chip is selected on the ith draw, i = 1,2 . Then P (W1 ∩ W2 ) . If both chips in the urn are white, P(W1) = 1; P(W2|W1) = P (W1 ) 1 if one is white and one is black, P(W1) = . Since each chip distribution is equally likely, 2 1 1 1 3 1 1 1 5 5/8 5 = . P(W1) = 1 ⋅ + ⋅ = . Similarly, P(W1 ∩ W2) = 1 ⋅ + ⋅ = , so P(W2|W1) = 2 2 2 4 2 4 2 8 3/ 4 6

P ( A ∩ B ) P ( A) + P( B ) − P( A ∪ B) a + b − P ( A ∪ B ) = = . P( B ) P( B) b a + b −1 . But P(A ∪ B) ≤ 1, so P(A|B) ≥ b

2.4.10 P[(A ∩ B)| (A ∪ B)C] =

P[( A ∩ B ) ∩ ( A ∪ B )C ] P(∅) = =0 C P[( A ∪ B ) ] P[( A ∪ B )C ]

2.4.11 (a) P(AC ∩ BC) = 1 − P(A ∪ B) = 1 − [P(a) + P(b) − P(A ∩ B)] = 1 − [0.65 + 0.55 − 0.25] = 0.05 (b) P[(AC ∩ B) ∪ (A ∩ BC)] = P(AC ∩ B) + P(A ∩ BC) = [P(a) − P(A ∩ B)] + [P(b) − P(A ∩ B)] = [0.65 − 0.25] + [0.55 − 0.25] = 0.70 (c) P(A ∪ B) = 0.95 (d) P[(A ∩ B)C] = 1 − P(A ∩ B) = 1 − 0.25 = 0.75 (e) P{[(AC ∩ B) ∪ (A ∩ BC)]| A ∪ B} =

P[( AC ∩ B ) ∪ ( A ∩ B C )] = 0.70/0.95 = 70/95 P( A ∪ B )

(f) P(A ∩ B)| A ∪ B) = P(A ∩ B)/P(A ∪ B) = 0.25/0.95 = 25/95 (g) P(B|AC) = P(AC ∩ B)/P(AC) ] = [P(b) − P(A ∩ B)]/[1 − P(a)] = [0.55 − 0.25]/[1 − 0.65] = 30/35 2.4.12 P(No. of heads ≥ 2| No. of heads ≤ 2) = P(No. of heads ≥ 2 and No. of heads ≤ 2)/P(No. of heads ≤ 2) = P(No. of heads = 2)/P(No. of heads ≤ 2) = (3/8)/(7/8) = 3/7

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Section 2.4: Conditional Probability

9

2.4.13 P(first die ≥ 4|sum = 8) = P(first die ≥ 4 and sum = 8)/P(sum = 8) = P({(4, 4), (5, 3), (6, 2)}/P({(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}) = 3/5 2.4.14 There are 4 ways to choose three aces (count which one is left out). There are 48 ways to choose the card that is not an ace, so there are 4 × 48 = 192 sets of cards where exactly three are aces. That gives 193 sets where there are at least three aces. The conditional probability is (1/270,725)/(193/270,725) = 1/193. 2.4.15 First note that P(A ∪ B) = 1 − P[(A ∪ B)C] = 1 − 0.2 = 0.8. Then P(b) = P(A ∪ B) − P(A ∩ BC) − P(A ∩ B) = 0.8 − 0.3 − 0.1 = 0.5. Finally P(A|B) = P(A∩ B)/P(b) = 0.1/0.5 = 1/5 2.4.16 P(A|B) = 0.5 implies P(A ∩ B) = 0.5P(b). P(B|A) = 0.4 implies P(A ∩ B) = (0.4)P(a). Thus, 0.5P(b) = 0.4P(a) or P(b) = 0.8P(a). Then, 0.9 = P(a) + P(b) = P(a) + 0.8P(a) or P(a) = 0.9/1.8 = 0.5. 2.4.17 P[(A ∩ B)C] = P[(A ∪ B)C] + P(A ∩ BC) + P(AC ∩ B) = 0.2 + 0.1 + 0.3 = 0.6 P(A ∪ B|(A ∩ B)C) = P[(A ∩ BC) ∪ (AC ∩ B)]/P((A ∩ B)C) = [0.1 + 0.3]/0.6 = 2/3 2.4.18 P(sum ≥ 8|at least one die shows 5) = P(sum ≥ 8 and at least one die shows 5)/P(at least one die shows 5) = P({(5, 3), (5, 4), (5, 6), (3, 5), (4, 5), (6, 5), (5, 5)})/(11/36) = 7/11 2.4.19 P(Outandout wins|Australian Doll and Dusty Stake don’t win) = P(Outandout wins and Australian Doll and Dusty Stake don’t win)/P(Australian Doll and Dusty Stake don’t win) = 0.20/0.55 = 20/55 2.4.20 Suppose the guard will randomly choose to name Bob or Charley if they are the two to go free. Then the probability the guard will name Bob, for example, is P(Andy, Bob) + (1/2)P(Bob, Charley) = 1/3 + (1/2)(1/3) = 1/2. The probability Andy will go free given the guard names Bob is P(Andy, Bob)/P(Guard names Bob) = (1/3)/(1/2) = 2/3. A similar argument holds for the guard naming Charley. Andy’s concern is not justified. 2.4.21 P(BBRWW) = P(b)P(B|B)P(R|BB)P(W|BBR)P(W|BBRW) = =.0050. P(2, 6, 4, 9, 13) =

4 3 5 6 5 ⋅ ⋅ ⋅ ⋅ 15 14 13 12 11

1 1 1 1 1 1 . ⋅ ⋅ ⋅ ⋅ = 15 14 13 12 11 360,360

2.4.22 Let Ki be the event that the ith key tried opens the door, i = 1, 2, …, n. Then P(door opens first time with 3rd key) = P ( K1C ∩ K 2C ∩ K 3 ) = P ( K1C ) ⋅ P ( K 2C K1C ) ⋅ P ( K 3 K1C ∩ K 2C ) = n −1 n − 2 1 1 ⋅ ⋅ = . n n −1 n − 2 n

2.4.23 (1/52)(1/51)(1/50)(1/49) = 1/6,497,400 2.4.24 (1/2)(1/2)(1/2)(2/3)(3/4) = 1/16

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10

Chapter 2: Probability

2.4.25 Let Ai be the event “Bearing came from supplier i”, i = 1, 2, 3. Let B be the event “Bearing in toy manufacturer’s inventory is defective.” Then P(A1) = 0.5, P(A2) = 0.3, P(A3 ) = 0.2 and P(B|A1) = 0.02, P(B|A2) = 0.03, P(B|A3) = 0.04 Combining these probabilities according to Theorem 2.4.1 gives P(b) = (0.02)(0.5) + (0.03)(0.3) + (0.04)(0.2) = 0.027 meaning that the manufacturer can expect 2.7% of her ball-bearing stock to be defective. 2.4.26 Let B be the event that the face (or sum of faces) equals 6. Let A1 be the event that a Head comes up and A2, the event that a Tail comes up. Then P(b) = P(B|A1)P(A1) + P(B|A2)P(A2) 1 1 5 1 = ⋅ + ⋅ = 0.15. 6 2 36 2 2.4.27 Let B be the event that the countries go to war. Let A be the event that terrorism increases. Then P(b) = P(B|A)P(a) + P(B|AC)P(AC) = (0.65)(0.30) + (0.05)(0.70) = 0.23. 2.4.28 Let B be the event that a donation is received; let A1, A2, and A3 denote the events that the call is placed to Belle Meade, Oak Hill, and Antioch, respectively. 3 1000 1000 2000 Then P(b) = P ( B Ai ) P ( Ai ) = (0.60) ⋅ + (0.55) ⋅ + (0.35) ⋅ = 0.46 . 4000 4000 4000 i =1

∑

2.4.29 Let B denote the event that the person interviewed answers truthfully, and let A be the event that the person interviewed is a man. Then P(b) = P(B|A)P(a) + P(B|AC)P(AC) = (0.78)(0.47) + (0.63)(0.53) = 0.70. 2.4.30 Let B be the event that a red chip is ultimately drawn from Urn I. Let ARW, for example, denote the event that a red is transferred from Urn I and a white is transferred from Urn II. Then P(b) = P(B|ARR)P(ARR) + P(B|ARW)P(ARW) + P(B|AWR)P(AWR) + P(B|AWW)P(AWW) 3 ⎛ 3 2 ⎞ 2 ⎛ 3 2 ⎞ ⎛ 1 2 ⎞ 3 ⎛ 1 2 ⎞ 11 = ⎜ ⋅ ⎟ + ⎜ ⋅ ⎟ + 1⎜ ⋅ ⎟ + ⎜ ⋅ ⎟ = . 4 ⎝ 4 4 ⎠ 4 ⎝ 4 4 ⎠ ⎝ 4 4 ⎠ 4 ⎝ 4 4 ⎠ 16 2.4.31 Let B denote the event that someone will test positive, and let A denote the event that someone is infected. Then P(b) = P(B|A)P(a) + P(B|AC)P(AC) = (0.999)(0.0001) + (0.0001)(0.9999) = 0.00019989. 2.4.32 The optimal allocation has 1 white chip in one urn and the other 19 chips (9 white and 10 black) 1 9 1 in the other urn. Then P(white is drawn) = 1 ⋅ + ⋅ = 0.74. 2 19 2 2.4.33 If B is the event that Backwater wins and A is the event that their first-string quarterback plays, then P(b) = P(B|A)P(a) + P(B|AC)P(AC) = (0.75)(0.70) + (0.40)(0.30) = 0.645. 2.4.34 Since the identities of the six chips drawn are not known, their selection does not affect any probability associated with the seventh chip. Therefore, 40 . P(seventh chip drawn is red) = P(first chip drawn is red) = 100

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Section 2.4: Conditional Probability

11

2.4.35 No. Let B denote the event that the person calling the toss is correct. Let AH be the event that the coin comes up Heads and let AT be the event that the coin comes up Tails. 1 ⎛1⎞ ⎛1⎞ Then P(b) = P(B|AH)P(AH) + P(B|AT)P(AT) = (0.7) ⎜ ⎟ + (0.3) ⎜ ⎟ = . ⎝2⎠ ⎝2⎠ 2 2.4.36 Let B be the event of a guilty verdict; let A be the event that the defense can discredit the police. Then P(b) = P(B|A)P(a) + P(B|AC)P(AC) = 0.15(0.70) + 0.80(0.30) = 0.345 2.4.37 Let A1 be the event of a 3.5-4.0 GPA; A2, of a 3.0-3.5 GPA; and A3, of a GPA less than 3.0. If B is the event of getting into medical school, then P(b) = P(B|A1)P(A1) + P(B|A2)P(A2) + P(B|A3)P(A3) = (0.8)(0.25) + (0.5)(0.35) + (0.1)(0.40) = 0.415 2.4.38 Let B be the event of early release; let A be the event that the prisoner is related to someone on the governor’s staff. Then P (b) = P(B|A)P(a) + P(B|AC)P(AC) = (0.90)(0.40) + (0.01)(0.60) = 0.366 2.4.39 Let A1 be the event of being a Humanities major; A2, of being a Natural Science major; A3, of being a History major; and A4, of being a Social Science major. If B is the event of a male student, then P(b) = P(B|A1)P(A1) + P(B|A2)P(A2) + P(B|A3)P(A3) + P(B|A4)P(A4) = (0.40)(0.4) + (0.85)(0.1) + (0.55)(0.3) + (0.25)(0.2) = 0.46 2.4.40 Let B denote the event that the chip drawn from Urn II is red; let AR and AW denote the events that the chips transferred are red and white, respectively. P( B | AW ) P ( AW ) (2 / 4)(2 / 3) 4 Then P ( AW | B ) = = = P ( B | AR ) P( AR ) + P ( B | AW ) P ( AW ) (3 / 4)(1/ 3) + (2 / 4)(2 / 3) 7 2.4.41 Let Ai be the event that Urn i is chosen, i = I, II, III. Then, P(Ai) = 1/3, i = I, II, III. Suppose B is the event a red chip is drawn. Note that P(B|A1) = 3/8, P(B|A2) = 1/2 and P(B|A3) = 5/8. P ( A3 | B ) =

=

P ( B | A3 ) P ( A3 ) P ( B | A1 ) P ( A1 ) + P ( B | A2 ) P ( A2 ) + P ( B | A3 ) P ( A3 )

(5 / 8)(1/ 3) = 5/12. (3 / 8)(1/ 3) + (1/ 2)(1/ 3) + (5 / 8)(1/ 3)

2.4.42 If B is the event that the warning light flashes and A is the event that the oil pressure is low, then P(A|B) =

P ( B | A) P ( A) (0.99)(0.10) = = 0.85 C C P ( B | A) P ( A) + P ( B | A ) P ( A ) (0.99)(0.10) + (0.02)(0.90)

2.4.43 Let B be the event that the basement leaks, and let AT, AW, and AH denote the events that the house was built by Tara, Westview, and Hearthstone, respectively. Then P(B|AT) = 0.60, P(B|AW) = 0.50, and P(B|AH) = 0.40. Also, P(AT) = 2/11, P(AW) = 3/11, and P(AH) = 6/11. Applying Bayes’ rule to each of the builders shows that P(AT|B) = 0.24, P(AW|B) = 0.29, and P(AH|B) = 0.47, implying that Hearthstone is the most likely contractor.

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12

Chapter 2: Probability

2.4.44 Let B denote the event that Francesca passed, and let AX and AY denote the events that she was enrolled in Professor X’s section and Professor Y’s section, respectively. Since P(B|AX) = 0.85, P(B|AY) = 0.60, P(AX) = 0.4, and P(AY) = 0.6, (0.85)(0.4) = 0.486 P(AX|B) = (0.85)(0.4) + (0.60)(0.6) 2.4.45 Let B denote the event that a check bounces, and let A be the event that a customer wears sunglasses. Then P(B|A) = 0.50, P(B|AC) = 1 − 0.98 = 0.02, and P(a) = 0.10, so P(A|B) =

(0.50)(0.10) = 0.74 (0.50)(0.10) + (0.02)(0.90)

2.4.46 Let B be the event that Basil dies, and define A1, A2, and A3 to be the events that he ordered cherries flambe, chocolate mousse, or no dessert, respectively. Then P(B|A1) = 0.60, P(B|A2) = 0.90, P(B|A3) = 0, P(A1) = 0.50, P(A2) = 0.40, and P(A3) = 0.10. Comparing P(A1|B) and P(A2|B) suggests that Margo should be considered the prime suspect: P(A1|B) =

(0.60)(0.50) = 0.45 (0.60)(0.50) + (0.90)(0.40) + (0)(0.10)

P(A2|B) =

(0.90)(0.40) = 0.55 (0.60)(0.50) + (0.90)(0.40) + (0)(0.10)

2.4.47 Define B to be the event that Josh answers a randomly selected question correctly, and let A1 and A2 denote the events that he was 1) unprepared for the question and 2) prepared for the question, respectively. Then P(B|A1) = 0.20, P(B|A2) = 1, P(A2) = p, P(A1) = 1 − p, and P ( B | A2 ) P( A2 ) 1⋅ p = P ( B | A1 ) P ( A1 ) + P ( B | A2 ) P ( A2 ) (0.20)(1 − p ) + (1 ⋅ p ) which implies that p = 0.70 (meaning that Josh was prepared for (0.70)(20) = 14 of the questions).

P(A2|B) = 0.92 =

2.4.48 Let B denote the event that the program diagnoses the child as abused, and let A be the event that the child is abused. Then P(a) = 1/90, P(B|A) = 0.90, and P(B|AC) = 0.03, so P(A|B) =

(0.90)(1/ 90) = 0.25 (0.90)(1/ 90) + (0.03)(89 / 90)

If P(a) = 1/1000, P(A|B) = 0.029; if P(a) = 1/50, P(A|B) = 0.38. 2.4.49 Let A1 be the event of being a Humanities major; A2, of being a History and Culture major; and A3, of being a Science major. If B is the event of being a woman, then (0.45)(0.5) = 225/510 P(A2|B) = (0.75)(0.3) + (0.45)(0.5) + (0.30)(0.2)

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Section 2.5: Independence

13

2.4.50 Let B be the event that a 1 is received. Let A be the event that a 1 was sent. Then (0.10)(0.3) = 30/695 P(AC|B) = (0.95)(0.7) + (0.10)(0.3) 2.4.51 Let B be the event that Zach’s girlfriend responds promptly. Let A be the event that Zach sent an e-mail, so AC is the event of leaving a message. Then (0.8)(2 / 3) = 16/25 P(A|B) = (0.8)(2 / 3) + (0.9)(1/ 3) 2.4.52 Let A be the event that the shipment came from Warehouse A with events B and C defined similarly. Let D be the event of a complaint. P ( D | C ) P (C ) P(C|D) = P ( D | A) P( A) + P ( D | B ) P ( B ) + P ( D | C ) P (C ) (0.02)(0.5) = = 10/29 (0.03)(0.3) + (0.05)(0.2) + (0.02)(0.5) 2.4.53 Let Ai be the event that Drawer i is chosen, i, = 1, 2, 3. If B is the event a silver coin is selected, (0.5)(1/ 3) = 1/3 then P(A3|B) = (0)(1/ 3) + (1)(1/ 3) + (0.5)(1/ 3)

Section 2.5: Independence 2.5.1

(a) No, because P(A ∩ B) > 0. (b) No, because P(A ∩ B) = 0.2 ≠ P(a) ⋅ P(b) = (0.6)(0.5) = 0.3 (c) P(AC ∪ BC) = P((A ∩ B)C) = 1 − P(A ∩ B) = 1 − 0.2 = 0.8.

2.5.2

Let C and M be the events that Spike passes chemistry and mathematics, respectively. Since P(C ∩ M) = 0.12 ≠ P(c) ⋅ P(M) = (0.35)(0.40) = 0.14, C and M are not independent. P(Spike fails both) = 1 − P(Spike passes at least one) = 1 − P(C ∪ M) = 1 − [P(c) + P(M) − P(C ∩ M)] = 0.37.

2.5.3

P(one face is twice the other face) = P((1, 2), (2, 1), (2, 4), (4, 2), (3, 6), (6, 3)) =

2.5.4

Let Ri, Bi, and Wi be the events that red, black, and white chips are drawn from urn i, i = 1, 2. Then P(both chips drawn are same color) = P((R1 ∩ R2) ∪ (B1 ∩ B2) ∪ (W1 ∩ W2)) = P(R1) ⋅ P(R2) + P(B1) ⋅ P(B2) + P(W1) ⋅ P(W2) [because the intersections are mutually exclusive and the individual draws are independent]. But P(R1) ⋅ P(R2) + P(B1) ⋅ P(B2) + P(W1) ⋅ P(W2) ⎛ 3 ⎞⎛ 2⎞ ⎛ 2 ⎞⎛ 4⎞ ⎛ 5 ⎞⎛3⎞ = ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = 0.32. ⎝ 10 ⎠ ⎝ 9 ⎠ ⎝ 10 ⎠ ⎝ 9 ⎠ ⎝ 10 ⎠ ⎝ 9 ⎠

2.5.5

P(Dana wins at least 1 game out of 2) = 0.3, which implies that P(Dana loses 2 games out of 2) = 0.7. Therefore, P(Dana wins at least 1 game out of 4) = 1 − P(Dana loses all 4 games) = 1 − P(Dana loses first 2 games and Dana loses second 2 games) = 1 − (0.7)(0.7) = 0.51.

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6 . 36

14

Chapter 2: Probability

2.5.6

Six equally-likely orderings are possible for any set of three distinct random numbers: x1 < x2 < x3, x1 < x3 < x2, x2 < x1 < x3, x2 < x3 < x1, x3 < x1 < x2, and x3 < x2 < x1. By inspection, 2 1 1 . P(a) = , and P(b) = , so P(A ∩ B) = P(a) ⋅ P(b) = 6 6 18

2.5.7

(a) 1. P(A ∪ B) = P(a) + P(b) − P(A ∩ B) = 1/4 + 1/8 + 0 = 3/8 2. P(A ∪ B) = P(a) + P(b) − P(a)P(b) = 1/4 + 1/8 − (1/4)(1/8) = 11/32 P( A ∩ B) 0 = =0 P( B ) P( B) P ( A ∩ B ) P( A) P( B ) = = P ( A) = 1/ 4 2. P(A|B) = P( B ) P ( B)

(b) 1. P(A|B) =

2.5.8

(a) P(A ∪ B ∪ C) = P(a) + P(b) + P(c) − P(a)P(b) − P(a)P(c) − P(b)P(c) + P(a)P(b)P(c) (b) P(A ∪ B ∪ C) = 1 − P[(A ∪ B ∪ C)C] = 1 − P(AC ∩ BC ∩ CC) = 1 − P(AC)P(BC)P(CC)

2.5.9

Let Ai be the event of i heads in the first two tosses, i = 0, 1, 2. Let Bi be the event of i heads in the last two tosses, i = 0, 1, 2. The A’s and B’s are independent. The event of interest is (A0 ∩ B0) ∪ (A1 ∩ B1) ∪ (A2 ∩ B2) and P[(A0 ∩ B0) ∪ (A1 ∩ B1) ∪ (A2 ∩ B2)] = P(A0)P(B0) + P(A1)P(B1) + P(A2)P(B2) = (1/4)(1/4) + (1/2)(1/2) + (1/4)(1/4) = 6/16

2.5.10 A and B are disjoint, so they cannot be independent. 2.5.11 Equation 2.5.3: P(A ∩ B ∩ C) = P({1, 3)}) = 1/36 = (2/6)(3/6)(6/36) = P(a)P(b)P(c) Equation 2.5.4: P(B ∩ C) = P({1, 3), (5,6)}) = 2/36 ≠ (3/6)(6/36) = P(b)P(c) 2.5.12 Equation 2.5 3: P(A ∩ B ∩ C) = P({2, 4, 10, 12)}) = 4/36 ≠ (1/2)(1/2)(1/2) = P(a)P(b)P(c) Equation 2.5.4: P(A ∩ B) = P({2, 4, 10, 12, 24, 26, 32, 34, 36)}) = 9/36 = 1/4 = (1/2)(1/2) = P(a)P(b) P(A ∩ C) = P({1, 2, 3, 4, 5, 10, 11, 12, 13)}) = 9/36 = 1/4 = (1/2)(1/2) = P(a)P(c) P(B ∩ C) = P({2, 4, 6, 8, 10, 12, 14, 16, 18)}) = 9/36 = 1/4 = (1/2)(1/2) = P(a)P(c) 2.5.13 11 [= 6 verifications of the form P(Ai ∩ Aj) = P(Ai) ⋅ P(Aj) + 4 verifications of the form P(Ai ∩ Aj ∩ Ak) = P(Ai) ⋅ P(Aj) ⋅ P(Ak) + 1 verification that P(A1 ∩ A2 ∩ A3 ∩ A4) = P(A1) ⋅ P(A2) ⋅ P(A3) ⋅ P(A4)]. 3 2 6 6 3 2 , P(b) = , P(c) = , P(A ∩ B) = , P(A ∩ C) = , P(B ∩ C) = , and 6 6 36 36 36 36 1 . It follows that A, B, and C are mutually independent because P(A ∩ B ∩ C) = 36 1 3 2 6 6 3 2 P(A ∩ B ∩ C) = = P(a) ⋅ P(b) ⋅ P(c) = ⋅ ⋅ , P(A ∩ B) = = P(a) ⋅ P(b) = ⋅ , P(A 36 6 6 36 36 6 6 3 3 6 2 2 6 ∩ C) = = P(a) ⋅ P(c) = ⋅ , and P(B ∩ C) = = P(b) ⋅ P(c) = ⋅ . 36 6 36 36 6 36

2.5.14 P(a) =

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Section 2.5: Independence

15

2.5.15 P(A ∩ B ∩ C) = 0 (since the sum of two odd numbers is necessarily even) ≠ P(a) ⋅ P(b) ⋅ P(c) 9 > 0, so A, B, and C are not mutually independent. However, P(A ∩ B) = 36 3 3 9 3 18 9 = P(a) ⋅ P(b) = ⋅ , P(A ∩ C) = = P(a) ⋅ P(c) = ⋅ , and P(B ∩ C) = = P(b) ⋅ P(c) = 6 6 36 6 36 36 3 18 , so A, B, and C are pairwise independent. ⋅ 6 36 2.5.16 Let Ri and Gi be the events that the ith light is red and green, respectively, i = 1, 2, 3, 4. Then 1 1 P(R1) = P(R2) = and P(R3) = P(R4) = . Because of the considerable distance between the 3 2 intersections, what happens from light to light can be considered independent events. P(driver stops at least 3 times) = P(driver stops exactly 3 times) + P(driver stops all 4 times) = P((R1 ∩ R2 ∩ R3 ∩ G4) ∪ (R1 ∩ R2 ∩ G3 ∩ R4) ∪ (R1 ∩ G2 ∩ R3 ∩ R4) ⎛1⎞⎛1⎞⎛ 1⎞⎛ 1⎞ ⎛1⎞⎛1⎞⎛ 1⎞⎛ 1⎞ ∪ (G1 ∩ R2 ∩ R3 ∩ R4) ∪ (R1 ∩ R2 ∩ R3 ∩ R4)) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛1⎞⎛ 2⎞⎛1 ⎞⎛1 ⎞ ⎛ 2⎞⎛1⎞⎛ 1⎞⎛ 1⎞ ⎛1⎞⎛1⎞⎛ 1⎞⎛ 1⎞ 7 +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = . ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 36 2.5.17 Let M, L, and G be the events that a student passes the mathematics, language, and general 6175 7600 8075 , P(L) = , and P(g) = . knowledge tests, respectively. Then P(M) = 9500 9500 9500 P(student fails to qualify) = P(student fails at least one exam) = 1 − P(student passes all three exams) = 1 − P(M ∩ L ∩ G) = 1 − P(M) ⋅ P(L) ⋅ P(g) = 0.56. 2.5.18 Let Ai denote the event that switch Ai closes, i = 1, 2, 3, 4. Since the Ai’s are independent events, P(circuit is completed) = P((A1 ∩ A2) ∪ (A3 ∩ A4)) = P(A1 ∩ A2) + P(A3 ∩ A4) − P((A1 ∩ A2) ∩ (A3 ∩ A4)) = 2p2 − p4. 2.5.19 Let p be the probability of having a winning game card. Then 0.32 = P(winning at least once in 5 tries) = 1 − P(not winning in 5 tries) = 1 − (1 − p)5, so p = 0.074 2.5.20 Let AH, AT, BH, BT, CH, and CT denote the events that players A, B, and C throw heads and tails on individual tosses. Then P(A throws first head) = P(AH ∪ (AT ∩ BT ∩ CT ∩ AH) ∪ ⋅⋅⋅) 2 ⎞ 4 1 1 ⎛1⎞ 1 ⎛1⎞ 1⎛ 1 = + ⎜ ⎟+ ⎜ ⎟ + = ⎜ = . Similarly, P(B throws first head) 2 2 ⎝8⎠ 2 ⎝8⎠ 2 ⎝ 1 − 1/ 8 ⎟⎠ 7 ⎞ 2 1 1 ⎛1⎞ 1 ⎛1⎞ 1⎛ 1 = P((AT ∩ BH) ∪ (AT ∩ BT ∩ CT ∩ AT ∩ BH) ∪ …) = + ⎜ ⎟ + ⎜ ⎟ + ... = ⎜ = . 4 4 ⎝8⎠ 4 ⎝8⎠ 4 ⎝ 1 − 1/ 8 ⎠⎟ 7 2

P(C throws first head) = 1 −

4 2 1 − = . 7 7 7

2.5.21 P(at least one child becomes adult) = 1 − P(no child becomes adult) = 1 − 0.2n . ln 0.25 or n ≥ 0.86 , so take n = 1 . Then 1 − 2n ≥ 0.75 implies n ≥ ln 0.2

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16

Chapter 2: Probability

2.5.22 P(at least one viewer can name actor) = 1 − P(no viewer can name actor) = 1 − (0.85)10 = 0.80. P(exactly one viewer can name actor) = 10 (0.15) (0.85)9 = 0.347. 2.5.23 Let B be the event that no heads appear, and let Ai be the event that i coins are tossed, i = 1, 2, …, 2

6

1 ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 63 6. Then P(b) = P ( B | Ai ) P ( Ai ) = ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ... + ⎜ ⎟ ⎜ ⎟ = . ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠ ⎝ 6 ⎠ 384 2 6 2 6 i =1 6

∑

2.5.24 P(at least one red chip is drawn from at least one urn) = 1 − P(all chips drawn are white) r r r rm ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ = 1 − ⎜ ⎟ ⋅⎜ ⎟ ⎜⎝ ⎟⎠ = 1 − ⎜⎝ ⎟⎠ . ⎝7⎠ ⎝7⎠ 7 7 n

⎛ 35 ⎞ 2.5.25 P(at least one double six in n throws) = 1 − P(no double sixes in n throws) = 1 − ⎜ ⎟ . By trial ⎝ 36 ⎠ and error, the smallest n for which P(at least one double six in n throws) exceeds 0.50 is 25 24

⎛ 35 ⎞ [1 − ⎜ ⎟ = 0.49; 1 − ⎝ 36 ⎠

⎛ 35 ⎞ ⎜⎝ ⎟⎠ 36

25

= 0.51].

2.5.26 Let A be the event that a sum of 8 appears before a sum of 7. Let B be the event that a sum of 8 appears on a given roll and let C be the event that the sum appearing on a given roll is neither 7 5 25 , P(c) = , and P(a) = P(b) + P(c)P(b) + P(c)P(c)P(b) nor 8. Then P(b) = 36 36 2 k ⎞ 5 5 25 5 ⎛ 25 ⎞ 5 5 ∞ ⎛ 25 ⎞ 5 ⎛ 1 + +⎜ ⎟ + = = . + ⋅⋅⋅ = ⎜⎝ ⎟⎠ = ⎜ ⎝ ⎠ 36 36 36 36 36 36 k = 0 36 36 ⎝ 1 − 25 / 36 ⎟⎠ 11

∑

2.5.27 Let W, B, and R denote the events of getting a white, black and red chip, respectively, on a given draw. Then P(white appears before red) = P(W ∪ (B ∩ W) ∪ (B ∩ B ∩ W) ∪ ⋅⋅⋅) 2

=

w b w b w ⎛ ⎞ + ⋅ +⎜ ⋅ + ⎟ w+b+ r w+b+ r w+b+ r ⎝w+b+ r ⎠ w+b+ r

=

⎛ ⎞ w 1 w ⋅⎜ = . ⎟ w + b + r ⎝ 1 − b /( w + b + r ) ⎠ w + r

2.5.28 P(B|A1) = 1 − P(all m I-teams fail) = 1 − (1 − r)m; similarly, P(B|A2) = 1 − (1 − r)n−m. From Theorem 2.4.1, P(b) = [1 − (1 − r)m]p + [1 − (1 − r)n−m](1 − p). Treating m as a continuous variable and differentiating P(b) gives dP( B ) dP( B ) = −p(1 − r)m⋅ln(1 − r) + (1 − p)(1 − r)n−m ⋅ln(1 − r). Setting = 0 implies that dm dm n ln[(1 − p ) / p] m= + . 2 2ln(1 − r ) 2.5.29 P(at least one four) = 1 − P(no fours) = 1 − (0.9)n. 1 − (0.9)n ≥ 0.7 implies n = 12

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Section 2.6: Combinatorics

17

2.5.30 Let B be the event that all n tosses come up heads. Let A1 be the event that the coin has two heads, and let A2 be the event the coin is fair. Then (1/ 2)n (8 / 9) 8(1/ 2) n P ( A2 | B ) = = 1(1/ 9) + (1/ 2) n (8/ 9) 1 + 8(1/ 2) n By inspection, the limit of P ( A2 | B ) as n goes to infinity is 0.

Section 2.6: Combinatorics 2.6.1

2 ⋅ 3 ⋅ 2 ⋅ 2 = 24

2.6.2

20 ⋅ 9 ⋅ 6 ⋅ 20 = 21,600

2.6.3

3 ⋅ 3 ⋅ 5 = 45. Included will be aeu and cdx.

2.6.4

(a) 262 ⋅ 104 = 6,760,000 (b) 262 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 = 3,407,040 (c) The total number of plates with four zeros is 26 ⋅ 26, so the total number not having four zeros must be 262 ⋅ 104 − 262 = 6,759,324.

2.6.5

There are 9 choices for the first digit (1 through 9), 9 choices for the second digit (0 + whichever eight digits are not appearing in the hundreds place), and 8 choices for the last digit. The number of admissible integers, then, is 9 ⋅ 9 ⋅ 8 = 648. For the integer to be odd, the last digit must be either 1, 3, 5, 7, or 9. That leaves 8 choices for the first digit and 8 choices for the second digit, making a total of 320 (= 8 ⋅ 8 ⋅ 5) odd integers.

2.6.6

For each topping, the customer has 2 choices: “add” or “do not add.” The eight available toppings, then, can produce a total of 28 = 256 different hamburgers.

2.6.7

The bases can be occupied in any of 27 ways (each of the seven can be either “empty” or “occupied”). Moreover, the batter can come to the plate facing any of five possible “out” situations (0 through 4). It follows that the number of base-out configurations is 5 ⋅ 27, or 640.

2.6.8

With 4 choices for the first digit, 1 for the third digit, 5 for the last digit, and 10 for each of the remaining six digits, the total number of admissible zip codes is 20,000,000(= 4 ⋅ 106 ⋅ 1 ⋅ 5).

2.6.9

4 ⋅ 14 ⋅ 6 + 4 ⋅ 6 ⋅ 5 + 14 ⋅ 6 ⋅ 5 + 4 ⋅ 14 ⋅ 5 = 1156

2.6.10 There are two mutually exclusive sets of ways for the black and white keys to alternate—the black keys can be 1st, 3rd, 5th, and 7th notes in the melody, or the 2nd, 4th 6th, and 8th. Since there are 5 black keys and 7 white keys, there are 5 ⋅ 7 ⋅ 5 ⋅ 7 ⋅ 5 ⋅ 7 ⋅ 5 ⋅ 7 variations in the first set and 7 ⋅ 5 ⋅ 7 ⋅ 5 ⋅ 7 ⋅ 5 ⋅ 7 ⋅ 5 in the second set. The total number of alternating melodies is the sum 54 7 4 + 7 4 54 = 3,001,250. 2.6.11 The number of usable garage codes is 28 − 1 = 255, because the “combination” where none of the buttons is pushed is inadmissible (recall Example 2.6.3). Five additional families can be added before the eight-button system becomes inadequate.

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18

Chapter 2: Probability

2.6.12 4, because 21 + 22 + 23 < 26 but 21 + 22 + 23 + 24 ≥ 26. 2.6.13 In order to exceed 256, the binary sequence of coins must have a head in the ninth position and at least one head somewhere in the first eight tosses. The number of sequences satisfying those conditions is 28 − 1, or 255. (The “1” corresponds to the sequences TTTTTTTTH, whose value would not exceed 256.) 2.6.14 There are 3 choices for the vowel and 4 choices for the consonant, so there are 3 ⋅ 4 = 12 choices, if order doesn’t matter. If we are taking ordered arrangements, then there are 24 ways, since each unordered selection can be written vowel first or consonant first. 2.6.15 There are 1 ⋅ 3 ways if the ace of clubs is the first card and 12 ⋅ 4 ways if it is not. The total is then 3 + 12 ⋅ 4 = 51 2.6.16 Monica has 3 ⋅ 5 ⋅ 2 = 30 routes from Nashville to Anchorage, so there are 30 ⋅ 30 = 900 choices of round trips. 2.6.17

6P3

= 6 ⋅ 5 ⋅ 4 = 120

2.6.18

4P4

= 4! = 24; 2P2 ⋅ 2P2 = 4

30! 2.6.20

9P9

(

)

1⎞ ⎛ 2π + ⎜ 30 + ⎟ log10(30) − 30log10e = 32.42246, which implies that ⎝ 2⎠ 1032.42246 = 2.645 × 1032.

2.6.19 log10(30!)

log10

= 9! = 362,880

2.6.21 There are 2 choices for the first digit, 6 choices for the middle digit, and 5 choices for the last digit, so the number of admissible integers that can be formed from the digits 1 through 7 is 60 (= 2 ⋅ 6 ⋅ 5). 2.6.22 (a) 8P8 = 8! = 40,320 (b) The men can be arranged in, say, the odd-numbered chairs in 4P4 ways; for each of those permutations, the women can be seated in the even-numbered chairs in 4P4 ways. But the men could also be in the even-numbered chairs. It follows that the total number of alternating seating arrangements is 4P4 ⋅ 4P4 + 4P4 ⋅ 4P4 = 1152. 2.6.23 There are 4 different sets of three semesters in which the electives could be taken. For each of those sets, the electives can be selected and arranged in 10P3 ways, which means that the number of possible schedules is 4 ⋅ 10P3, or 2880. 2.6.24

= 720; 6P6 ⋅ 6P6 = 518,400; 6!6!26 is the number of ways six male/female cheerleading teams can be positioned along a sideline if each team has the option of putting the male in front or the female in front; 6!6!26212 is the number of arrangements subject to the conditions of the previous answer but with the additional option that each cheerleader can face either forwards or backwards. 6P6

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Section 2.6: Combinatorics

19

2.6.25 The number of playing sequences where at least one side is out of order = total number of playing sequences − number of correct playing sequences = 6P6 − 1 = 719. 2.6.26 Within each of the n families, members can be lined up in mPm = m! ways. Since the n families can be permuted in nPn = n! ways, the total number of admissible ways to arrange the nm people is n! ⋅ (m!)n. 2.6.27 There are 2P2 = 2 ways for you and a friend to be arranged, 8P8 ways for the other eight to be permuted, and six ways for you and a friend to be in consecutive positions in line. By the multiplication rule, the number of admissible arrangements is 2P2 ⋅ 8P8 ⋅ 6 = 483,840. 2.6.28 By inspection, nP1 = n. Assume that nPk = n(n − 1) ⋅⋅⋅ (n − k + 1) is the number of ways to arrange k distinct objects without repetition. Notice that n − k options would be available for a (k + 1)st object added to the sequences. By the multiplication rule, the number of sequences of length k + 1 must be n(n − 1) ⋅⋅⋅ (n − k + 1)(n − k). But the latter is the formula for nPk+1. 2.6.29 (13!)4 2.6.30 By definition, (n + 1)! = (n + 1) ⋅ n!; let n = 0. 2.6.31

9 P2

⋅ 4C1 = 288

2.6.32 Two people between them: 4 ⋅ 2 ⋅ 5! = 960 Three people between them: 3 ⋅ 2 ⋅ 5! = 720 Four people between them: 2 ⋅ 2 ⋅ 5! = 480 Five people between them: 1 ⋅ 2 ⋅ 5! = 240 Total number of ways: 2400 2.6.33 (a) (4!)(5!) = 2880 (c) (4!)(5!) = 2880

(b) 6(4!)(5!) = 17, 280 ⎛9 ⎞ (d) ⎜ ⎟ (2)(5!) = 30, 240 ⎝4⎠

9! = 3780 ways; 4!2!2!1! FLORIDA can be permuted in 7! = 5040 ways.

2.6.34 TENNESSEE can be permuted in

2.6.35 If the first digit is a 4, the remaining six digits can be arranged in digit is a 5, the remaining six digits can be arranged in

6! = 120 ways; if the first 3!(1!)3

6! = 180 ways. The total number 2!2!(1!)2

of admissible numbers, then, is 120 + 180 = 300. 2.6.36 (a) 8!/3!3!2! = 560

(b) 8! = 40,320

(c) 8!/3!(1!)5 = 6720

2.6.37 (a) 4! ⋅ 3! ⋅ 3! = 864 (b) 3! ⋅ 4!3!3! = 5184 (each of the 3! permutations of the three nationalities can generate 4!3!3! arrangements of the ten people in line)

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20

Chapter 2: Probability

(c) 10! = 3,628,800 (d) 10!/4!3!3! = 4200 11! ways. The seven 3!2!(1!)6 consonants can be arranged in 7!/3!(1!)4 ways, of which 4! have the property that the three L’s come first. By the reasoning used in Example 2.6.13, it follows that the number of admissible 11! , or 95,040. arrangements is 4!/(7!/3!) ⋅ 3!2!

2.6.38 Altogether, the letters in S L U M G U L L I O N can be permuted in

2.6.39 Imagine a field of 4 entrants (A, B, C, D) assigned to positions 1 through 4, where positions 1 and 2 correspond to the opponents for game 1 and positions 3 and 4 correspond to the opponents for game 2. Although the four players can be assigned to the four positions in 4! ways, not all of BC AD ADBC and produce those permutations yield different tournaments. For example, 1 2 3 4 1 2 3 4 BC AD CB AD the same set of games, as do and . In general, n games can be arranged in n! 1 2 3 4 1 2 3 4 ways, and the two players in each game can be permuted in 2! ways. Given a field of 2n entrants, then, the number of distinct pairings is (2n)!/n!(2!)n, or 1 ⋅ 3 ⋅ 5 ⋅⋅⋅ (2n − 1). 2.6.40 Since x12 can be the result of the factors x6 ⋅ x6 ⋅ 1 ⋅⋅⋅ 1 or x3 ⋅ x3 ⋅ x3 ⋅ x3 ⋅ 1 ⋅⋅⋅ 1 or x6 ⋅ x3 ⋅ x3 ⋅ 1 ⋅⋅⋅ 1, the analysis described in Example 2.6.16 implies that the coefficient of x12 is 18! 18! 18! = 5661. + + 2!16! 4!14! 1!2!15! 2.6.41 The letters in E L E E M O S Y N A R Y minus the pair S Y can be permuted in 10!/3! ways. Since S Y can be positioned in front of, within, or behind those ten letters in 11 ways, the number of admissible arrangements is 11 ⋅ 10!/3! = 6,652,800. 2.6.42 Each admissible spelling of ABRACADABRA can be viewed as a path consisting of 10 steps, five to the right (R) and five to the left (L). Thus, each spelling corresponds to a permutation of 10! = 252 such permutations. the five R’s and five L’s. There are 5!5! 2.6.43 Six, because the first four pitches must include two balls and two strikes, which can occur in 4!/2!2! = 6 ways. 2.6.44 9!/2!3!1!3! = 5040 (recall Example 2.6.16) 2.6.45 Think of the six points being numbered 1 through 6. Any permutation of three A’s and three AABB AB —corresponds to the three vertices chosen for triangle A and B’s—for example, 1 2 3 4 5 6 the three for triangle B. It follows that 6!/3!3! = 20 different sets of two triangles can be drawn. 2.6.46 Consider k! objects categorized into (k − 1)! groups, each group being of size k. By Theorem 2.6.2, the number of ways to arrange the k! objects is (k!)!/(k!)(k − 1)!, but the latter must be an integer.

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Section 2.6: Combinatorics

21

14! 5! total permutations of the letters. There are = 30 arrangements 2!2!1!2!2!3!1!1! 2!2!1! of the vowels, only one of which leaves the vowels in their original position. Thus, there are 1 14! ⋅ = 30,270,240 arrangements of the word leaving the vowels in their original 30 2!2!1!2!1!3!1!1! position.

2.6.47 There are

2.6.48

15! = 1, 513, 512, 000 4!3!1!3!1!1!1!1!

2.6.49 The three courses with A grades can be: emf, emp, emh, efp, efh, eph, mfp, mfh, mph, fph, or 10 possibilities. From the point of view of Theorem 2.6.2, the grade assignments correspond to the 5! = 10. set of permutations of three A’s and two B’s, which equals 3!2! 2.6.50 Since every (unordered) set of two letters describes a different line, the number of possible lines ⎛5 ⎞ is ⎜ ⎟ = 10. ⎝2⎠ 2.6.51 To achieve the two-to-one ratio, six pledges need to be chosen from the set of 10 and three from ⎛10 ⎞ ⎛15 ⎞ the set of 15, so the number of admissible classes is ⎜ ⎟ ⋅ ⎜ ⎟ = 95,550. ⎝6⎠ ⎝3⎠ 2.6.52 Of the eight crew members, five need to be on a given side of the boat. Clearly, the remaining three can be assigned to the sides in 3 ways. Moreover, the rowers on each side can be permuted in 4! ways. By the multiplication rule, then, the number of ways to arrange the crew is 1728 (= 3 ⋅ 4! ⋅ 4!). ⎛9 ⎞ 2.6.53 (a) ⎜ ⎟ = 126 ⎝4⎠

⎛5 ⎞ ⎛ 4⎞ (b) ⎜ ⎟ ⎜ ⎟ = 60 ⎝ 2⎠ ⎝ 2⎠

⎛9 ⎞ ⎛5 ⎞ ⎛ 4⎞ (c) ⎜ ⎟ − ⎜ ⎟ − ⎜ ⎟ = 120 ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠

⎛7⎞ 2.6.54 ⎜ ⎟ = 21; order does not matter. ⎝5 ⎠

2.6.55 Consider a simpler problem: Two teams of two each are to be chosen from a set of four ⎛4⎞ players—A, B, C, and D. Although a single team can be chosen in ⎜ ⎟ ways, the number of ⎝2⎠ ⎛4⎞ pairs of teams is only ⎜ ⎟ 2 , because [( A B), (C D)] and [(C D), (A B)] would correspond to ⎝2⎠ the same matchup. Applying that reasoning here means that the ten players can split up in ⎛10 ⎞ ⎜⎝ 5 ⎟⎠ 2 = 126 ways.

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22

Chapter 2: Probability

2.6.56 Number the spaces between the twenty pages from 1 to 19. Choosing any two of these spaces partitions the reading assignment into three non-zero, numbers, x1 , x2 , and x3 , corresponding to the numbers of pages read on Monday, Tuesday, and Wednesday, respectively. Therefore, the ⎛19 ⎞ number of ways to complete the reading assignment is ⎜ ⎟ = 171. ⎝2⎠ ⎛8 ⎞ 2.6.57 The four I’s need to occupy any of the ⎜ ⎟ sets of four spaces between and around the other ⎝ 4⎠ 7! seven letters. Since the latter can be permuted in ways, the total number of admissible 2!4!1! ⎛ 8 ⎞ 7! = 7350. arrangements is ⎜ ⎟ ⋅ ⎝ 4 ⎠ 2!4!1! n

2.6.58 Let x = y = 1 in the expansion (x+ y)n =

⎛n⎞

∑ ⎜⎝ k ⎟⎠x

y n − k . The total number of hamburgers referred

k

k =0

to in Question 2.6.6 (= 28) must also be equal to the number of ways to choose k condiments, k = ⎛8 ⎞ ⎛8⎞ ⎛8⎞ 0, 1, 2, …, 8—that is, ⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟ . ⎝ 0 ⎠ ⎝1 ⎠ ⎝8⎠ 2.6.59 Consider the problem of selecting an unordered sample of n objects from a set of 2n objects, where the 2n have been divided into two groups, each of size n. Clearly, we could choose n from the first group and 0 from the second group, or n − 1 from the first group and 1 from the second ⎛ 2n ⎞ ⎛n⎞ ⎛n⎞ ⎛ n ⎞ ⎛n⎞ ⎛ n⎞ ⎛ n⎞ group, and so on. Altogether, ⎜ ⎟ must equal ⎜ ⎟ ⎜ ⎟ + ⎜ + ... + ⎜ ⎟ ⎜ ⎟ . But ⎟ ⎜ ⎟ ⎝n⎠ ⎝ n ⎠ ⎝ 0 ⎠ ⎝ n − 1⎠ ⎝1 ⎠ ⎝0 ⎠ ⎝ n⎠ ⎛ 2n ⎞ ⎛n⎞ ⎛ n ⎞ ⎜⎝ j ⎠⎟ = ⎝⎜ n − j ⎠⎟ , j = 0, 1, …, n so ⎜⎝ n ⎟⎠ =

∑ n

2.6.60 Let x = y = 1 in the expansion (x − y)n =

2

⎛ n⎞ ⎜⎝ j ⎟⎠ . j =0 n

⎛n⎞

∑ ⎜⎝ k ⎠⎟ x

k

(−y)n−k. Then x − y = 0 and the sum reduces to

k =0

n

0=

⎛n⎞

∑ ⎜⎝ k ⎟⎠ (−1) k =0

n− k

⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛ n⎞ , or equivalently, ⎜ ⎟ + ⎜ ⎟ + ... = ⎜ ⎟ + ⎜ ⎟ + … . ⎝1 ⎠ ⎝ 3 ⎠ ⎝0 ⎠ ⎝ 2⎠

⎛ n ⎞ 2.6.61 The ratio of two successive terms in the sequence is ⎜ ⎝ j + 1⎟⎠

⎛n⎞ n − j ⎜⎝ j ⎟⎠ = j + 1 . For small j, n −1 , though, the ratio is less than 1, n − j > j + 1, implying that the terms are increasing. For j > 2 meaning the terms are decreasing.

2.6.62 Four months of daily performance create a need for roughly 120 different sets of jokes. If n denotes the number of different jokes that Mitch has to learn, the question is asking for the ⎛n⎞ smallest n for which ⎜ ⎟ ≥ 120 . By trial and error, n = 9. ⎝4⎠

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Section 2.7: Combinatorial Probability

23

2.6.63 Using Newton’s binomial expansion, the equation (1 + t)d ⋅ (1 + t)e = (1 + t)d+e can be written ⎛ d ⎛ d ⎞ j ⎞ ⎛ e ⎛ e ⎞ j ⎞ d +e ⎛ d + e ⎞ j ⎜ ⎜ ⎟t ⎟ ⋅ ⎜ ⎜ ⎟t ⎟ = ⎜ ⎟t ⎝ j =0 ⎝ j ⎠ ⎠ ⎝ j =0 ⎝ j ⎠ ⎠ j =0 ⎝ j ⎠

∑

∑

∑

Since the exponent k can arise as t 0 ⋅ t k , t1 ⋅ t k −1 , … , or ⎛d ⎞ ⎛e ⎞ ⎛d ⎞ ⎛ e ⎞ ⎛d ⎞ ⎛e ⎞ ⎛d + e⎞ ⎜⎝ 0 ⎟⎠ ⎜⎝ k ⎟⎠ + ⎜⎝1 ⎟⎠ ⎜⎝ k − 1⎟⎠ + ... + ⎜⎝ k ⎟⎠ ⎜⎝ 0 ⎟⎠ = ⎜⎝ k ⎠⎟ . That is,

t k ⋅ t 0 , it follows that k ⎛d ⎞ ⎛ e ⎞ ⎛d + e⎞ = ⎜ ⎟⎜ ⎟. ⎜⎝ k ⎠⎟ j =0 ⎝ j ⎠ ⎝ k − j ⎠

∑

Section 2.7: Combinatorial Probability 2.7.1

⎛7⎞ ⎛3 ⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

2.7.2

P(sum = 5) =

2.7.3

P(numbers differ by more than 2) = 1 − P(numbers differ by one) − P(numbers differ by 2) ⎛ 20 ⎞ ⎛ 20 ⎞ 153 = 0.81. = 1 − 19 ⎜ ⎟ − 18 ⎜ ⎟ = ⎝2⎠ ⎝ 2 ⎠ 190

2.7.4 2.7.5

⎛10 ⎞ ⎜⎝ 4 ⎟⎠ ⎛6⎞ 2 Number of pairs that sum to 5 =2 ⎜ ⎟= . Total number of pairs ⎝ 2 ⎠ 15

⎛ 4 ⎞ ⎛ 48 ⎞ P(A ∪ B) = P(a) + P(b) − P(A ∩ B) = ⎜ ⎟ ⎜ ⎟ ⎝ 4⎠ ⎝ 9 ⎠

⎛ 52 ⎞ ⎛ 4 ⎞ ⎛ 48 ⎞ + ⎝⎜13 ⎠⎟ ⎜⎝ 4 ⎠⎟ ⎝⎜ 9 ⎠⎟

⎛ 52 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ − ⎝⎜13 ⎠⎟ ⎝⎜ 4 ⎠⎟ ⎝⎜ 4 ⎠⎟ ⎝⎜ 5 ⎠⎟

⎛ 52 ⎞ ⎝⎜13 ⎠⎟

Let A1 be the event that an urn with 3W and 3R is sampled; let A2 be the event that the urn with 5W and 1R is sampled. Let B be the event that the three chips drawn are white. By Bayes’ rule, P ( A2 | B ) =

=

⎡ ⎛ 3⎞ ⎛ 3 ⎞ ⎢⎜ ⎟ ⎜ ⎟ ⎣ ⎝ 3⎠ ⎝ 0 ⎠ 50

2.7.6

⎛2⎞ ⎜⎝1 ⎟⎠

2.7.7

6/6n = 1/6n−1

P ( B | A2 ) P ( A2 ) P ( B | A1 ) P ( A1 ) + P ( B | A2 ) P ( A2 )

⎡ ⎛ 5 ⎞ ⎛1 ⎞ ⎛ 6 ⎞ ⎤ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⋅ (1/10) 10 ⎣ ⎝ 3⎠ ⎝ 0 ⎠ ⎝ 3 ⎠ ⎦ = 19 ⎡ ⎛ 5 ⎞ ⎛1 ⎞ ⎛ 6 ⎞ ⎤ ⎛6⎞⎤ ⎜⎝ 3 ⎟⎠ ⎥ ⋅ (9 /10) + ⎢ ⎜⎝ 3 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎥ ⋅ (1/10) ⎦ ⎣ ⎦

⎛100 ⎞ ⎜⎝ 50 ⎟⎠

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24

2.7.8

Chapter 2: Probability

There are 6 faces that could be the “three-of-a-kind” and 5 faces that could be the “two-of-a⎛5 ⎞ kind.” Moreover, the five dice bearing those two numbers could occur in any of 5!/2!3! = ⎜ ⎟ ⎝2⎠ ⎛5 ⎞ orders. It follows that P(“full house”) = 6 ⋅ 5 ⋅ ⎜ ⎟ 65 = 50 64 ⎝2⎠

2.7.9

By Theorem, 2.6.2, the 2n grains of sand can be arranged in (2n)!/n!n! ways. Two of those arrangements have the property that the colors will completely separate. Therefore, the probability of the latter is 2(n!)2/(2n)!

2.7.10 P(monkey spells CALCULUS) = 1/[8!/(2!)3(1!)2] = 1/5040; P(monkey spells ALGEBRA) = 1/[7!/2!(1!)5] = 2/5040. 2.7.11 P(different floors) = 7!/77; P(same floor) = 7/77 = 1/76. The assumption being made is that all possible departure patterns are equally likely, which is probably not true, since residents living on lower floors would be less inclined to wait for the elevator than would those living on the top floors. 23! . The 2!2!4!2!1!3!2!4!2!2!1!1! number of permutations where all of the S’s are adjacent is counted by treating the S’s as a single letter that appears once. The denominator above will have one of the 4! replaced by 1!. The 23! number of such permutations, then, is . The probability that the S’s are 2!2!4!2!1!3!2!1!2!2!1!1! adjacent is then the ratio of these two terms or 4!23!/26! = 1/650. The requested probability is then the complement, 649/650.

2.7.12 The total number of distinguishable permutations of the phrase is

2.7.13 The 10 short pieces and 10 long pieces can be lined up in a row in 20!/(10)!(10)! ways. Consider each of the 10 pairs of consecutive pieces as defining the reconstructed sticks. Each of those pairs could combine a short piece (S) and a long piece (L) in two ways: SL or LS. Therefore, the number of permutations that would produce 10 sticks, each having a short and a long component ⎛ 20 ⎞ is 210, so the desired probability is 210 ⎜ ⎟ . ⎝10 ⎠ 2.7.14 6!/66 ⎛k ⎞ 2.7.15 Any of ⎜ ⎟ people could share any of 365 possible birthdays. The remaining k − 2 people can ⎝2⎠ generate 364 ⋅ 363 ⋅⋅⋅ (365 − k + 2) sequences of distinct birthdays. Therefore, P(exactly one ⎛k ⎞ match) = ⎜ ⎟ ⋅ 365 ⋅ 364 ⋅⋅⋅ (365 − k + 2)/365k. ⎝2⎠

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Section 2.7: Combinatorial Probability

25

⎛12 ⎞ ⎛11⎞ ⎛10 ⎞ 2.7.16 The expression ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ orders the denominations of the three single cards—in effect, each ⎝ 1 ⎠⎝ 1 ⎠⎝ 1 ⎠ ⎛ 52 ⎞ set of three denominations would be counted 3! times. The denominator (= ⎜ ⎟ ) in that ⎝5⎠ particular probability calculation, though, does not consider the cards to be ordered. To be consistent, the denominations for the three single cards must be treated as a combination, ⎛12 ⎞ meaning the number of choices is ⎜ ⎟ . ⎝3⎠

2.7.17 To get a flush, Dana needs to draw any three of the remaining eleven diamonds. Since only fortyseven cards are effectively left in the deck (others may already have been dealt, but their ⎛11⎞ ⎛ 47 ⎞ identities are unknown), P(Dana draws to flush) = ⎜ ⎟ ⎜ ⎟ . ⎝3⎠ ⎝ 3 ⎠ 2.7.18 P(draws to full house or four-of-a-kind) = P(draws to full house) + P(draws to four-of-a-kind) 3 1 4 + = . = 47 47 47 2.7.19 There are two pairs of cards that would give Tim a straight flush (5 of clubs and 7 of clubs or 7 of ⎛ 47 ⎞ clubs and 10 of clubs). Therefore, P(Tim draws to straight flush) = 2 ⎜ ⎟ . A flush, by ⎝2⎠ definition, consists of five cards in the same suit whose denominations are not all consecutive. It ⎡ ⎛10 ⎞ ⎤ ⎛ 47 ⎞ follows that P(Tim draws to flush) = ⎢ ⎜ ⎟ − 2⎥ ⎜ ⎟ , where the “2” refers to the straight ⎣⎝ 2 ⎠ ⎦ ⎝ 2 ⎠ flushes cited earlier. 2.7.20 A sum of 48 requires four 10’s and an 8 or three 10’s and two 9’s; a sum of 49 requires four 10’s and a 9; no sums higher than 49 are possible. Therefore, ⎡ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎤ ⎛ 52 ⎞ ⎛ 52 ⎞ P(sum ≥ 48) = ⎢ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ = 32 ⎜ ⎟ . ⎝5⎠ ⎣ ⎝ 4 ⎠ ⎝1 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 4 ⎠ ⎝1 ⎠ ⎦ ⎝ 5 ⎠ ⎛5⎞ ⎛ 4⎞ 2.7.21 ⎜ ⎟ ⎜ ⎟ ⎝3⎠ ⎝ 2⎠ ⎛ 32 ⎞ 2.7.22 ⎜ ⎟ ⎝13 ⎠

3

⎛ 3⎞ ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞ ⎜⎝1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝1 ⎟⎠

⎛ 52 ⎞ ⎜⎝ 9 ⎟⎠

⎛ 52 ⎞ ⎜⎝13 ⎟⎠ 4

⎡ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎤ ⎛ 32 ⎞ 2.7.23 ⎢ ⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ ⎣ ⎝1 ⎠ ⎝1 ⎠ ⎦ ⎝ 4 ⎠

⎛ 48 ⎞ ⎜⎝12 ⎟⎠

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26

Chapter 2: Probability

n+r n−r steps forward and steps backward will result in a net gain of r 2 2 steps forward. Since the total number of (equally-likely) paths is 2n,

2.7.24 Any permutation of

P(conventioneer ends up r steps forward) =

⎛n + r ⎞ ⎛n − r ⎞ n! ⎜ ! ! ⎝ 2 ⎠⎟ ⎝⎜ 2 ⎠⎟ 2n

.

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Chapter 3: Random Variables Section 3.2: Binomial and Hypergeometric Probabilities 3.2.1

The number of days, k, the stock rises is binomial with n = 4 and p = 0.25. The stock will be the ⎛4⎞ same after four days if k = 2. The probability that k = 2 is ⎜ ⎟ (0.25) 2 (0.75) 2 = 0.211 ⎝2⎠

3.2.2

Let k be the number of control rods properly inserted. The system fails if k ≤ 4. The probability of that occurrence is given by the binomial probability sum 4 ⎛10 ⎞ 10 − k k = 0.0064 ⎜⎝ k ⎟⎠ (0.8) (0.2) k =0

∑

3.2.3

The probability of 12 female presidents is 0.2312 , which is approximately 1/50,000,000.

3.2.4

⎛ 6⎞ ⎛ 6⎞ 1 − ⎜ ⎟ (0.153)0 (0.847)6 − ⎜ ⎟ (0.153)1 (0.847)5 = 0.231 ⎝ 0⎠ ⎝1⎠

3.2.5

1−

11

⎛11⎞

∑ ⎜⎝ k ⎟⎠ (0.9)

k

(0.1)11− k = 0.0185

k =8

3.2.6

The probability of k sightings is given by the binomial probability model with n = 10,000 and p = 1/100,000. The probability of at least one genuine sighting is the probability that k ≥ 1. The probability of the complementary event, k = 0, is (99,999 /100, 000)10,000 = 0.905. Thus, the probability that k ≥ 1 is 1 − 0.905 = 0.095.

3.2.7

For the two-engine plane, P(Flight lands safely) = P(One or two engines work properly) ⎛2⎞ ⎛ 2⎞ = ⎜ ⎟ (0.6)1 (0.4)1 + ⎜ ⎟ (0.6) 2 (0.4)0 = 0.84 ⎝1⎠ ⎝ 2⎠ For the four-engine plane, P(Flight lands safely) = P(Two or more engines work properly) ⎛4⎞ ⎛ 4⎞ ⎛ 4⎞ = ⎜ ⎟ (0.6) 2 (0.4) 2 + ⎜ ⎟ (0.6)3 (0.4)1 + ⎜ ⎟ (0.6) 4 (0.4)0 = 0.8208 ⎝2⎠ ⎝ 3⎠ ⎝ 4⎠ The two-engine plane is a better choice.

3.2.8

Probabilities for the first system are binomial with n = 50 and p = 0.05. The probability that k ≥ 1 is 1 − (0.95)50 = 1 − 0.077 = 0.923. Probabilities for the second system are binomial with n = 100 and p = 0.02. The probability that k ≥ 1 is 1 − (0.98)100 = 1 − 0.133 = 0.867 System 2 is superior from a bulb replacement perspective.

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28

3.2.9

Chapter 3: Random Variables

The number of 6’s obtained in n tosses is binomial with p = 1/6. The first probability in question has n = 6. The probability that k ≥ 1 is 1 − (5 / 6)6 = 1 − 0.33 = 0.67. For the second situation, n = 12. The probability that k ≥ 2 one minus the probability that k = 0 or 1, which is 1 − (5 / 6)12 − 12(1/ 6)(5 / 6)11 = 0.62. Finally, take n = 18. The probability that k ≥ 3 is one minus the probability that k = 0, 1, or 2, which is 1 − (5 / 6)18 − 18(1/ 6)(5 / 6)17 − 153(1/ 6)2 (5 / 6)16 = 0.60.

3.2.10 The number of missile hits on the plane is binomial with n = 6 and p = 0.2. The probability that the plane will crash is the probability that k ≥ 2. This event is the complement of the event that k = 0 or 1, so the probability is 1 − (0.8)6 − 6(0.2)(0.8)5 = 0.345 The number of rocket hits on the plane is also binomial, but with n = 10 and p = 0.05. The probability that the boat will be disabled is P(k ≥ 1), which is 1 − (0.95)10 = 0.401 3.2.11 The number of girls is binomial with n = 4 and p = 1/2. The probability of two girls and two boys ⎛4⎞ ⎛ 4⎞ is ⎜ ⎟ (0.5)4 = 0.375. The probability of three and one is 2 ⎜ ⎟ (0.5) 4 = 0.5, so the latter is more ⎝2⎠ ⎝ 3⎠ likely. 3.2.12 The number of recoveries if the drug is effective is binomial with n = 12 and p = 1/2. The drug will discredited if the number of recoveries is 6 or less. The probability of this is 6 ⎛12 ⎞ 12 ⎜⎝ k ⎟⎠ (0.5) = 0.613. k =0

∑

⎛ k − 1⎞ 3.2.13 The probability it takes k calls to get four drivers is ⎜ 0.804 0.20k − 4 . We seek the smallest ⎟ 3 ⎝ ⎠ ⎛ k − 1⎞ k −4 4 ≥ 0.95. By trial and error, n = 7. ⎟⎠ 0.80 0.20 3 k =4 n

number n so that

∑ ⎜⎝

3.2.14 The probability of any shell hitting the bunker is 30/500 = 0.06. The probability of exactly k ⎛ 25 ⎞ shells hitting the bunker is p(k) = ⎜ ⎟ (0.06)k (0.94) 25− k . The probability the bunker is destroyed ⎝k⎠ is 1 − p(0) − p(1) − p(2) = 0.187. 3.2.15 (1) The probability that any one of the seven measurements will be in the interval (1/2, 1) is 0.50. The probability that exactly three will fall in the interval is ⎛7⎞ 7 ⎜⎝ 3 ⎟⎠ 0.5 = 0.273 (2) The probability that any one of the seven measurements will be in the interval (3/4, 1) is 0.25. The probability that fewer than 3 will fall in the interval is 2 ⎛7 ⎞ k 7−k = 0.756 ⎜⎝ k ⎟⎠ (0.25) (0.75) k =0

∑

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Section 3.2: Binomial and Hypergeometric Probabilities

29

3.2.16 Use the methods of Example 3.2.3 for p = 0.5. Then the probabilities of Team A winning the series in 4, 5, 6, and 7 games are 0.0625, 0.125, 0.156, and 0.156, respectively. Since Team B has the same set of probabilities, the probability of the series ending in 4, 5, 6, and 7 games is double that for A, or 0.125, 0.250, 0.312, and 0.312, respectively. The “expected” frequencies are the number of years, 58, times the probability of each length. For example, we would “expect” 58(0.126) = 7.3 series of 4 games. The table below gives the comparison of observed and expected frequencies. Number of games 4 5 6 7

Observed Number of years 12 10 12 24

Expected Number of years 58(0.125) = 7.3 58(0.250) = 14.5 58(0.312) = 18.1 58(0.312) = 18.1

Note that the model has equal expected frequencies for 6 and 7 length series, but the observed numbers are quite different. This model does not fit the data well. 3.2.17 By the binomial theorem, ( x + y ) n =

n

⎛n⎞

∑ ⎜⎝ k ⎟⎠ x

k

y n − k . Let x = p and y = 1 − p.

k =0

⎛n⎞ k n− k Then 1 = [p + (1 − p)]n = ⎜⎝ k ⎟⎠ p (1 − p ) k =0 n

∑

3.2.18 Any particular sequence having k1 of Outcome 1 and k2 of Outcome 2, must have n − k1 − k2 of Outcome 3. The probability of such a sequence is p1k1 p2k2 (1 − p2 − p2 ) n − k1 − k2 . The number of such sequences depends on the number of ways to choose the k1 positions in the sequence for Outcome 1 and the k2 positions for Outcome 2. The k1 positions can be chosen in ⎛ n − k1 ⎞ ⎛n ⎞ ⎜⎝ k ⎟⎠ ways. For each such choice, the k2 positions can be chosen in ⎜⎝ k ⎠⎟ ways. Thus, P(k1) of 1 2 ⎛ n ⎞ ⎛ n − k1 ⎞ k1 k2 p1 p2 (1 − p1 − p2 ) n − k1 − k2 Outcome 1 and k2 of Outcome 2) = ⎜ ⎟ ⎜ ⎝ k1 ⎠ ⎝ k2 ⎟⎠ (n − k1 )! n! p1k1 p2k2 (1 − p1 − p2 ) n − k1 − k2 k1 !(n − k1 )! k2 !(n − k1 − k2 )! n! = p1k1 p2k2 (1 − p1 − p2 ) n − k1 − k2 k1 !k2 !(n − k1 − k2 )!

=

3.2.19 In the notation of Question 3.2.18, p1 = 0.5 and p2 = 0.3, with n = 10. Then the probability of 3 of 10! Outcome 1 and 5 of Outcome 2 is (0.5)3 (0.3)5 (0.2)2 = 0.031 3!5!2!

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30

Chapter 3: Random Variables

3.2.20 Use the hypergeometric model with N = 12, n = 5, r = 4, and w = 12 − 4 = 8. The probability that ⎛ 4⎞ ⎛8⎞ ⎜⎝ 2 ⎠⎟ ⎝⎜ 3 ⎠⎟ the committee will contain two accountants (k = 2) is = 14/33 ⎛12 ⎞ ⎜⎝ 5 ⎠⎟ 3.2.21 “At least twice as many black bears as tan-colored” translates into spotting 4, 5, or 6 black bears. ⎛ 6 ⎞ ⎛ 3 ⎞ ⎛ 6 ⎞ ⎛ 3⎞ ⎛ 6 ⎞ ⎛ 3 ⎞ ⎜⎝ 4 ⎠⎟ ⎝⎜ 2 ⎠⎟ ⎝⎜ 5 ⎠⎟ ⎝⎜1 ⎠⎟ ⎝⎜ 6 ⎠⎟ ⎝⎜ 0 ⎠⎟ + + = 64/84 The probability is ⎛9⎞ ⎛9 ⎞ ⎛9⎞ ⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠ 3.2.22 The probabilities are hypergeometric with N = 4050, n = 65, r = 514 , and w = 4050 − 514 = 3536. The probability that k children have not been vaccinated is ⎛ 514 ⎞ ⎛ 3536 ⎞ ⎜⎝ k ⎟⎠ ⎜⎝ 65 − k ⎟⎠ , k = 0, 1, 2, …, 65 ⎛ 4050 ⎞ ⎜⎝ 65 ⎟⎠ 3.2.23 The probability that k nuclear missiles will be destroyed by the anti-ballistic missiles is hypergeometric with N = 10, n = 7, r = 6, and w = 10 − 6 = 4. The probability the Country B will be hit by at least one nuclear missile is one minus the probability that k = 6, or ⎛6⎞ ⎛ 4⎞ ⎜⎝ 6 ⎟⎠ ⎜⎝1 ⎟⎠ = 0.967 1− ⎛10 ⎞ ⎜⎝ 7 ⎟⎠ 3.2.24 Let k be the number of questions chosen that Anne has studied. Then the probabilities for k are hypergeometric with N = 10, n = 5, r = 8, and w = 10 − 8 = 2. The probability of her getting at least four correct is the probability that k = 4 or 5, which is ⎛8 ⎞ ⎛ 2 ⎞ ⎛8 ⎞ ⎛ 2 ⎞ ⎜⎝ 4 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎜⎝ 0 ⎟⎠ 140 56 + = 0.778 + = 252 252 ⎛10 ⎞ ⎛10 ⎞ ⎜⎝ 5 ⎟⎠ ⎜⎝ 5 ⎟⎠

3.2.25 The probabilities for the number of men chosen are hypergeometric with N = 18, n = 5, r = 8, and w = 10. The event that both men and women are represented is the complement of the event that 0 ⎛ 8 ⎞ ⎛10 ⎞ ⎛ 8 ⎞ ⎛10 ⎞ ⎜⎝ 0 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎜⎝ 0 ⎟⎠ 252 56 + or 5 men will be chosen, or 1 − = 0.964 + =1− 8568 8568 ⎛18 ⎞ ⎛18 ⎞ ⎜⎝ 5 ⎠⎟ ⎝⎜ 5 ⎠⎟

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Section 3.2: Binomial and Hypergeometric Probabilities

31

3.2.26 The probability is hypergeometric with N = 80, n = 10, r = 20, and w = 60, and equals ⎛ 20 ⎞ ⎛ 60 ⎞ ⎜⎝ 6 ⎟⎠ ⎜⎝ 4 ⎟⎠ = 0.0115 ⎛ 80 ⎞ ⎜⎝10 ⎟⎠ 3.2.27 First, calculate the probability that exactly one real diamond is taken during the first three grabs. There are three possible positions in the sequence for the real diamond, so this probability is 3(10)(25)(24) . (35)(34)(33)

The probability of a real diamond being taken on the fourth removal is 9/32. Thus, the desired 3(10)(25)(24) 9 162,000 probability is × = = 0.129. (35)(34)(33) 32 1, 256,640

3.2.28

⎛ 2⎞ ⎛8 ⎞ ⎛ 6 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛8 ⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ + ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ + ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ 1 + 15 + 1 17 = = = 0.378 45 45 ⎛10 ⎞ ⎜⎝ 2 ⎟⎠

⎛N ⎞ 3.2.29 The k-th term of (1 + µ)N = ⎜ ⎟ µ k ⎝k ⎠

(1 + µ) (1 + µ) r

N−r

⎛ r ⎛ r ⎞ i ⎞ ⎛ N −r ⎛ N − r ⎞ j ⎞ = ⎜ ⎜ ⎟µ ⎟ ⎜ ⎜ ⎟µ ⎟ ⎝ i =1 ⎝ i ⎠ ⎠ ⎝ j =1 ⎝ j ⎠ ⎠

∑

∑

⎛r⎞ ⎛ N − r ⎞ k µ ⎠⎟ i =1 r

The k-th term of this product is

∑ ⎜⎝ i ⎠⎟ ⎝⎜ k − i

⎛N ⎞ Equating coefficients gives ⎜ ⎟ = ⎝k⎠

k

⎛r⎞ ⎛ N − r ⎞

∑ ⎜⎝ i ⎟⎠ ⎜⎝ k − i ⎟⎠ . i =1

⎛N ⎞ Dividing through by ⎜ ⎟ shows that the hypergeometric terms sum to 1. ⎝k⎠

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32

3.2.30

Chapter 3: Random Variables

⎛r ⎞ ⎛w ⎞ ⎛r ⎞ ⎛w ⎞ ⎜⎝ k + 1⎟⎠ ⎜⎝ n − k − 1⎟⎠ ⎜⎝ k ⎟⎠ ⎜⎝ n − k ⎟⎠ ⎛ r ⎞ ⎛ w ⎞ ⎛r ⎞ ⎛w ⎞ ÷ =⎜ ÷ ⎟ ⎜ ⎝ k + 1⎠ ⎝ n − k − 1⎠⎟ ⎝⎜ k ⎠⎟ ⎝⎜ n − k ⎠⎟ ⎛N⎞ ⎛N⎞ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎛r ⎞ ⎛w ⎞ ⎛r ⎞ ⎛w ⎞ = ⎜ ÷ ⎟ ⎜ ⎟ ⎝ k + 1⎠ ⎝ n − k − 1⎠ ⎜⎝ k ⎟⎠ ⎜⎝ n − k ⎟⎠ =

r! w! k !(r − k )! (n − k )!( w − n + k )! ⋅ ⋅ ⋅ (k + 1)!(r − k − 1)! (n − k − 1)!( w − n + k + 1)! r! w!

=

n−k r−k ⋅ ⋅ (k + 1) ( w − n + k + 1)

3.2.31 Let W0, W1 and W2 be the events of drawing zero, one, or two white chips, respectively, from Urn I. Let A be the event of drawing a white chip from Urn II. Then P(a) = P(A|W0)P(W0) + P(A|W1)P(W1) + P(A|W2)P(W2) ⎛5 ⎞ ⎛ 4⎞ ⎛ 5⎞ ⎛ 4 ⎞ ⎛5 ⎞ ⎛ 4⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 5 ⎝ 2 ⎠ ⎝ 0 ⎠ 6 ⎝ 1 ⎠ ⎝ 1 ⎠ 7 ⎜⎝ 0 ⎟⎠ ⎜⎝ 2 ⎟⎠ = + + = 53/99 11 ⎛ 9 ⎞ 11 ⎛ 9 ⎞ 11 ⎛ 9 ⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

3.2.32 For any value of r = number of defective items, the probability of accepting the sample is ⎛ r ⎞ ⎛100 − r ⎞ ⎛ r ⎞ ⎛100 − r ⎞ ⎜⎝ 0 ⎟⎠ ⎜⎝ 10 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝ 9 ⎟⎠ pr = + ⎛100 ⎞ ⎛100 ⎞ ⎜⎝ 10 ⎟⎠ ⎜⎝ 10 ⎟⎠ Then the operating characteristic curve is the plot of the presumed percent defective versus the probability of accepting the shipment, or 100(r/100) = r on the x-axis and pr on the y-axis. If there are 16 defective, you will accept the shipment approximately 50% of the time.

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Section 3.2: Binomial and Hypergeometric Probabilities

33

r! ways to divide the red chips into three groups of the given sizes. There are r1 !r2 !r3 ! ( N − r )! ways to divide the white chips into the three groups of the required (n1 − r1 )!(n2 − r2 )!(n3 − r3 )! sizes. The total number of ways to divide the N objects into groups of n1, n2, and n3 objects is ⎛ n1 ⎞ ⎛ n2 ⎞ ⎛ n3 ⎞ ( N − r )! r! r !r !r ! (n − r )!(n2 − r2 )!(n3 − r3 )! ⎜⎝ r1 ⎟⎠ ⎜⎝ r2 ⎟⎠ ⎜⎝ r3 ⎠⎟ N! . Thus, the desired probability is 1 2 3 1 1 = . N! n1 !n2 !n3 ! ⎛N ⎞ ⎜⎝ r ⎟⎠ n1 !n2 !n3 !

3.2.33 There are

3.2.34 First, calculate the probability that the first group contains two disease carriers and the others ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎜⎝ 2 ⎠⎟ ⎝⎜ 1 ⎠⎟ ⎝⎜ 1 ⎠⎟ = 49/285. have one each. The probability of this, according to Question 3.2.33, is ⎛ 21⎞ ⎜⎝ 4 ⎟⎠ The probability that either of the other two groups has 2 carriers and the others have one is the same. Thus, the probability that each group has at least one diseased member is 49 49 3 = = 0.516. Then the probability that at least one group is disease free is 285 95 1 − 0.516 = 0.484. ⎛ ni ⎞ ⎛N ⎞ 3.2.35 There are ⎜ ⎟ total ways to choose the sample. There are ⎜ ⎟ ways to arrange for ki of the ni ⎝n⎠ ⎝ ki ⎠ objects to be chosen, for each i. Using the multiplication rule shows that the probability of getting k1 objects of the first kind, k2 objects of the second kind, …, kt objects of the t-th kind is ⎛ n1 ⎞ ⎛ n2 ⎞ ⎛ nt ⎞ ⎜⎝ k ⎟⎠ ⎜⎝ k ⎟⎠ " ⎝⎜ k ⎠⎟ 1 2 t N ⎛ ⎞ ⎜⎝ n ⎟⎠

3.2.36 In the notation of Question 3.2.33, let n1 = 5, n2 = 4, n3 = 4, n4 = 3, so N = 16. The sample size is given to be n = 8, and k1 = k2 = k3 = k4 = 2. Then the probability that each class has two ⎛5 ⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛3 ⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ (10)(6)(6)(3) 1080 representatives is = = = 0.084. 12,870 12,870 ⎛16 ⎞ ⎜⎝ 8 ⎟⎠

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34

Chapter 3: Random Variables

Section 3.3: Discrete Random Variables 3.3.1

(a) Each outcome has probability 1/10 Outcome 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5 3, 4 3, 5 4, 5

X = larger no. drawn 2 3 4 5 3 4 5 4 5 5

Counting the number of each value of the larger of the two and multiplying by 1/10 gives the pdf: pX(k) 1/10 2/10 3/10 4/10

k 2 3 4 5 (b) Outcome 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5 3, 4 3, 5 4, 5

k 3 4 5 6 7 8 9

X = larger no. drawn 2 3 4 5 3 4 5 4 5 5

V = sum of two nos. 3 4 5 6 5 6 7 7 8 9

pX(k) 1/10 1/10 2/10 2/10 2/10 1/10 1/10

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Section 3.3: Discrete Random Variables

3.3.2

35

(a) There are 5 × 5 = 25 total outcomes. The set of outcomes leading to a maximum of k is (X = k) = {(j, k)|1 ≤ j ≤ k − 1} ∪ {(k, j)|1 ≤ j ≤ k − 1} ∪ {(k, k}, which has 2(k − 1) + 1 = 2k − 1 elements. Thus, pX(k) = (2k − 1)/25 (b) Outcomes (1, 1) (1, 2) (2, 1) (1, 3) (2, 2) (3, 1) (1, 4) (2, 3) (3, 2) (4, 1) (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (2, 5) (3, 4) (4, ,3) (5, 2) (3, 5) (4, 4), (5, 3) (4, 5) (5, 4) (5, 5)

V = sum of two nos. 2 3 4 5 6 7 8 9 10

pV(k) = (k − 1)/25 for k = 1, 2, 3, 4, 5, 6 and pV(k) = (11 − k)/25 for k > 6 3.3.3

pX(k) = P(X = k) = P(X ≤ k) − P(X ≤ k − 1). But the event (X ≤ k) occurs when all three dice are ≤ k and that can occur in k3 ways. Thus P(X ≤ k) = k3/216. Similarly, P(X ≤ k − 1) = (k − 1)3/216. Thus pX(k) = k3/216 − (k − 1)3/216.

3.3.4

pX(1) = 6/63 = 6/216 = 1/36 pX(2) = 3(6)(5)/63 = 90/216 = 15/36 pX(3) = (6)(5)(4)/63 = 120/216 = 20/36

3.3.5 Outcomes (H, H, H) (H, H, T) (H, T, H) (T, H, H) (T, T, H) (T, H, T) (T, H, H) (T, T, T)

V = no. heads – no. tails 3 1 −1 −3

pX(3) = 1/8, pX(1) = 3/8, pX(−1) = 3/8, pX(−3) = 1/8 3.3.6 Outcomes (1, 1) (2, 1) (1, 3) (3, 1) (1, 4) (2, 3) (4, 1) (1, 5) (2, 4) (3, 3) (1, 6) (2, 5) (3, 4) (4, 3) (2, 6) (3, 5) (4, 4) (1, 8) (3, 6) (4, 5) (2, 8) (4, 6) (3, 8) (4, 8)

k 2 3 4 5 6 7 8 9 10 11 12

pX(k) (1/6)(1/6) = 1/36 (2/6)(1/6) = 2/36 (1/6)(1/6) + (2/6)(1/6) = 3/36 (1/6)(1/6) + (2/6)(1/6) + (1/6)(1/6) = 4/36 (1/6)(1/6) + (2/6)(1/6) + (2/6)(1/6) = 5/36 (1/6)(1/6) +(2/6)(1/6)+ (2/6)(1/6)+ (1/6)(1/6) = 6/36 (2/6)(1/6) + (2/6)(1/6) + (1/6)(1/6) = 5/36 (1/6)(1/6) + (2/6)(1/6) + (1/6)(1/6) = 4/36 (2/6)(1/6) + (1/6)(1/6) = 3/36 (2/6)(1/6) = 2/36 (1/6)(1/6) = 1/36

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36

3.3.7

Chapter 3: Random Variables

This is similar to Question 3.3.5. If there are k steps to the right (heads), then there are 4 − k steps to the left (tails). The final position X is number of heads − number of tails = k − (4 − k) = 2k − 4. ⎛4⎞ 1 The probability of this is the binomial of getting k heads in 4 tosses = ⎜ ⎟ . ⎝ k ⎠ 16 ⎛4⎞ 1 Thus, pX(2k − 4) = ⎜ ⎟ , k = 0, 1, 2, 3, 4 ⎝ k ⎠ 16

3.3.8 3.3.9

3.3.10

k 4− k ⎛4⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ pX(2k − 4) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ , k = 0, 1, 2, 3, 4 ⎝k ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠

Consider the case k = 0 as an example. If you are on the left, with your friend on your immediate right, you can stand in positions 1, 2, 3, or 4. The remaining people can stand in 3! ways. Each of these must be multiplied by 2, since your friend could be the one on the left. The total number of permutations of the five people is 5! Thus, pX(0) = (2)(4)(3!)/5! = 48/120 = 4/10. In a similar manner pX(1) = (2)(3)(3!)/5! = 36/120 = 3/10 pX(2) = (2)(2)(3!)/5! = 24/120 = 2/10 pX(3) = (2)(1)(3!)/5! = 12/120 = 1/10 ⎛2⎞ ⎛ 2 ⎞ ⎜⎝ k ⎠⎟ ⎝⎜ 2 − k ⎠⎟ , k = 0, 1, 2. For X1 + X2 = m, let p X1 ( k ) = p X 2 ( k ) = ⎛4⎞ ⎜⎝ 2 ⎟⎠

X1 = k and X2 = m − k, for k = 0, 1, …, m. Then p X 3 (m) =

m

∑p k =0

X1 ( k ) p X 2 ( m

− k) ,

m = 0, 1, 2, 3, 4, or m

p X 3 ( m)

0 1 2 3 4

1/36 2/9 1/2 2/9 1/36

k −1 k −1 ⎛ 4 ⎞ 4− 2 ⎛1⎞ 2 2 ⎛ k −1⎞ ⎜ ⎛ ⎞ 3.3.11 P (2 X + 1 = k ) = P[ X = (k − 1) / 2] , so p2 X +1 (k ) = p X ⎜ , = k − 1⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎜ ⎟⎝ 3⎠ ⎝3⎠ ⎝ 2 ⎠ k = 1, 3, 5, 7, 9

3.3.12 FX(k) = P(X ≤ k) = k3, as explained in the solution to Question 3.3.3. ⎛4⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ 3.3.13 FX(k) = P(X≤k) = P( X = j ) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝6⎠ j =0 j =0 ⎝ j ⎠ 6 k

∑

k

∑

j

4− j

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Section 3.4: Continuous Random Variables

3.3.14 px(k) = FX(k) − FX(k − 1) =

37

k ( k + 1) ( k − 1)k k − = 42 42 21

3.3.15 See the solution to Question 3.3.3.

Section 3.4: Continuous Random Variables 3.4.1

3.4.2

3.4.3

P(0 ≤ Y ≤ 1/2) =

∫

1/ 2 0

4 y 3 dy = y 4

1/ 2

= 1/16 0

2 2 2 y y2 P(3/4 ≤ Y ≤ 1) = + y dy = + 3/ 4 3 3 3 3

∫

1

P ( Y − 1/ 2 < 1/ 4) = P(1/4 < Y < 3/4) =

3.4.4

P(Y > 1) =

3.4.5

(a)

∫

∞ 10

∫

3 1

(1/ 9) y 2 dy = (1/ 27) y 3 ∞

0.2e−0.2 y dy = − e −0.2 y

3 1

∫

1

= 1− 3/ 4

3/ 4 1/ 4

11 5 = 16 16

3 2 y3 y dy = 2 2

3/ 4

= 1/ 4

27 1 26 13 − = = 128 128 128 64

= 1 − 1/27 = 26/27

= e−2 = 0.135

10

(b) If A = probability customer leaves on first trip, and B = probability customer leaves on second trip, then P(a) = P(b) = 0.135. In this notation, pX(1) = P(a)P(BC) + P(AC)P(b) = 2(0.865)(0.135) = 0.23355 3.4.6

Clearly the function given is non-negative and continuous. We must show that it integrates to 1. 1

∫

1 0

(n + 2)(n + 1) y (1 − y )dy = n

∫

1 0

( n + 2)(n + 1)( y − y n

n +1

⎛ y n +1 y n + 2 ⎞ )dy = (n + 2)(n + 1) ⎜ − ⎝ n + 1 n + 2 ⎠⎟ 0

1

= ⎡⎣ (n + 2) y n +1 − (n + 1) y n + 2 ⎤⎦ = 1 0

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38

3.4.7

3.4.8 3.4.9

Chapter 3: Random Variables

FY(y) = P(Y ≤ y) =

FY(y) = P(Y ≤ y) =

∫ ∫

y

0

y

0

4t 3 dt = t 4

y

= y 4 . Then P(Y ≤ 1/2) = FY(1/2) = (1/2)4 = 1/16

0 y

λ e − λt dt = − e− λt = 1 − e − λ y 0

For y < −1, FY(y) = 0 1 1 + y + y2 −1 2 2 y y 1 1 1 1 − t ) dt = + (1 − t )dt = + y − y 2 For 0 ≤ y ≤ 1, FY(y) = ( 0 −1 2 2 2 For y > 1, FY(y) = 1

For −1 ≤ y < 0, FY(y) =

y

∫

(1 + t )dt =

∫

∫

FY(y)

3.4.10 (1) P(1/2 < Y ≤ 3/4) = FY(3/4) − FY(1/2) = (3/4)2 − (1/2)2 = 0.3125 d d 2 (2) fY(y) = FY = y = 2y, 0 ≤ y < 1 dy dy P(1/2 < Y ≤ 3/4) =

∫

3/ 4 1/ 2

2ydy = y 2

3/ 4 1/ 2

= 0.3125

3.4.11 (a) P(Y < 2) = FY(2), since FY is continuous over [0, 2]. Then FY(2) = ln 2 = 0.693 (b) P(2 < Y ≤ 2.5) = FY(2.5) − FY(2) = ln 2.5 − ln2 = 0.223 (c) The probability is the same as (b) since FY is continuous over [0, e] (d) fY(y) =

d d 1 FY ( y ) = ln y = , 1 ≤ y ≤ e dy dy y

3.4.12 First note that fY(y) =

d d FY ( y ) = (4 y 3 − 3 y 4 ) = 12y2 − 12y3, 0 ≤ y ≤ 1. dy dy

Then P(1/4 < Y ≤ 3/4) =

∫

3/ 4 1/ 4

(12 y 2 − 12 y 3 )dy = (4 y 3 − 3 y 4 )

3/ 4 1/ 4

= 0.6875.

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Section 3.5: Expected Values

39

d 1 2 1 1 ( y + y 3 ) = y + y 2 ,0 ≤ y ≤ 2 dy 12 6 4

3.4.13 fY(y) =

3.4.14 Integrating by parts, we find that FY(y) =

3.4.15 F ′ ( y ) = −1(1 + e − y )−2 (− e − y ) =

∫

y 0

te − t dt = −te − t − e − t

y 0

= 1 − (1 + y )e − y .

e− y > 0 , so F(y) is increasing. The other two assertions (1 + e− y ) 2

follow from the facts that lim e − y = ∞ and lim e − y = 0 . y → −∞

y→ ∞

3.4.16 FW ( w) = P (W ≤ w) = P (2Y ≤ w) = P (Y ≤ w / 2) = FY ( w / 2) d d 1 1 fW ( w) = F ( w) = FY ( w / 2) = fW ( w / 2) ⋅ ( w / 2) ′ = 4( w / 2)3 = w3 dw dw 2 4 where 2(0) ≤ w ≤ 2(1) or 0 ≤ w ≤ 2 3.4.17 P(−a < Y < a) = P(−a < Y ≤ 0) + P(0 < Y < a) =

∫

=

∫

0

fY ( y )dy +

−a a 0

fY ( y )dy +

∫

∫

a 0 a

0

fY ( y )dy = −

∫

0 a

fY (− y )dy +

∫

a 0

fY ( y )dy

fY ( y )dy = 2[ FY (a ) − FY (0)]

But by the symmetry of fY, FY(0) = 1/2. Thus, 2[FY(a) − FY(0)] = 2[FY(a) − 1/2] = 2FY(a) − 1 3.4.18 FY(y) =

∫

y 0

(1/ λ )e − t / λ dt = 1 − e−t/λ, so

(1/ λ )e − y / λ = 1/λ 1 − (1 − e− y / λ ) Since the hazard rate is constant, the item does not age. Its reliability does not decrease over time.

h(y) =

Section 3.5: Expected Values 3.5.1

E(X) = −1(0.935) + 2(0.0514) + 18(0.0115) + 180(0.0016) + 1,300(1.35 × 10−4) + 2,600(6.12 × 10−6) + 10,000(1.12 × 10−7) = −0.144668

3.5.2

Let X be the winnings of betting on red in Monte Carlo. Then E ( X ) =

3.5.3

E(X) = $30,000(0.857375) + $18,000(0.135375) + $6,000(0.007125) + (−$6,000)(0.000125) = $28,200.00

18 19 −1 − = . 37 37 37 18 20 −2 Let X* be the winnings of betting on red in Las Vegas. Then E ( X ) = − = . 38 38 38 ⎛ −1 −2 ⎞ The amount bet, M, is the solution to the equation M ⎜ − ⎟ = $3,000 or M is approximately ⎝ 37 38 ⎠ equal to $117, 167.

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40

3.5.4

Chapter 3: Random Variables

⎛2⎞ ⎛ 4⎞ ⎛ 2⎞ ⎛ 4⎞ ⎛ 2⎞ ⎛ 4⎞ ⎜⎝ 0 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ + 2⋅ + 10 ⋅ Rule A: Expected value = −5 + 0 ⋅ = −49/15 ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎛ 2⎞ ⎛ 4⎞ ⎛ 2⎞ ⎛ 4⎞ ⎛ 2⎞ ⎛ 4⎞ ⎜⎝ 0 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ + 1⋅ + 20 ⋅ = −47/15 Rule B: Expected value = −5 + 0 ⋅ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

Neither game is fair to the player, but Rule B has the better payoff. 5

3.5.5

P is the solution to the equation

5

5

∑[kP(1 − p ) − 50,000 p ] = P∑ k (1 − p ) − 50,000∑ p k

k

k =1

k

k =1

k

k =1

= 1000 , where pk is the probability of death in year k, k = 1, 2, 3, 4, 5. 5

Since

∑

5

∑ k (1 − p ) = 4.99164 , the equation becomes

pk = 0.00272 and

k

k =1

k =1

4.99164P – 50,000(0.00272) = 1000, or P = $227.58. 4(20) 4 = . 4 + 96 5

3.5.6

The random variable X is hypergeometric, where r = 4, w = 96, n = 20. Then E(X) =

3.5.7

This is a hypergeometric problem where r = number of students needing vaccinations = 125 and w = number of students already vaccinated = 642 − 125 = 517. An absenteeism rate of 12% corresponds to a sample n = (0.12)(642)  77 missing students. The expected number of 125(77) unvaccinated students who are absent when the physician visits is  15 . 125 + 517

3.5.8

(a) E(Y) =

∫

1 0

y ⋅ 3(1 − y )2 dy =

∫

1 0

3( y − 2 y 2 + y 3 )dy

1

2 1 ⎤ 1 ⎡1 = 3 ⎢ y 2 − y3 + y 4 ⎥ = 3 4 ⎦0 4 ⎣2

(b) E(Y) =

∫

∞ 0

∞

1 1 ⎡ 1 ⎤ y ⋅ 4 ye −2 y dy = 4 ⎢ − y 2 e−2 y − ye−2 y − e−2 y ⎥ = 1 2 4 ⎣ 2 ⎦0

⎛3⎞ (c) E(Y) = y ⋅ ⎜ ⎟ dy + 0 ⎝4⎠

(d) E(Y) =

3.5.9

E(Y) =

∫

3 0

∫

1

∫

π /2 0

∫

3 2

3y2 ⎛1⎞ y ⋅ ⎜ ⎟ dy = ⎝4⎠ 8

1

y2 + 8 0 π /2

y ⋅ sin y dy = (− y cos y + sin y ) 0

1 ⎛1 ⎞ y ⎜ y 2 ⎟ dy = ⎝9 ⎠ 9

∫

3 0

3

=1 2

=1

3

y4 9 y dy = = years 36 0 4 3

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Section 3.5: Expected Values

3.5.10 E(Y) =

∫

b a

41

1 y2 y dy = b−a 2(b − a )

b

= a

b2 a2 b+a − = . This simply says that a uniform 2(b − a) 2(b − a ) 2

bar will balance at its middle. 3.5.11 E(Y) =

∫

∞ 0

y ⋅ λe

−λ y

1 ⎛ ⎞ dy = ⎜ − ye− λ y − e − λ y ⎟ ⎝ ⎠ λ

∞

= 0

1

λ

1 3.5.12 Since 2 ≥ 0, this function will be a pdf if its integral is 1, and y

what would be its expected value is

∫

∞ 1

y

1 dy = y2

∫

∞ 1

∫

∞ 1

∞

1 1 dy = − = 1. However, 2 y1 y

1 ∞ dy = ln y 1 , but this last quantity is infinite. y

3.5.13 Let X be the number of cars passing the emissions test. Then X is binomial with n = 200 and p = 0.80. Two formulas for E(X) are: n 200 ⎛n⎞ ⎛ 200 ⎞ (1) E(X) = k ⎜ ⎟ p k (1 − p ) n − k = k⎜ (0.80)k (0.20) 200− k ⎟ k k ⎠ k =1 ⎝ ⎠ k =1 ⎝ (2) E(X) = np = 200(0.80) = 160

∑

∑

3.5.14 The probability that an observation of Y lies in the interval (1/2, 1) is

∫

1 1/ 2

3 y 2 dy = y 3

1 1/ 2

=

7 8

Then X is binomial with n = 15 and p = 7/8. E(X) = 15(7/8) = 105/8. 3.5.15 If birthdays are randomly distributed throughout the year, the city should expect revenue of ($50)(74,806)(30/365) or $307,421.92. 3.5.16 If we assume that the probability of bankruptcy due to fraud is 23/68, then we can expect 9(23/68) = 3.04, or roughly 3 of the 9 additional bankruptcies will be due to fraud. 3.5.17 For the experiment described, construct the table: Sample 1, 2 1, 3 1, 4 2, 3 2, 4 3, 4

Larger of the two, k 2 3 4 3 4 4

Each of the six samples is equally likely to be drawn, so pX(2) = 1/6, pX(3) = 2/6, and pX(4) = 3/6. Then E(X) = 2(1/6) + 3(2/6) + 4(3/6) = 20/6 = 10/3.

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42

Chapter 3: Random Variables

3.5.18 Outcome HHH HHT HTH HTT THH THT TTH TTT

X 6 2 4 1 2 0 1 0

From the table, we can calculate pX(0) = 1/4, pX(1) = 1/4, pX(2) = 1/4, pX(4) = 1/8, pX(6) = 1/8. Then E(X) = 0 ⋅ (1/4) + 1 ⋅ (1/4) + 2 ⋅ (1/4) + 4 ⋅ (1/8) + 6 ⋅ (1/8) = 2. 3.5.19 The “fair” ante is the expected value of X, which is 9 ∞ 1000 ∞ ⎛ 1 ⎞ 1000 1 1000 5608 ⎛ 1 ⎞ k ⎛ 1 ⎞ = $10.95 =9+ = 2 ⎜ k ⎟+ 1000 ⎜ k ⎟ = 9 + 10 ⎜ k ⎠⎟ = 9 + 10 1 ⎝ ⎠ ⎝ ⎠ ⎝ 512 512 2 2 2 k =0 2 2 1− k =1 k =10 2

∑

∑

∑

∞

3.5.20 (a) E(X) =

k

⎛1⎞ ck ⎜ ⎟ = ⎝2⎠ k =1

∑

∞

(b)

∞

c ∞ ⎛c⎞ c ⎛c⎞ = ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = c 2 2 2 2 − k =1 k =0

∑

k

∑

k

∞ ⎛1⎞ ⎛1⎞ log 2k ⎜ ⎟ = log 2 k ⎜ ⎟ ⎝2⎠ ⎝2⎠ k =1 k =1 To evaluate the sum requires a special technique: For a parameter t, 0 < t < 1, note that ∞ t . Differentiate both sides of the equation with respect to t to obtain tk = 1− t k =1

k

∑

∑

k

∑ ∞

∑ kt

k −1

k =1 ∞

∑

kt k =

k =1

3.5.21 pX(1) =

=

1 . Multiplying both sides by t gives the desired equation: (1 − t ) 2

t . In the case of interest, t = 1/2, so (1 − t ) 2

∞

k

⎛1⎞ k ⎜ ⎟ = 2, and E(X) = 2 ⋅ log 2. ⎝2⎠ k =1

∑

6 1 = 216 36

pX(2) =

3(6)(5) 15 = 216 36

pX(3) =

6(5)(4) 20 = 216 36

E(X) = 1 ⋅

1 15 20 91 + 2 ⋅ + 3⋅ = 36 36 36 36

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Section 3.5: Expected Values

43

3.5.22 For the experiment described, construct the table Sample 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5 3, 4 3, 5 4, 5

Absolute value of difference 1 2 3 4 1 2 3 1 2 1

If X denotes the absolute value of the difference, then from the table: pX(1) = 4/10, pX(2) = 3/10, pX(3) = 2/10, pX(4) = 1/10 E(X) = 1(4/10) + 2(3/10) + 3(2/10) + 4(1/10) = 2 k ⎛ k − 1⎞ ⎛ 1 ⎞ 3.5.23 Let X be the length of the series. Then pX(k) = 2 ⎜ ⎜ ⎟ , k = 4, 5, 6, 7. ⎝ 3 ⎟⎠ ⎝ 2 ⎠

⎛ k − 1⎞ ⎛ 1 ⎞ ⎛2⎞ ⎛4⎞ ⎛5⎞ ⎛ 5 ⎞ 93 E(X) = = 5.8125 (k )(2) ⎜ ⎜⎝ ⎟⎠ = 4 ⎜⎝ ⎟⎠ + 5 ⎜⎝ ⎟⎠ + 6 ⎜⎝ ⎟⎠ + 7 ⎜⎝ ⎟⎠ = ⎟ 16 16 16 16 16 ⎝ 3 ⎠ 2 k =4 k

7

∑

3.5.24 Let X = number of drawings to obtain a white chip. Then 1 1 , k = 1, 2, … pX(k) = ⋅ k k +1 ∞ ⎛ 1 ⎞ ∞ 1 E(X) = . k⎜ ⎟= k =1 ⎝ k ( k + 1) ⎠ k =1 k + 1

∑

∑

i = 2 n+1

For each n, let Tn =

∑

i = 2n ∞

∞

1

∑ k + 1 ≥ ∑T

n

k =1

n =1

=

2n 1 1 . Then Tn ≥ n +1 = . i 2 2

1 1 1 + + + " This last sum is infinite, so E(X) does not exist. 2 2 2

⎛r ⎞ ⎛ w ⎞ ⎛ w ⎞ r! ⎜⎝ k ⎟⎠ ⎜⎝ n − k ⎟⎠ r k !(r − k )! ⎜⎝ n − k ⎟⎠ 3.5.25 E(X) = = k k (r + w)! ⎛ r + w⎞ k =1 k =1 ⎜⎝ n ⎟⎠ n !(r + w − n)! Factor out the presumed value of E(X) = rn/(r + w): ⎛ w ⎞ ⎛ r − 1⎞ ⎛ w ⎞ (r − 1)! ⎜ ⎟ r r ⎜ (k − 1)!(r − k )! ⎝ n − k ⎠ ⎝ k − 1⎟⎠ ⎜⎝ n − k ⎟⎠ rn rn E(X) = = (r − 1 + w)! r + w k =1 r + w k =1 ⎛ r − 1 + w ⎞ ⎜⎝ n − 1 ⎠⎟ ( n − 1)!(r + w − n)! r

∑

∑

∑

∑

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44

Chapter 3: Random Variables

Change the index of summation to begin at 0, which gives ⎛ r − 1⎞ ⎛ w ⎞ r −1 ⎜ ⎝ k ⎠⎟ ⎝⎜ n − 1 − k ⎠⎟ rn E(X) = . The terms of the summation are urn probabilities where r + w k =0 ⎛ r − 1 + w⎞ ⎜⎝ n − 1 ⎟⎠

∑

there are r − 1 red balls, w white balls, and a sample size of n − 1 is drawn. Since these are the probabilities of a hypergeometric pdf, the sum is one. This leaves us with the desired equality rn . E(X) = r+w ∞

∑ jp

3.5.26 E(X) =

X

( j) =

j =1

3.5.27 (a) 0.5 = (b) 0.5 =

∫

m

0

∫

∞

j

∑∑ p

( j) =

j =1 k =1

∞

∞

∑∑ p

X

( j) =

k =1 j = k

(θ + 1) yθ dy = yθ +1

m 0

∞

∑ P( X ≥ k ) k =1

1

= mθ +1 , so m = (0.5)θ +1

⎛y 1⎞ y⎞ ⎜⎝ y + ⎟⎠ dy = ⎜ + ⎟ 2 ⎝ 2 2⎠ 2

m⎛

0

X

m

= 0

m2 m + . Solving the quadratic equation 2 2

−1 + 5 1 2 . (m + m − 1) = 0 gives m = 2 2

3.5.28 E(3X − 4) = 3E(X) − 4 = 3(10)(2/5) − 4 = 8 3.5.29 $100(12)(0.11) = $132 1

⎛2 ⎛ 1 ⎞ w2 ⎞ 3.5.30 E (W ) = w ⎜ − 1⎟ dw = ⎜ w3/ 2 − = 1/ 6 0 ⎝ w ⎠ 2 ⎟⎠ ⎝3 0

∫

1

Also, E (W ) = E (Y 2 ) =

3.5.31 E(Q) =

∫

∞ 0

1

2 ⎞ ⎛2 y 2 [2(1 − y )] dy = ⎜ y 3 − y 4 ⎟ = 1/ 6 . 0 ⎝3 4 ⎠0

∫

1

2(1 − e −2 y )6e −6 y dy = 12

∫

∞

(e −6 y − e −8 y )dy

0

∞

1 1 ⎡ 1 ⎤ = 12 ⎢ − e−6 y + e −8 y ⎥ = , or $50,000 8 ⎣ 6 ⎦0 2 1

1 ⎤ ⎡1 3.5.32 E(Volume) = 5 y 6 y (1 − y )dy = 30 ( y − y )dy = 30 ⎢ y 4 − y 5 ⎥ = 1.5 in3 0 0 5 ⎦0 ⎣4

∫

1

∫

2

1

3

4

1 (100 − y )dy 0 5000 100 1 1 1 ⎡2 2 5/ 2 ⎤ 1/ 2 3/ 2 3/ 2 = (100 y − y )dy = 100 y − y ⎥ 500 0 500 ⎢⎣ 3 5 ⎦0 = 53.3, so the professor’s “curve” did not work.

3.5.33 Class average = E(g(Y)) =

∫

100

10 y1/ 2

∫

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Section 3.6: The Variance

45 1

2

2⎞ 4 2 ⎞ ⎛1 3.5.34 E (W ) = ⎜ y − ⎟ (2 y ) dy = 2 ⎜ y 4 − y 3 + y 2 ⎟ = 1/ 36 0⎝ ⎝4 3⎠ 9 9 ⎠0

∫

1⎛

1 3.5.35 The area of the triangle = y 2 , so E(Area) = 4 n

3.5.36 1 =

∑ ki = k i =1

⎛1 E⎜ ⎝X

⎞ ⎟⎠ =

n

∫

10 6

1 2 1 1 y3 y dy = 4 10 − 6 16 3

10

= 16.33. 6

n(n + 1) 2 implies k = n(n + 1) 2 1

2

∑ i n(n + 1)i = 2/(n + 1) i =1

Section 3.6: The Variance 3.6.1

If sampling is done with replacement, X is binomial with n = 2 and p = 2/5. By Theorem 3.5.1, µ = 2(2/5) = 4/5. E(X 2) = 0 ⋅ (9/25) + 1⋅ (12/25) + 4 ⋅ (4/25) = 28/25. Then Var(X) = 28/25 − (4/5)2 = 12/25.

3.6.2

µ=

∫

⎛3⎞ y ⎜ ⎟ dy + 0 ⎝4⎠ 1

∫

3 2

⎛1⎞ y ⎜ ⎟ dy = 1 ⎝4⎠

⎛3⎞ y 2 ⎜ ⎟ dy + 0 ⎝4⎠ 11 5 Var(X) = −1 = 6 6

E(X 2) =

3.6.3

1

∫

∫

3 2

11 ⎛1⎞ y 2 ⎜ ⎟ dy = ⎝4⎠ 6

Since X is hypergeometric, µ =

3(6) 9 = 10 5

⎛6 ⎞ ⎛ 4 ⎞ ⎜⎝ k ⎟⎠ ⎜⎝ 3 − k ⎟⎠ E(X 2) = k 2 = 0 ⋅ (4/120) + 1 ⋅ (36/120) + 4 ⋅ (60/12) + 9 ⋅ (20/120) = ⎛10 ⎞ k =0 ⎜⎝ 3 ⎠⎟ 456/120 = 38/10 Var(X) = 38/10 − (9/5)2 = 28/50 = 0.56, and σ = 0.748 3

∑

3.6.4

µ = 1/2. E(Y 2) =

3.6.5

µ=

∫

1 0

∫

1 0

y 2 (1)dy = 1/3. Var(Y) = 1/3 − (1/2)2 = 1/12

∫

1

y3(1 − y ) 2 dy = 3 ( y − 2 y 2 + y 3 ) dy = 1/4

E(Y 2) =

∫

1 0

0

∫

1

y 2 3(1 − y ) 2 dy = 3 ( y 2 − 2 y 3 + y 4 )dy = 1/10 0

Var(Y) = 1/10 − (1/4) = 3/80 2

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46

3.6.6

Chapter 3: Random Variables

µ=

∫

k

y

0

2y 2k dy = . E(Y 2) = 2 3 k

∫

k

y2

0

2y k2 dy = 2 k2

2

k 2 ⎛ 2k ⎞ k2 k2 −⎜ ⎟ = . Var(Y) = 2 implies = 2 or k = 6. 2 ⎝ 3 ⎠ 18 18

Var(Y) =

3.6.7

⎧1 − y, ⎪ fY(y) = ⎨1/ 2, ⎪0, ⎩

µ=

∫

1 0

0 ≤ y ≤1 2≤ y≤3 elsewhere

y (1 − y )dy +

∫

E(Y 2) =

1 0

∫

3 2

⎛1⎞ y ⎜ ⎟dy = 17/12 ⎝2⎠ ⎛1⎞ y 2 ⎜ ⎟ dy = 13/4 ⎝2⎠

3

∫

y 2 (1 − y )dy +

2

σ = 13/ 4 − (17 /12) 2 = 179 /12 = 1.115

3.6.8

(a)

∫

∞

2 −1 dy = 2 = 1 3 y y 1

∞ 1

(b) E(Y) =

∫

(c) E(Y 2) = 3.6.9

∞

2 −2 =2 y 3 dy = y 1 y

∞ 1

∫

∞ 1

y2

2 dy = 2ln y y3

∞ 1

, which is infinite.

Let Y = Frankie’s selection. Johnny wants to choose k so that E[(Y − k)2] is minimized. The minimum occurs when k = E(Y) = (a + b)/2 (see Question 3.6.13).

3.6.10 E(Y) = E(Y2) =

∫

1 0

∫

y (5 y )dy = 5 4

1 0

∫

1 0

y 2 (5 y 4 )dy = 5

1

5 5 y dy = y 6 = 6 0 6 5

∫

1 0

1

y 6 dy =

5 7 5 y = 7 0 7

2

Var(Y) = E(Y2) − E(Y)2 =

5 ⎛5⎞ 5 25 5 −⎜ ⎟ = − = 7 ⎝6⎠ 7 36 252

3.6.11 Using integration by parts, we find that E(Y2) =

∫

∞ 0

y 2 λ e− λ y dy = − y 2 e− λ y

The right hand term is 2

∫

∞ 0

∞ 0

+

ye − λ y dy =

Then Var(Y) = E(Y2) − E(Y)2 =

∫

∞ 0

2

2 ye− λ y dy = 0 +

λ∫

∞ 0

y λ e − λ y dy =

∫ 2

λ

∞ 0

2 ye − λ y dy,

E (Y ) =

2 1

λλ

=

2

λ2

.

2

1 ⎛1⎞ −⎜ ⎟ = 2. 2 λ ⎝λ⎠ λ 2

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Section 3.6: The Variance

47

3.6.12 For the given Y, E(Y) = 1/2 and Var(Y) = 1/4. Then ⎛ 1 1⎞ 3⎞ ⎛ P Y > E (Y ) + 2 Var(Y ) = P ⎜Y > + 2 = P ⎜Y > ⎟ ⎟ ⎝ 2 4⎠ 2⎠ ⎝

(

=

∫

)

∞ 3/ 2

2e−2 y dy = 1 −

∫

3/ 2 ∞

2e −2 y dy = 1 − (1 − e −2(3/ 2) ) = e−3

= 0.0498 3.6.13 E[(X − a)2] = E[((X − µ) + (µ − a))2] = E[(X − µ)2] + E[(µ − a)2] + 2(µ − a)E(X − µ) = Var(X) + (µ − a)2, since E(X − µ) = 0. This is minimized when a = µ, so the minimum of g(a) = Var(X). 3.6.14 Var(−5Y + 12) = (−5)2Var(Y) = 25(3/80) = 15/16 2

5 ⎛5 ⎞ ⎛5⎞ ⎛5 ⎞ ⎛5⎞ 3.6.15 Var ⎜ (Y − 32) ⎟ = ⎜ ⎟ Var(Y), by Theorem 3.6.2. So σ ⎜ (Y − 32) ⎟ = ⎜ ⎟ σ (Y ) = (15.7) = ⎝9 ⎠ ⎝9⎠ ⎝9 ⎠ ⎝9⎠ 9 8.7°C.

3.6.16 (1) E[(W − µ)/σ] = (1/σ)E[(W − µ)] = 0 (2) Var[(W − µ)/σ] = (1/σ2)Var[(W − µ)] = (1/σ2)σ2 = 1 1 1 ⎛ y−a⎞ fU ⎜ , (b − a)(0) + a ≤ y ≤ (b − a)(1) + a, or = ⎟ b−a ⎝b−a⎠ b−a 1 fY(y) = , a ≤ y ≤ b, which is the uniform pdf over [a, b] b−a

3.6.17 (a) fY(y) =

(b) Var(Y) = Var[(b − a)U + a] = (b − a)2 Var(U) = (b − a)2/12 3.6.18 E(Y) = 5.5 and Var(Y) = 0.75. E (W1 ) = 0.2281 + (0.9948) E (Y ) + E ( E1 ) = 0.2281 + (0.9948)(5.5) + 0 = 5.6995 E (W2 ) = −0.0748 + (1.0024) E (Y ) + E ( E2 ) = −0.0748 + (1.0024)(5.5) + 0 = 5.4384 Var(W1 ) = (0.9948)2 Var(Y ) + Var( E1 ) = (0.9948)2 (0.75) + 0.0427 = 0.7849 Var(W2 ) = (1.0024) 2 Var(Y ) + Var( E2 ) = (1.0024) 2 (0.75) + 0.0159 = 0.7695

The above two equalities follow from the second corollary to Theorem 3.9.5. So the second procedure is better, since the mean of W2 is closer to the true mean, and it has smaller variance. r

3.6.19 E(Y ) =

∫

2 0

2

1 1 y r +1 2r y dy = = 2 2 r +1 0 r +1 r

E[(Y − 1)6] =

⎛6 ⎞ j 6− j = ⎜⎝ j ⎟⎠E (Y )(−1) j =0 6

∑

⎛ 6 ⎞ 2r 6− j ⎜⎝ j ⎟⎠ r + 1 (−1) j =0 6

∑

= (1)( 1) + (−6)(1) + 15(4/3) + (−20)(2) + (15)(16/5) + (−6)(32/6) + (1)(64/7) = 1/7

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48

Chapter 3: Random Variables

3.6.20 For the given fY, µ = 1 and σ = 1.

γ1 =

E ⎡⎣ (Y − 1)3 ⎤⎦ 1

=

⎛3 ⎞ 3− j j ⎜⎝ j ⎟⎠E (Y )(−1) = j =0 3

∑

3

⎛3 ⎞

∑ ⎜⎝ j ⎟⎠ ( j !)(−1)

3− j

j =0

(1)(1)(−1) + (3)(1)(1) + (3)(2)(−1) + (1)(6)(1) = 2 3.6.21 For the uniform random variable U, E(U) = 1/2 and Var(U) = 1/12. Also the k-th moment of U is E (U k ) =

∫

1 0

u k du = 1/(k + 1). Then the coefficient of kurtosis is

4 ⎛ 4⎞ 1 ⎛ 1 ⎞ E[(U − µ )4 ] E[(U − 1/ 2) 4 ] = = (144) ⎜⎝ − ⎠⎟ 4 2 2 ⎜ ⎟ 2 σ Var (U ) (1/12) k =0 ⎝ k ⎠ k + 1 = (144)(1/80) = 9/5

γ2 =

E[(U − µ ) 4 ]

∑

=

3.6.22 10 = E ⎡⎣ (W − 2)3 ⎤⎦ =

⎛3 ⎞

3

∑ ⎜⎝ j ⎟⎠E (W

j

4− k

)(−2)3− j = (1)(1)(−8) + (3)(2)(4) + (3)E(W 2)(−2) + (1)(4)(1)

j =0

This would imply that E(W 2) = 5/3. In that case, Var(W) = 5/3 − (2)2 < 0, which is not possible. 3.6.23 Let E(X) = µ; let σ be the standard deviation of X. Then E(aX + b) = aµ + b. Also, Var(aX + b) = a2 σ2, so the standard deviation of aX + b = aσ. Then 3 E ⎡((aX + b) − (a µ + b)) ⎤ ⎦ γ1 = ⎣ (aσ )3

a 3 E ⎡⎣ ( X − µ )3 ⎤⎦

E ⎣⎡ ( X − µ )3 ⎦⎤

= γ 1( X ) σ3 a 3σ 3 The demonstration for γ 2 is similar.

=

=

3.6.24 (a) Question 3.4.6 established that Y is a pdf for any positive integer n. As a corollary, we know that 1 =

∫

1 0

∫

1 0

(n + 2)(n + 1 y n (1 − y )dy or equivalently, for any positive integer n,

Then E (Y 2 ) = =

1 (n + 2)(n + 1)

y n (1 − y )dy =

∫

1 0

y n (n + 2)(n + 1) y n (1 − y )dy = (n + 2)(n + 1)

1 0

y n + 2 (1 − y )dy

(n + 2)(n + 1) (n + 2)(n + 1) (n + 1) . By a similar argument, E(Y) = = . (n + 4)(n + 3) (n + 4)(n + 3) (n + 3)

Thus, Var(Y) = E (Y 2 ) − E (Y ) 2 =

(b)

∫

E (Y k ) = =

∫

1 0

(n + 2)(n + 1) (n + 1) 2 2(n + 1) − = (n + 4)(n + 3) (n + 3) 2 (n + 4)(n + 3) 2

y k ( n + 2)(n + 1) y n (1 − y )dy = (n + 2)(n + 1)

∫

1 0

y n + k (1 − y )dy

(n + 2)(n + 1) (n + k + 2)(n + k + 1)

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 3.7: Joint Densities

∫

1=

3.6.25 (a)

∞

1

49 ∞

⎡ y −5 ⎤ 1 cy dy = c ⎢ ⎥ = c , so c = 5. 5 ⎣ −5 ⎦1 −6

∞

⎡ y r −5 ⎤ (b) E(Y ) = 5 y y dy = 5 ⎢ ⎥ . For this last expression to be finite, r must be < 5. 1 ⎣ r − 5 ⎦1 The highest integral moment is r = 4.

∫

r

∞

r

−6

Section 3.7: Joint Densities 3.7.1

1=

∑ p( x, y) = c∑ xy = c[(1)(1) + (2)(1) + (2)(2) + (3)(1)] = 10c, so c = 1/10 x, y

3.7.2

1= = c

x, y

1

1

0

0

∫∫

∫

1 0

c( x 2 + y 2 )dxdy = c 1

⎡ x3 ⎤ ⎢ ⎥ dy + c ⎣ 3 ⎦0

1

∫∫

y

3.7.3

1=

3.7.4

1= c ⎛ 0⎝

0

1

0

∫

1 0

1

1

0

0

∫∫

x 2 dxdy + c

1

⎡ y3 ⎤ ⎢ ⎥ dx = c ⎣ 3 ⎦0

∫

1 0

1

1

0

0

∫

11

∫∫

1 dy + c 3 y

⎤ c( x + y )dxdy = c ⎢ + xy ⎥ dy = c 0 2 ⎣ ⎦0

∫ ∫

∫

y 0

xydx ⎞dy = c ⎠

∫

1 ⎡ x2

1 ⎛ ⎡ x2 y ⎤

0

⎞ ⎜⎢ ⎟ ⎥ dy = c 0⎜ ⎝ ⎣ 2 ⎦ 0 ⎟⎠ y

∫

1 0

∫

y 2 dydx

3

dx =

1

⎡ y3 ⎤ c dy = c ⎢ ⎥ = , so c = 2. 2 ⎣ 2 ⎦0 2

1 3y2 0

2 c , c = 3/2. 3

y3 dy 2

1

⎡ y4 ⎤ ⎛1⎞ c ⎢ ⎥ = c ⎜ ⎟ , so c = 8 ⎝8⎠ ⎣ 8 ⎦0

3.7.5

4 ⎛3 ⎞ ⎛ 2 ⎞ ⎛ ⎞ ⎜⎝ x ⎠⎟ ⎝⎜ y ⎠⎟ ⎝⎜ 3 − x − y ⎠⎟ , 0 ≤ x ≤ 3, 0 ≤ y ≤ 2, x + y ≤ 3 P(X = x, Y = y) = ⎛9⎞ ⎜⎝ 3 ⎟⎠

3.7.6

⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ ⎜⎝ x ⎠⎟ ⎝⎜ y ⎠⎟ ⎝⎜ 4 − x − y ⎠⎟ P(X = x, Y = y) = , 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, x + y ≤ 4 ⎛ 52 ⎞ ⎜⎝ 4 ⎟⎠

3.7.7

P(X > Y) = pX,Y(1, 0) + pX,Y(2, 0) + pX,Y(2, 1) = 6/50 + 4/50 + 3/50 = 13/50

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50

Chapter 3: Random Variables

3.7.8 Outcome HHH HHT HTH HTT THH THT TTH TTT

X 1 0 1 0 1 0 1 0

(x, y) (0, 0) (0, 1) (0, 2) (0, 3) (1, 0) (1, 1) (1, 2) (1, 3)

pX,Y(x, y) 1/8 2/8 1/8 0 0 1/8 2/8 1/8

Y 3 2 2 1 2 1 1 0

3.7.9 0 1 2

Number of 3’s, Y

Number of 2’s, X 1 2 8/36 1/36 2/36 0 0 0

0 16/36 8/36 1/36

From the matrix above, we calculate pZ(0) = pX,Y(0, 0) = 16/36 pZ(1) = pX,Y(0, 1) + pX,Y(1, 0) = 2(8/36) = 16/36 pZ(2) = pX,Y(0, 2) + pX,Y(2, 0) + pX,Y(1, 1) = 4/36 3.7.10 (a) 1 =

1

1

0

0

∫∫

3.7.11 P(Y< 3X) = = 2 = 2

∫

∞

∫

∞

0

0

(

c dxdy = c, so c = 1

∞

∫ ∫ 0

3x x

2e− ( x + y ) dydx =

3x

)

e − x ⎡⎣ −e − y ⎤⎦ dx = 2 x

∫

∞

0

(b) P(0 < X < 1/2, 0 < Y < 1/4) =

∫

∞

0

e− x ⎛ ⎝

∫

3x x

1/ 4

1/ 2

0

0

∫ ∫

1 dxdy = 1/8

2e − y dy ⎞dx ⎠

e − x ⎡⎣e − x − e −3 x ⎤⎦ dx ∞

1 ⎡ 1 −2 x 1 − 4 x ⎤ −4 x ⎡ −2 x ⎤ ⎣ e − e ⎦ dx = 2 ⎢⎣ − 2 e + 4 e ⎥⎦ = 2 0

3.7.12 The density is the bivariate uniform over a circle of radius 2. The area of the circle is π(2)2 = 4π. Thus, fX,Y(x, y) = 1/4π for x2 + y2 ≤ 4.

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Section 3.7: Joint Densities

3.7.13 P(X < 2Y) =

=

=

1

1

0

x/2

∫∫

∫

1 0

51

1

1

0

x/2

∫∫

x dydx +

( x + y )dydx

1

1

0

x/2

∫∫

⎡ x2 ⎤ x − ⎢ ⎥ dx + 2⎦ ⎣ 1

∫

1 0

y dydx

⎡ 1 x2 ⎤ ⎢ − ⎥ dx ⎣2 8 ⎦ 1

⎡ x 2 x3 ⎤ ⎡ x x3 ⎤ 19 = ⎢ − ⎥ +⎢ − ⎥ = 6 ⎦ 0 ⎣ 2 24 ⎦ 0 24 ⎣2

3.7.14 The probability of an observation falling into the interval (0, 1/3) is probability of an observation falling into the interval (1/3, 2/3) is

∫

∫

1/ 3 0

2/3 1/ 3

2t dt = 1/9. The

2t dt = 1/3. Assume

without any loss of generality that the five observations are done in order. To calculate pX,Y(1, 2), ⎛5⎞ ⎛4⎞ note that there are ⎜ ⎟ places where the observation in (0, 1/3) could occur, and ⎜ ⎟ choices for ⎝1 ⎠ ⎝2⎠ ⎛5⎞ ⎛ 4⎞ the location of the observations in (1/3, 2/3). Then pX,Y(1, 2) = ⎜ ⎟ ⎜ ⎟ (1/ 9)1 (1/ 3) 2 (5 / 9) 2 ⎝1 ⎠ ⎝ 2 ⎠ = 750/6561.

3.7.15 The set where y > h/2 is a triangle with height h/2 and base b/2. Its area is bh/8. Thus the area of the set where y < h/2 is bh/2 − bh/8 = 3bh/8. The probability that a randomly chosen point will fall in the lower half of the triangle is (3bh/8)/(bh/2) = 3/4. ⎛3 ⎞ ⎛3 ⎞ ⎛ 6 ⎞ ⎜⎝ x ⎟⎠ min(2,3− x ) ⎛ 2 ⎞ ⎛ 4 ⎞ ⎜⎝ x ⎟⎠ ⎜⎝ 3 − x ⎟⎠ = 3.7.16 pX(x) = , x = 0,1, 2,3 ⎛ 9 ⎞ y = 0 ⎜⎝ y ⎟⎠ ⎜⎝ 3 − x − y ⎟⎠ ⎛9⎞ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠

∑

3.7.17 From the solution to Question 3.7.8, pX(x) = 1/8 + 2/8 + 1/8 = 1/2, x = 0, 1, pY(0) = 1/8, pY(1) = 3/8, pY(2) = 3/8, pY(3) = 1/8. 3.7.18 Let X1 be the number in the upper quarter; X2, the number in the middle half. From Question 6! 3.2.18, we know that P(X1 = 2, X2 = 2) = (0.25) 2 (0.50)2 (0.25) 2 = 0.088. The simplest way 2!2!2! to deal with the marginal probability is to recognize that the probability of belonging to the ⎛6⎞ middle half is binomial with n = 6 and p = .05. This probability is ⎜ ⎟ (0.5) 2 (0.5) 4 = 0.234. ⎝2⎠

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52

Chapter 3: Random Variables

3.7.19 (a) fX(x) = fY(y) =

∫

∫

fY(y) =

∫

(c) fX(x) =

0

∫

(b) fX(x) =

∫

1

1 y 1 dy = = ,0≤x≤2 2 20 2

1

2

1 x dx = = 1, 0 ≤ y ≤ 1 2 20

2 0

3 2 1 1 y dy = y 3 = 1 /2, 0 ≤ x ≤ 2 2 2 0 2 3 2 3 y dx = y 2 x = 3 y 2 , 0 ≤ y ≤ 1 2 2 0

1 0 2 0

12 0

3

(

2 xy + y 2 3

( x + 2 y )dy =

)

1 0

⎞ 2 ⎛ x2 ( x + 2 y )dx = ⎜ + 2 xy ⎟ fY(y) = 0 3 3⎝ 2 ⎠

∫

2 ( x + 1) , 0 ≤ x ≤ 1 3

=

12

1

= 0

4 1 y+ ,0≤y≤1 3 3

1

⎛ y2 ⎞ 1⎞ ⎛ (d) fX(x) = c ( x + y ) dy = c ⎜ xy + ⎟ = c ⎜ x + ⎟ , 0 ≤ x ≤ 1 0 ⎝ 2 ⎠ 2⎠ ⎝ 0

∫

1

1

⎛ x2 x ⎞ 1⎞ ⎛ In order for the above to be a density, 1 = c ⎜ x + ⎟ dx = c ⎜ + ⎟ = c, so 0 ⎝ 2⎠ ⎝ 2 2⎠ 0

∫

1

1 ,0≤x≤1 2 1 fY(y) = y + , 0 ≤ y ≤ 1, by symmetry of the joint pdf 2

fX(x) = x +

(e) fX(x) =

∫

1 0

4 xy dy = 2 xy 2

1 0

= 2x, 0 ≤ x ≤ 1

fY(y) = 2y, 0 ≤ y ≤ 1, by the symmetry of the joint pdf (f)

fX(x) =

∫

∞

0

xye − ( x + y ) dy = xe − x

= xe−x (− ye

−y

− e− y )

∞ 0

∫

∞

0

ye− y dy

= xe−x, 0 ≤ x

fY(y) = ye−y, 0 ≤ y, by symmetry of the joint pdf (g) fX(x) =

∫

∞ 0

ye − xy − y dy =

∫

∞ 0

ye− ( x +1) y dy

Integrating by parts gives 2 ⎛ y − ( x +1) y ⎛ 1 ⎞ − ( x +1) y ⎞ e − − ⎜ ⎟ ⎜⎝ ⎟ e x +1⎠ ⎝ x +1 ⎠

fY(y) =

∫

∞

0

ye − xy − y dx =

∫

∞

0

∞

0

2

⎛ 1 ⎞ =⎜ ,0 1 − u FX,Y(u, v) =

0

⎤ ⎥ dx = ⎥ 0⎦ v

⎢ y ⎣⎢ 2

2 ( x + 2 y ) dy dx = 3

3.7.28 (a) For 0 ≤ u ≤ v ≤ 2, FX ,Y (u , v) = (b) FX,Y(u, v) =

u ⎡1

∫

∫

v 0

∫

u

∫ ∫

v

1− v 1− x

0

u 0

2 1 2 (vx + v 2 dx) = u 2 v + uv 2 3 3 3

xdx = u 2 v 2

v

y 1 dx = 2x 2

u

0

1 (v − x)dx = (2uv − u 2 ) 0 4

∫

u

ln u − ln ydy = v ln u − v ln v + v

⎡3 x 2 u ⎤ dy = ⎢ 0⎥ 0 ⎣ ⎦ v

∫

u

∫

∫

v 0

3u 2 dy = 3u 2 v

6 xdy dx

= 3u2v − [3u2v − 3u2 + 2u3 − 3(1 − v)2v + 3(1 − v)2 − 2(1 − v)3] = 3u2 − 2u3 + 3(1 − v)2v − 3(1 − v)2 + 2(1 − v)3 = 3u2 − 2u3 − (1 − v)3 3.7.29 By Theorem 3.7.3, fX,Y =

∂2 ∂2 ∂ ⎛ ∂ ⎞ FX ,Y = ( xy ) = xy ∂x∂y ∂x∂y ∂x ⎜⎝ ∂y ⎠⎟

∂ ( x) = 1, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. ∂x The graph of fX,Y is a square plate of height one over the unit square.

=

∂2 ∂2 FX ,Y = [(1 − e − λ y )(1 − e − λ x )] ∂x∂y ∂x∂y ∂ ∂ ∂ = [(1 − e− λ y )(1 − e− λ x )] = [( λ e − λ y )(1 − e − λ x )] ∂x ∂y ∂x

3.7.30 By Theorem, 3.7.3, fX,Y =

= λ e − λ y λ e − λ x , x ≥ 0, y ≥ 0

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Section 3.7: Joint Densities

55

3.7.31 First note that 1 = FX,Y(1, 1) = k[4(12)(12) + 5(1)(14)] = 9k, so k = 1/9. ∂2 ∂2 ⎛ 4 2 2 5 4 ⎞ Then fX,Y = FX ,Y = ⎜ x y + xy ⎟⎠ ∂x∂y ∂x∂y ⎝ 9 9 =

∂ ∂ ⎛4 2 2 5 4⎞ ∂ ⎛8 2 20 3 ⎞ 16 20 3 xy ⎟ = xy + y ⎜⎝ x y + xy ⎟⎠ = ⎜⎝ x y + ⎠ 9 ∂x ∂y 9 ∂x 9 9 9 9

P(0 < X < 1/2, 1/2 < Y < 1) =

=

∫

1/ 2 8

9

0

xy 2 +

5 41 y dx = 9 1/ 2

∫

1/ 2

∫ ∫ 0

20 3 ⎞ ⎛ 16 xy + y ⎟ dydx ⎜ 1/ 2 ⎝ 9 ⎠ 9 1

1/ 2 ⎛ 2

25 ⎞ 1 2 25 x ⎜⎝ x + ⎟⎠ dx = x + 3 48 3 48

1/ 2

0

0

= 11/32

3.7.32 P(a < X ≤ b, Y ≤ d) = FX,Y(b, d) − FX,Y(a, d) P(a < X ≤ b, Y ≤ c) = FX,Y(b, c) − FX,Y(a, c) P(a < X ≤ b, c < Y ≤ d) = P(a < X ≤ b, Y ≤ d) − P(a < X ≤ b, Y ≤ c) = (FX,Y(b, d) − FX,Y(a, d)) − (FX,Y(b, c) − FX,Y(a, c)) = FX,Y(b, d) − FX,Y(a, d) − FX,Y(b, c) + FX,Y(a, c) 3.7.33 P(X1 ≥ 1050, X2 ≥ 1050, X3 ≥ 1050, X4 ≥ 1050) =

∞

∞

∞

∞

4

1

∫ ∫ ∫ ∫ ∏ 1000e

⎛ = ⎜ ⎝

1050 1050 1050 1050

∫

− xi /1000

dx1dx2 dx3 dx4

i =1

4

1 − x /1000 ⎞ e dx ⎟ = (e −1.05 )4 = 0.015 1050 1000 ⎠ ∞

40 ⎛ 4⎞ ⎛ 4 ⎞ ⎛ 4⎞ ⎛ ⎞ ⎜⎝ x ⎟⎠ ⎜⎝ y ⎟⎠ ⎜⎝ z ⎟⎠ ⎜⎝ 6 − x − y − z ⎟⎠ where 0 ≤ x, y, z ≤ 4, x + y + z ≤ 6 3.7.34 (a) pX,Y,Z(x, y, z) = ⎛ 52 ⎞ ⎜⎝ 6 ⎟⎠ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ ⎜⎝ x ⎠⎟ ⎝⎜ y ⎠⎟ ⎝⎜ 6 − x − y ⎠⎟ (b) pX,Y(x, y) = where 0 ≤ x, y ≤ 4, x + y ≤ 6 ⎛ 52 ⎞ ⎜⎝ 6 ⎟⎠ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ ⎜⎝ x ⎟⎠ ⎜⎝ z ⎟⎠ ⎜⎝ 6 − x − z ⎟⎠ where 0 ≤ x, z ≤ 4, x + z ≤ 6 pX,Z(x, z) = ⎛ 52 ⎞ ⎜⎝ 6 ⎟⎠ 2

∑

3.7.35 pX,Y(0, 1) =

p X ,Y , Z (0,1, z ) =

z =0

0

=

1

2

0

1 2

z

z =0 2− z

3! ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ 0!1! ⎝ 2 ⎠ ⎝ 12 ⎠

3! ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 2! ⎛1⎞ ⎛1⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜ ⎟ ⎜ ⎟ 0!1! 2 12 2 z = 0 z !(2 − z )! ⎝ 6 ⎠ ⎝ 4 ⎠

∑

∑ =

z

1 ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ z !(2 − z )! ⎝ 6 ⎠ ⎝ 4 ⎠ 0

2− z

1

2

3! ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 1 ⎞ 25 ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ + ⎟⎠ = 0!1! 2 12 2 6 4 576

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56

Chapter 3: Random Variables

∫

3.7.36 (a) fX,Y(x, y) =

∞ 0

f X ,Y , Z ( x, y, z )dz =

∫

∞ 0

( x + y )e− z dz

∞

= ( x + y ) ⎡⎣ − e− z ⎤⎦ = (x + y), 0 ≤ x, y ≤ 1 0

∫

(b) fY,Z(y, z) =

1 0

f X ,Y , Z ( x, y, z ) dx =

∫

1

⎡ x2 ⎤ ⎛1 ( x + y )e− z dx = e − z ⎢ + xy ⎥ = ⎜ + 0 ⎣2 ⎦0 ⎝ 2 1

⎞ y ⎟ e− z , ⎠

0 ≤ y ≤ 1, z ≥ 0

∫

(c) fZ(z) =

1 0

∫

1

1

0

0

∫∫

3.7.37 fW,X(w, x) =

1

⎡ y y2 ⎤ ⎞ fY , Z ( y , z )dy = ⎜ + y ⎟ e − z dy = e − z ⎢ + ⎥ = e−z, z ≥ 0 0⎝2 ⎠ ⎣ 2 2 ⎦0 1⎛ 1

fW , X ,Y , Z ( w, x, y, z )dydz =

1

1

0

0

∫∫

16 wxyz dydz =

∫

1

1

⎡8wxy 2 z ⎤ dz = ⎦0 0⎣

1

∫ [8wxz]dz 0

1

= ⎡⎣ 4wxz 2 ⎤⎦ = 4wx, 0 < w, x < 1 0 P(0 < W < 1/2, 1/2 < X < 1) =

∫

1/ 2 0

2 w ⎡⎣ x 2 ⎤⎦

1 1/ 2

dx =

∫

1/ 2 0

1/ 2

1

0

1/ 2

∫ ∫

4wx dxdw =

3 3 1/ 2 3 w dw = w2 = 0 2 4 16

3.7.38 We must show that pX,Y(j, k) = pX(j)pY(k). But for any pair (j, k), pX,Y(j, k) = 1/36 = (1/6)(1/6) = pX(j)pY(k). 3.7.39 The marginal pdfs for fX,Y are fX(x) = λe−λx and fY(y) = λe−λy (Hint: see the solution to 3.7.19(f)). Their product is fX,Y, so X and Y are independent. The probability that one component fails to last 1000 hours is 1 − e−1000λ. Because of independence of the two components, the probability that two components both fail is the square of that, or (1 − e−1000λ)2. ⎧1 2 1 ⎪⎪ 4 ⋅ 5 = 10 3.7.40 (a) pX,Y(x, y) = ⎨ ⎪1 ⋅ 1 = 1 ⎪⎩ 4 5 20

y=x y≠x

(b) pX(x) = 1/4, since each ball in Urn I is equally likely to be drawn pY(y) = 1/10 + 3(1/20) = 1/4 (c) P(X = 1, Y = 1) = 1/10, but P(X = 1)P(Y = 1) = 1/16 3.7.41 First, note k = 2. Then, 2 times area of A = P(Y ≥ 3/4). Also, 2 times area of B = P(X ≥ 3/4). The square C is the set (X ≥ 3/4) ∩ (Y ≥ 3/4). However, C is in the region where the density is 0. Thus, P((X ≥ 3/4) ∩ (Y ≥ 3/4)) is zero, but the product P(X ≥ 3/4)P(Y ≥ 3/4) is not zero.

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Section 3.7: Joint Densities

2 ( x + 1) 3 3 12 2⎛ 1⎞ fY(y) = ( x + 2 y )dx = ⎜ 2 y + ⎟ 0 3 3⎝ 2⎠

3.7.42 fX(x) =

∫

12

57

0

( x + 2 y )dy =

∫

2 2⎛ 1⎞ 2 ( x + 1) ⎜ 2 y + ⎟ ≠ ( x + 2 y ) . 3 3⎝ 2⎠ 3

But

3.7.43 P(Y < X) =

1

x

0

0

∫∫

f X ,Y ( x, y ) dydx =

1

x

0

0

∫∫

∫

(2 x)(3 y 2 ) dydx =

1 0

2 x 4 dx =

2 5

y t x2 dt = . FY(y) = 2t dt = y2 0 2 0 4 x2 y2 FX,Y(x, y) = FX(x)FY(y) = , 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 4

3.7.44 FX(x) =

∫

x

∫

⎛Y ⎞ 3.7.45 P ⎜ > 2 ⎟ = P (Y > 2 X ) = ⎝X ⎠

3.7.46 P(a < X < b, c < Y < d) = =

∫

b a

xe − x dx

∫

d c

1

∫∫ 0

y/2

0

b

d

a

c

∫ ∫

(2 x)(1)dxdy =

xye − ( x + y ) dydx =

∫

∫

1

y/2 y3 1 ⎡ x 2 ⎤ dy = = ⎦0 0⎣ 12 0 12

b a

1

xe − x ⎛ ⎝

∫

d c

ye − y dy ⎞ dx ⎠

ye − y dy = P(a < X < b) P(c < Y < d)

3.7.47 Take a = c = 0, b = d = 1/2. Then P(0 < X < 1/2, 0 < Y < 1/2) =

1/ 2

1/ 2

0

0

∫ ∫

(2 x + y − 2 xy )dydx

= 5/32. fX(x) = fY(y) =

∫ ∫

1 0 1 0

(2 x + y − 2 xy )dy = x + 1/2, so P(0 < X < 1/2) =

∫

1/ 2 ⎛ 0

1⎞ 3 ⎜⎝ x + ⎟⎠ dx = 2 8

(2 x + y − 2 xy )dx = 1, so P(0 < X < 1/2) = 1/2. But, 5/32 ≠ (3/8)(1/2)

3.7.48 We proceed by showing that the events g(X) ε A and h(Y) εB are independent, for sets of real numbers, A and B. Note that P(g(X) εA and h(Y) εB) = P(X ε g−1(a) and Y ε g−1(b)). Since X and Y are independent, P(X ε g−1(a) and Y ε g−1(b)) = P(X ε g−1(a))P(Y ε g−1(b)) = P(g(X) ε A)P(h(Y) ε B)

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58

Chapter 3: Random Variables

3.7.49 Let K be the region of the plane where fX,Y ≠ 0. If K is not a rectangle with sides parallel to the coordinate axes, there exists a rectangle A = {(x, y)| a ≤ x ≤ b, c ≤ y ≤ d} with A ∩ K = ∅, but for A1 = {(x, y)| a ≤ x ≤ b, all y} and A2 = {(x, y)| all x, c ≤ y ≤ d}, A1 ∩ K ≠ ∅ and A2 ∩ K ≠ ∅. Then P(a) = 0, but P(A1) ≠ 0 and P(A2) ≠ 0. However, A = A1 ∩ A2, so P(A1 ∩ A2) ≠ P(A1)P(A2). n

3.7.50

f X1 , X 2 ,..., X n (x1, x2, … xn) =

3.7.51 (a) P(X1 < 1/2) =

∫

1/ 2 0

∏ (1/ λ )e

− xj / λ

= (1/ λ ) n e

−

1

λ

n

∑xj j =1

j =1

4x3 dx = x 4

1/ 2 0

= 1/16

(b) This asks for the probability of exactly one success in a binomial experiment with n = 4 and ⎛4⎞ p = 1/16, so the probability is ⎜ ⎟ (1/16)1 (15 /16)3 = 0.206. ⎝1 ⎠ (c) (d)

f X1 , X 2 , X 3 , X 4 ( x1 , x2 , x3 , x4 ) = FX 2 , X 3 ( x2 , x3 ) =

x3

∫ ∫ 0

x2 0

4

∏ 4x

3 j

= 256( x1 x2 x3 x4 )3 , 0 ≤ x1 , x2 , x3 , x4 ≤ 1

j =1

(4s 3 )(4t 3 ) dsdt =

∫

x2 0

4 s 2 ds

∫

x3 0

4t 3 dt = x24 x34 , 0 ≤ x2, x3 ≤ 1.

3.7.52 P(X1 < 1/2, X2 > 1/2, X3 < 1/2, X4 > 1/2, …, X2k > 1/2) = P(X1 < 1/2)P(X2 > 1/2)P(X3 < 1/2)P(X4 > 1/2), …, P(X2k > 1/2) because the Xi are independent. Since the Xi are uniform over the unit interval, P(Xi < 1/2) = P(Xi > 1/2) = 1/2. Thus the desired probability is (1/2)2k.

Section 3.8: Transforming and Combining Random Variables 3.8.1

(a) pX+Y(w) =

∑p

X

( x) pY ( w − x). Since pX(x) = 0 for negative x, we can take the lower limit of

all x

the sum to be 0. Since pY(w − x) = 0 for w − x < 0, or x > w, we can take the upper limit of the sum to be w. Then we obtain w w k w− k 1 −λ λ −µ µ −(λ + µ ) λ k µ w− k =e p X +Y ( w) = e e k! ( w − k )! k =0 k = 0 k !( w − k )!

∑

= e−( λ + µ )

∑

1 w 1 w! λ k µ w− k = e − ( λ + µ ) ( λ + µ ) w , w = 0, 1, 2, … w! k = 0 k !( w − k )! w!

∑

This pdf has the same form as the ones for X and Y, but with parameter λ + µ . (b) pX+Y(w) =

∑p

X

( x) pY ( w − x) . The lower limit of the sum is 1.

all x

For this pdf, we must have w − k ≥ 1 so the upper limit of the sum is w− 1. Then p X +Y ( w) =

w −1

∑ k =1

(1 − p ) k −1 p (1 − p ) w− k −1 p = (1 − p ) w− 2 p 2

w−1

∑1 = ( w − 1)(1 − p)

w− 2

p 2 , w = 2, 3, 4,…

k =1

The pdf for X + Y does not have the same form as those for X and Y, but Section 4.5 will show that they all belong to the same family—the negative binomial.

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Section 3.8: Transforming and Combining Random Variables

∫

∞

f X ( x) fY ( w − x)dx =

∫

w

59

( xe− x )(e − ( w− x ) )dx = e − w

∫

w

x dx =

w2 − w e , w≥ 0 2

3.8.2

fX+Y(w) =

3.8.3

First suppose that 0 ≤ w ≤ 1. As in the previous problem the upper limit of the integral is w, and fX+Y(w) =

∫

−∞

w 0

0

0

(1)(1)dx = w. Now consider the case 1 ≤ w ≤ 2. Here, the first integrand vanishes

unless x is ≤ 1. Also, the second pdf is 0 unless w − x ≤ 1 or x ≥ w − 1. Then fX+Y(w) =

∫

1 w −1

(1)(1)dx = 2 − w.

0 ≤ w ≤1 ⎧w In summary, fX+Y(w) = ⎨ ⎩2 − w 1 ≤ w ≤ 2

3.8.4

Consider the continuous case. It suffices to show that FV,X+Y = FVFX+Y. FV,X+Y(v,w) = P(V ≤ v, X + Y ≤ w) = v ∞ w− x v ∞ w− x fV (v) f X ( x) fY ( y )dydxdv = fV (v) ⎛ f X ( x) fY ( y )dydx ⎞dv ⎝ ⎠ −∞ −∞ −∞ −∞ −∞ −∞ = FV(v)FX+Y(w)

∫ ∫ ∫

3.8.5

∫

(

∫ ∫

)

FW(w) = P(W ≤ w) = P(Y2 ≤ w) = P Y ≤ w = f = FY Now differentiate both sides to obtain fW ( w) = 1

( w)

d d FW ( w) = FY dw dw

( w ) = 2 1w f ( w ) . Y

( w ) . Since f ( w ) = 1 ,

, 0 ≤ w ≤ 1.

From Question 3.8.5, fW ( w) =

3.8.7

From Question 3.8.5, 1 1 fW ( w) = fY w . Thus fW ( w) = 6 w 1 − w = 3 1 − w where 0 ≤ w ≤ 1. 2 w 2 w

2 w

fY

( )

3.8.8

Y

(

) (

2 w

)

From Question 3.8.5 2 − b ( u )2 1 1 1 fY 2 (u ) = fY u = a u e = a ue − bu . 2 2 u 2 u 2 2 21 2 − b (2 w / m ) 2 Then fW ( w) = fu ( w) = a we = 3/ 2 a we − b (2 w / m ) , 0 ≤ w. m m m2 m m

( )

3.8.9

fW ( w) =

1

3.8.6

(a) Let W = XY. Then fW(w) =

( )

∫

∞ −∞

1 f X ( x) fY ( w / x) dx x

Since fY(w/x) ≠ 0 when 0 ≤ w/x ≤ 1, then we need only consider w ≤ x. 11 1 Similarly, fX(x) ≠ 0 implies x ≤ 1. Thus the integral becomes dx = ln x w = − ln w, wx 0 ≤ w ≤ 1.

∫

(b) Again let W = XY. Since the range of integration here is the same as in Part (a), we can write ∞ 1 11 11 f X ( x) fY ( w / x)dx = dx = −4 w ln w, 0 ≤ w ≤ 1. fV(v) = 2( x)2( w / x) dx = 4 w −∞ x w x wx

∫

∫

∫

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60

Chapter 3: Random Variables

∫

3.8.10 (a) Let W = Y/X. Then fW(w) =

∞ −∞

x f X ( x) fY ( wx)dx

Since fX(x) = 0 for x < 0, the lower limit of the integral is 0. Since fY(wx) = 0 for wx > 1, we must have wx ≤ 1 or x ≤ 1/w. Case I: 0 ≤ w ≤ 1: In this case 1/w > 1, so the upper limit of the integral is 1.

∫

FW(w) =

∞ −∞

x f X ( x) fY ( xw)dx =

∫

1 0

x(1)(1)dx =1/ 2

Case II: w > 1: In this case 1/w ≤ 1, so the upper limit of the integral is 1/w. ∞ 1/ w 1 x f X ( x) fY ( xw)dx = x(1)(1)dx = 2 fW(w) = −∞ 0 2w

∫

∫

(b) Case I: 0 ≤ w ≤ 1: The limits of the integral are 0 and 1.

∫

FW(w) =

∞ −∞

x f X ( x) fY ( xw)dx =

∫

1 0

x(2 x)(2wx)dx = w

Case II: w > 1: The limits of the integral are 0 and 1/w. ∞ 1/ w 1 fW(w) = x f X ( x) fY ( xw)dx = x(2 x)(2 xw)dx = 3 −∞ 0 w

∫

∫

3.8.11 Let W = Y/X. Then fW(w) = =

1 ⎛ 1+ w ⎝

∫

∞ 0

∫

∞ −∞

x f X ( x) fY ( wx)dx =

∫

∞ 0

x( xe − x )e − wx dx =

∫

∞ 0

x 2 e − (1+ w) x dx

x 2 (1 + w)e− (1+ w) x dx ⎞ ⎠

Let V be the exponential random variable with parameter 1 + w. Then the quantity in parentheses above is E(V2). 1 1 2 (See Question 3.6.11) But E (V 2 ) = Var(V ) + E 2 (V ) = + = 2 2 (1 + w) (1 + w) (1 + w)2 Thus, fW(w) =

1 ⎛ 2 ⎞ 2 = , 0 ≤ w. 1 + w ⎝⎜ (1 + w) 2 ⎠⎟ (1 + w)3

Section 3.9: Further Properties of the Mean and Variance 3.9.1

Let Xi be the number from the i-th draw, i = 1, …, r. Then for each i, r 1 + 2 + ... + n n + 1 = . The sum of the numbers drawn is E(Xi) = X i , so the expected value of n 2 i =1

∑

r

the sum is

∑ E( X ) = i

i =1

3.9.2

r (n + 1) . 2

(

)(

)

fX,Y(x, y) = λ 2 e − λ ( x + y ) = λ e − λ x λ e − λ y implies that fX(x) = λ e − λ x and fY(y) = λ e − λ y . Then E(X + Y) = E(X) + E(Y) =

1

λ

+

1

λ

=

2

λ

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 3.9: Further Properties of the Mean and Variance

3.9.3

From Question 3.7.19(c), fX(x) =

2 ( x + 1) , 0 ≤ x ≤ 1, so E(X) = 3

5 0 9 1 ⎛4 4 1 1⎞ Also, fY(y) = y + , 0 ≤ y ≤ 1, so E(Y) = y ⎜ y + ⎟dy = 0 ⎝3 3⎠ 3 3 5 11 7 Then E(X + Y) = E(X) + E(Y) = + = . 9 18 6 =

2 3

61

∫

1

∫

1⎛ 4 0

1 ⎞ 11 2 ⎜⎝ y + y ⎟⎠ dy = . 3 3 8

Let Xi = 1 if a shot with the first gun is a bull’s eye and 0 otherwise, i = 1, …, 10. E(Xi) = 0.30. Let Vi = 1 if a shot with the second gun is a bull’s-eye and 0 otherwise, i = 1, …, 10. E(Vi) = 0.40. 10 10 10 ⎛ 10 ⎞ Cathie’s score is 4 X i + 6 Vi , and her expected score is E ⎜ 4 X i + 6 Vi ⎟ ⎝ i =1 ⎠ i =1 i =1 i =1

∑

10

= 4

∑

E( X i ) + 6

i =1

3.9.5

2 x ( x + 1)dx 0 3 1

( x 2 + x)dx =

∫

3.9.4

∫

∑

∑

∑

10

∑ E (V ) = 4(10)(0.30) + 6(10)(0.40) = 36. i

i =1

⎛

n

⎞

n

n

n

i =1

i =1

i =1

µ = E ⎜ ∑ ai X i ⎟ = ∑ ai E ( X i ) = ∑ ai µ = µ ∑ ai , so the given equality occurs if and only if ⎝ ⎠ i =1

n

∑a

i

= 1.

i =1

3.9.6

Let Xi be the daily closing price of the stock on day i. The daily expected gain is E(Xi) = (1/8)p − (1/8)q = (1/8)(p − q). After n days the expected gain is (n/8)(p − q).

3.9.7

(a) E(Xi) is the probability that the i-th ball drawn is red, 1 ≤ i ≤ n. Draw the balls in order without replacement, but do not note the colors. Then look at the i-th ball first. The probability that it is red is surely independent of when it is drawn. Thus, all of these expected values are the same and each equals r/(r + w). n

n

(b) Let X be the number of red balls drawn. Then X =

∑X

i

and E(X) =

i =1

∑ E ( X ) = nr/(r + w). i

i =1

3.9.8

Let X1 = number showing on face 1; X2 = number showing on face 2. Since X1 and X2 are independent, E(X1X2) = E(X1)E(X2) = (3.5)(3.5) = 12.25.

3.9.9

First note that 1 =

20

∫ ∫ 10

20

10

k ( x + y )dydx = k ⋅ 3000 , so k =

1 1 1 XY = + , then R = . R X Y X +Y 20 20 xy 1 1 E(R) = ( x + y )dydx = 3000 10 10 x + y 3000

1 . 3000

If

∫ ∫

3.9.10 From Question 3.8.5, f X 2 ( w) =

1

20

∫ ∫ 10

, so E ( X 2 ) =

2 w result holding for Y2. Then E(X 2 + Y 2) = 2/3.

20 10

∫

xydydx = 7.5.

1 0

w

1 2 w

dw =

1 2

∫

1 0

wdw =

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

1 ,with a similar 3

62

Chapter 3: Random Variables

3.9.11 The area of the triangle is the random variable W =

1 XY . Then 2

1 1 1 1 1 ⎛1 ⎞ 1 E ⎜ XY ⎟ = E ( XY ) = E ( X ) E (Y ) = ⋅ ⋅ = ⎝2 ⎠ 2 2 2 2 2 8

3.9.12 The Yi are independent for i = 1, 2, …, n. Thus, E

(

n

) ( Y ) ⋅ E ( Y ) ⋅ ...E ( Y )

Y1 ⋅ Y2 ⋅ ... ⋅ Yn = E

n

1

n

n

2

n

The Yi all have the same uniform pdf, so it suffices to calculate E

( Y ) , which is n

1

n

n ⎛ n ⎞ . . Thus, the expected value of the geometric mean is ⎜ 0 ⎝ n + 1 ⎟⎠ n +1 Note that the arithmetic mean is constant at 1/2 and does not depend on the sample size.

∫

1

n

y ⋅ 1 dy =

3.9.13 x 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6

y 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6

fX,Y 1/36 1/36 1/36 1/36 1/36 1/36 2/36 1/36 1/36 1/36 1/36 3/36 1/36 1/36 1/36 4/36 1/36 1/36 5/36 1/36 6/36

xy 1 2 3 4 5 6 4 6 8 10 12 9 12 15 18 16 20 24 25 30 36

xyfX,Y 1/36 2/36 3/36 4/36 5/36 6/36 8/36 6/36 8/36 10/36 12/36 27/36 12/36 15/36 18/36 64/36 20/36 24/36 125/36 30/36 216/36

616 . Clearly E(X) = 7/2. 36 1 2 5 7 9 11 161 E(Y) = 1 + 2 + 3 + 4 + 5 + 6 = 36 36 36 36 36 36 36 616 7 161 105 Cov(X,Y) = E(XY) − E(X)E(Y) = − ⋅ = 36 2 36 72

E(XY) is the sum of the last column =

3.9.14 Cov(aX + b, cY + d) = E[(aX + b)(cY + d)] − E(aX + b)E(cY + d) = E(acXY + adX + bcY + bd) − [aE(X) + b][cE(Y) + d] = acE(XY) + adE(X) + bcE(Y) + bd − acE(X)E(Y) − adE(X) − bcE(Y) − bd = ac[E(XY) − E(X)E(Y)] = acCov(X, Y) Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 3.9: Further Properties of the Mean and Variance

3.9.15

∫

2π 0

cos xdx =

∫

2π

sin xdx =

0

∫

2π 0

63

(cos x)(sin x) dx = 0, so E(X) = E(Y) = E(XY) = 0.

Then Cov(X, Y) = 0. But X and Y are functionally dependent, Y = 1 − X 2 , so they are probabilistically dependent. y

⎡ x2 ⎤ 3.9.16 E(XY) = y x dxdy = y ⎢ ⎥ dy = 0 −y 0 ⎣ 2 ⎦− y 1

∫ ∫

E(X) =

1

y

0

−y

∫∫

y

∫

1

∫

1

y ⋅ 0 dy = 0

0

x dxdy = 0, so Cov(X, Y) = 0. However, X and Y are dependent since

P(−1/2 < x < 1/2, 0 < Y < 1/2) = P(0 < Y < 1/2) ≠ P(−1/2 < x < 1/2)P(0 < Y < 1/2) 3.9.17 The random variables are independent and have the same exponential pdf, so Var(X + Y) = 1 2 Var(X) + Var(Y). By Question 3.6.11, Var(X) = Var(Y) = 2 , so Var(X + Y) = 2 .

λ

λ

3.9.18 From Question 3.9.3, we have E(X + Y) = E(X) + E(Y) = 5/9 + 11/18 = 21/18 = 7/6. 1 1 2 2 1 1 2 E[(X + Y)2] = ( x + y ) 2 ( x + 2 y ) dxdy = ( x + 2 xy + y 2 )( x + 2 y )dxdy 0 0 3 3 0 0 2 1 1 3 ( x + 2 x 2 y + xy 2 + 2 x 2 y + 4 xy 2 + 2 y 3 )dxdy 3 0 0

∫∫

∫∫

∫∫

=

2 3

∫∫

=

2 3

∫

1

1

0

0

( x3 + 4 x 2 y + 5 xy 2 + 2 y 3 )dxdy =

2 3

∫

1

1⎛ 1

4 3 5 2 2 4 3⎞ ⎜ x + x y + x y + 2 xy ⎟⎠ dy 0⎝4 3 2 0

1⎛ 1

4 5 2 ⎛1 4 5 2⎞ 3 ⎞ + y + y 2 + 2 y 3 ⎟dy = ⎜ + + + ⎟ = ⎜ 0⎝4 ⎠ 3 2 3 ⎝4 6 6 4⎠ 2 2

3 ⎛7⎞ 5 −⎜ ⎟ = . ⎝ ⎠ 2 6 36

Then Var(X + Y) = E[(X + Y)2] − E(X + Y)2 =

(

3.9.19 First note that E ⎡ Y1Y2 ⎢⎣ for i = 1, 2. E

(

) ⎤⎥⎦ = E[Y Y ] = E(Y )E(Y ) = ⎛⎜⎝ 12 ⎞⎟⎠ ⎛⎜⎝ 12 ⎞⎟⎠ = 14 , since the Y are independent, 2

1 2

1

2

i

) ( Y ) E ( Y ) = 23 ⋅ 23 = 94 , since E(Y ) = ∫

Y1Y2 = E

Then Var

(

1

)

Y1Y2 =

i

2

1 0

y ⋅ 1 dy =

2 , i = 1, 2. 3

2

1 ⎛4⎞ 17 . −⎜ ⎟ = ⎝ ⎠ 4 9 324

3.9.20 E(W) = E(4X + 6Y) = 4E(X) + 6E(Y) = 4npX + 6mpY Var(W) = Var(4X + 6Y) = 16Var(X) + 36Var(Y) = 16npX(1 − pX) + 36mpY(1 − pY) 3.9.21 Let Ui be the number of calls during the i-th hour in the normal nine hour work day. Then U = U1 + U2 + … + U9 is the number of calls during this nine hour period. E(U) = 9(7) = 63. For a Poisson random variable, the variance is equal to the mean, so Var(U) = 9(7) = 63. Similarly, if V is the number of calls during the off hours, E(V) = Var(V) = 15(4) = 60. Let the total cost be the random variable W = 50U + 60 V. Then E(W) = E(50U + 60V) = 50E(U) + 60E(V) = 50(63) + 60(60) = 6750; Var(W) = Var(50U + 60V) = 502 Var(U) + 602Var(V) = 502(63) + 602(60) = 373,500.

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64

Chapter 3: Random Variables 50

3.9.22 L =

49

∑B + ∑M i

i =1

i

, where Bi is the length of the i-th brick, and Mi is the thickness of the i-th

i =1

mortar separation. Assume all of the Bi and Mi are independent. By Theorem 3.9.5, 2

2

⎛ 1 ⎞ ⎛1⎞ Var(L) = 50Var(B1) +49Var(M1) = 50 ⎜ ⎟ + 49 ⎜ ⎟ = 0.240. Thus, the standard deviation of ⎝ 32 ⎠ ⎝ 16 ⎠ L is 0.490.

3.9.23 Let Ri be the resistance of the i-th resistor, 1 ≤ i ≤ 6. Assume the Ri are independent and each has ⎛ 6 ⎞ standard deviation σ . Then the variance of the circuit resistance is Var ⎜ Ri ⎟ = 6σ 2 . The ⎝ ⎠

∑ i =1

circuit must have 6σ ≤ (0.4) or σ ≤ 0.163. 2

2

3.9.24 Let p be the probability the gambler wins a hand. Let Tk be his winnings on the k-th hand. Then

( )

E(Tk) = kp. Also, E Tk2 = k2p, so Var(Tk) = k2p − (kp)2 = k2(p − p2). n

The total winnings T =

∑

n

Tk , so E(T) =

k =1

n

Var(T) =

∑k

2

( p − p2 ) =

k =1

∑ kp = k =1

n(n + 1) p. 2

n(n + 1)(2n + 1) ( p − p2 ) 6

Section 3.10: Order Statistics 3.10.1 P (Y3′ < 5) = 12 = 4 10

∫

5 0

∫

5 0

fY3′ ( y )dy =

∫

5 0

4! y (3 − 1)!(4 − 3)!10

3−1

y⎞ ⎛ ⎜⎝1 − ⎟⎠ 10

4− 3

1 dy 10

5

12 ⎡ 10 1 ⎤ 12 ⎡ 10 1 ⎤ y (10 − y )dy = 4 ⎢ y 3 − y 4 ⎥ = 4 ⎢ 53 − 54 ⎥ = 5/16 4 ⎦ 4 ⎦ 0 10 ⎣ 3 10 ⎣ 3 2

3.10.2 First find FY: FY(y) =

∫

y

0

3t 2 dt = y 3 . Then P (Y5′ > 0.75) = 1 − P (Y5′ < 0.75) .

6! ( y 3 )5−1 (1 − y 3 )6−5 3 y 2 dy 0 (5 − 1)! (6 − 5)! 0.75 6! 0.75 6! = ( y 3 )4 (1 − y 3 )3 y 2 dy = ( y 3 ) 4 (1 − y 3 )3 y 2 dy 0 0 4! 4!

But P (Y5′ < 0.75) =

∫

∫

=

∫

0.75

∫

0.75 0

0.75

⎡ y15 y18 ⎤ 90( y )(1 − y )dy = 90 ⎢ − ⎥ = 0.052, ⎣ 15 18 ⎦ 0 14

3

so P (Y5′ > 0.75) = 1 − P (Y5′ < 0.75) = 1 − 0.052 = 0.948. 3.10.3 P (Y2′ > y60 ) = 1 − P (Y2′ < y60 ) = 1 − P(Y1 < y60, Y2 < y60) = 1 − P(Y1 < y60)P(Y2 < y60) = 1 − (0.60)(0.60) = 0.64 3.10.4 The complement of the event is P (Y1′ > 0.6) ∪ P (Y5′ < 0.6) .

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 3.10: Order Statistics

65

These are disjoint events, so their probability is P (Y1′ > 0.6) + P (Y5′ < 0.6) . But P (Y1′ > 0.6) = P(Y1, Y2, Y3, Y4, Y5 > 0.6) = [P(Y > 0.6)]5 = ⎛ ⎝

∫

Also, P (Y5′ < 0.6) = P(Y1, Y2, Y3, Y4, Y5 < 0.6) = [P(Y < 0.6)]5 = ⎛ ⎝ The desired probability is 1 − 0.107 − 0.006 = 0.887.

5

2 ydy ⎞ = (0.64)5 = 0.107 ⎠ 0.6 1

∫

0.6

.0

5

2 ydy ⎞ = (0.36)5 = 0.006 ⎠

n

⎛1⎞ 3.10.5 P (Y1′ > m) = P(Y1, …, Yn > m) = ⎜ ⎟ ⎝2⎠ P (Yn′ > m) = 1 − P (Yn′ < m) = 1 − P(Y1, …, Yn < m) ⎛1⎞ = 1 − P(Y1 < m) ⋅ …⋅ P(Yn < m) = 1 − ⎜ ⎟ ⎝2⎠ If n ≥ 2, the latter probability is greater.

3.10.6 P(Ymin < 0.2) =

∫

0.2 0

nfY ( y )[1 − FY ( y )]n −1 dy by Theorem 3.10.1.

Since FY(y) = 1 − e−y, =

∫

0.2 0

n

ne − ny dy = − e− ny

∫

0.2 0

0.2

nfY ( y )[1 − FY ( y )]n −1 dy =

∫

0.2 0

ne − y [1 − (1 − e − y )]n −1

= 1 − e−0.2n

0

But (1 − e−0.2n) > 0.9 if e−0.2n < 0.1, which is equivalent to n > −

1 ln 0.1 = 11.513. The smallest 0.2

n satisfying this inequality is 12. 3.10.7 P (0.6 < Y4′ < 0.7) = FY4′ (0.7) − FY4′ (0.6) =

3.10.2) =

∫

0.7 0.6

60 y 3 (1 − y )2 dy =

∫

0.7 0.6

∫

0.7 0.6

6! y 4 −1 (1 − y )6− 4 (1)dy (by Theorem (4 − 1)!(6 − 4)! 0.7

4 5 6 60( y 3 − 2 y 4 + y 5 )dy = (15y − 24y + 10 y ) 0.6

= 0.74431 − 0.54432 = 0.19999 3.10.8 First note that FY(y) =

∫

y 0

2t dt = y2. Then by Theorem 3.10.2,

fY1′ ( y ) = 5(2 y )(1 − y 2 )5−1 = 10 y (1 − y 2 ) 4 . By this same result, fY5′ ( y ) = 5(2y)(y2)5−1 = 10y9.

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66

Chapter 3: Random Variables

3.10.9 P(Ymin > 20) = P(Y1 > 20, Y2 > 20, …, Yn > 20) = P(Y1 > 20)P(Y2 > 20) … P(Yn > 20) = [P(Y > 20)]n. But 20 is the median of Y, so P(Y > 20) = 1/2. Thus, P(Ymin > 20) = (1/2)n. ⎛1⎞ 3.10.10 P(Ymin = Yn) = P(Yn < Y1, Yn < Y2, …, Yn < Yn−1) = P(Yn < Y1)P(Yn < Y2) … P(Yn < Yn−1) = ⎜ ⎟ ⎝2⎠

n

3.10.11 The graphed pdf is the function fY(y) = 2y, so FY(y) = y2 Then fY4′ ( y ) = 20y6(1 − y2)2y = 40y7(1 − y2) and FY4′ ( y ) = 5y8 − 4y10. P (Y4′ > 0.75) = 1 − FY4′ (0.75) = 1 − 0.275 = 0.725

The probability that none of the schools will have fewer than 10% of their students bused is P(Ymin > 0.1) = 1 − FYmin (0.1) = 1 −

∫

0.1 0

0.1

10 y (1 − y 2 ) 4 dy = 1 − ⎡⎣ − (1 − y 2 )5 ⎤⎦ = 0.951 (see Question 0

3.10.8). 3.10.12 Using the solution to Question 3.10.6, we can, in a similar manner, establish that fY1′ ( y ) = nλ e − nλ y . The mean of such an exponential random variable is the inverse of its

parameter, or 1/nλ. 3.10.13 If Y1, Y2, …Yn is a random sample from the uniform distribution over [0, 1], then by Theorem n! i −1 n −i 3.10.2, the quantity FY ( y )] [1 − FY ( y ) ] f y ( y ) [ (i − 1)!(n − i)! n! = y i −1 (1 − y )n − i is the pdf of the i-th order statistic. (i − 1)!( n − i)! 1 1 n! n! Thus, 1 = y i −1 (1 − y )n − i dy = y i −1 (1 − y ) n −i dy 0 (i − 1)!( n − i )! (i − 1)!(n − i )! 0 1 (i − 1)!(n − i)! or, equivalently, y i −1 (1 − y ) n −i dy = . 0 n!

∫

∫

∫

1 n! n! y i −1 (1 − y )n − i dy = y i (1 − y ) n −i dy 0 (i − 1)!(n − i )! (i − 1)!(n − i )! 0 1 n! n! [(i + 1) − 1]![(n + 1) − (i + 1)]! = y (i +1)−1 (1 − y )( n +1) − (i +1) dy = (i − 1)!(n − i )! 0 (i − 1)!(n − i )! ( n + 1)! where this last equality comes from the result in Question 3.10.13. n! i !(n − 1)! i Thus, E (Yi ′) = = (i − 1)!(n − i )! (n + i)! (n + 1)

3.10.14 E (Yi ′) =

∫

1

∫

y⋅

∫

3.10.15 This question translates to asking for the probability that a random sample of three independent uniform random variables on [0, 1] has range R ≤ 1/2. Example 3.10.6 establishes that FR(r) = 3r2 − 2r3. The desired probability is FR(1/2) = 3(1/2)2 − 2(1/2)3 = 0.5. 3.10.16 This question requires finding the probability that a random sample of three independent exponential random variables on [0, 10] has range R ≤ 2. From Equation 3.10.5, we find the joint pdf of Ymin and Ymax to be 3[FY(v) − FY(u)]fY(u)fY(v) = 3[(1 − e−v) − (1 − e−u)]e−ue−v = 3(e−2u−v − e−u−2v), u ≤ v.

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Section 3.11: Conditional Densities

67

P(R ≤ 2) is obtained by integrating the joint pdf over a strip such as pictured in Figure 3.10.4, but the strip is infinite in extent. Thus, P(R ≤ 2) =

∫

u +2 u

(e

−2 u − v

∞

u +2

0

u

∫ ∫ −e

3(e −2u − v − e − u − 2 v ) dvdu . The inner integral is

− u − 2v

)dv = e

−2 u

−v

(− e )

u +2

u +2

−e

−u

u

⎛ 1 −2 v ⎞ ⎜⎝ − e ⎟⎠ 2 u

1 1 ⎛1 ⎞ = e −2u (e − u − e − u − 2 ) − e − u (e −2u − e −2u − 4 ) = e −3u ⎜ − e −2 + e−4 ⎟ ⎝2 ⎠ 2 2 1 ⎛1 ⎞ Then P(R ≤ 2) = 3 ⎜ − e −2 + e −4 ⎟ ⎝2 ⎠ 2

∫

∞

0

e −3u du =

1 1 − e −2 + e −4 = 0.374. 2 2

Section 3.11: Conditional Densities x + 1 + x ⋅1 x + 2 + x ⋅ 2 3 + 5x + = , x = 1, 2 21 21 21 p X ,Y ( x, y ) x + y + xy = , y = 1, 2 pY x ( y ) = 3 + 5x p X ( x)

3.11.1 pX(x) =

3.11.2 The probability that X = x and Y = y is the probability of y 4’s on the first two rolls and x − y rolls on the last four rolls. These events are independent, so y 2− y x− y 4− x + y ⎛ 2⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ pX,Y(x, y) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ for y ≤ x ⎜ ⎟ ⎜ ⎟ ⎝ y ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ x − y ⎟⎠ ⎝ 6 ⎠ ⎝ 6 ⎠

⎛2⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ ⎜⎝ y ⎟⎠ ⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠ y

Then pY x ( y ) =

p X ,Y ( x, y ) p X ( x)

=

2− y

x− y 4− x + y ⎛ 4 ⎞⎛1⎞ ⎛5⎞ ⎜⎝ x − y ⎟⎠ ⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠

⎛6⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝⎜ x ⎠⎟ ⎝ 6 ⎠ ⎝ 6 ⎠ x

6− x

⎛ 2⎞ ⎛ 4 ⎞ ⎜⎝ y ⎟⎠ ⎜⎝ x − y ⎟⎠ = , ⎛6⎞ ⎝⎜ x ⎠⎟

0 ≤ y ≤ min (2, x), which we recognize as a hypergeometric distribution. 4 4 ⎞ ⎛8 ⎞ ⎛ 6 ⎞ ⎛ ⎞ ⎛ 8 ⎞ ⎛ 10 ⎞ ⎛ 6 ⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ p X ,Y ( x, y ) ⎝ x ⎠ ⎝ y ⎠ ⎝ 3 − x − y ⎠ ⎝ x ⎠ ⎝ 3 − x ⎠ ⎝ y ⎠ ⎝ 3 − x − y ⎟⎠ , with 0 ≤ y ≤ 3 − x = ÷ = 3.11.3 pY⏐x(y) = p X ( x) ⎛18 ⎞ ⎛18 ⎞ ⎛ 10 ⎞ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 − x ⎟⎠ P ( X = 2, Y = 2) P (Y = 2) ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ ⎛ 4 ⎞ ⎛ 48 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 44 ⎞ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 1 ⎟⎠ = ÷ = = 0.015 ⎛ 52 ⎞ ⎛ 52 ⎞ ⎛ 4 ⎞ ⎛ 48 ⎞ ⎜⎝ 5 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠

3.11.4 P(X = 2|Y = 2) =

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68

Chapter 3: Random Variables 3

3.11.5 (a) 1/k =

3

∑∑ ( x + y) = 36, so k = 1/36 x =1 y =1

1 (b) pX(x) = 36

pY | x (1) =

3

1

∑ ( x + y) = 36 (3x + 6) y =1

1 ( x + 1) x +1 = 36 = , x = 1, 2, 3 1 p X ( x) 3x + 6 (3x + 6) 36

p X ,Y ( x,1)

⎛x⎞⎛1 ⎞ ⎛1⎞ 3.11.6 (a) pX,Y(x, y) = pY x ( y ) p X ( x) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ , y ≤ x ⎝ y⎠ ⎝ 2 ⎠ ⎝ 3⎠ x

3 7 ⎛ 1 ⎞ ⎛ x⎞ ⎛ 1 ⎞ (b) pY(0) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ 3 ⎠ x =1 ⎝ 0 ⎠ ⎝ 2 ⎠ 24 x

∑

3 11 ⎛ 1 ⎞ ⎛ x⎞ ⎛ 1 ⎞ pY(1) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ 3 ⎠ x =1 ⎝ 1 ⎠ ⎝ 2 ⎠ 24 x

∑

3 5 ⎛ 1 ⎞ ⎛ x⎞ ⎛ 1 ⎞ pY(2) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ 3 ⎠ x=2 ⎝ 2 ⎠ ⎝ 2 ⎠ 24 x

∑

3

1 ⎛ 1 ⎞ ⎛ 3⎞ ⎛ 1 ⎞ pY(3) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 2 ⎠ 24 1 ⋅1 + 1 ⋅ z + 1⋅ z 1 ⋅ 2 + 1 ⋅ z + 2 ⋅ z 2 ⋅1 + 2 ⋅ z + 1 ⋅ z 2 ⋅ 2 + 2 ⋅ z + 2 ⋅ z + + + 54 54 54 54 9 + 12 z , z = 1, 2 = 54 xy + xz + yz Then p X ,Y z ( x, y ) = , x = 1, 2 y = 1, 2 z = 1, 2 9 + 12 z

3.11.7 pZ(z) =

3.11.8 pW,X(1, 1) = P({(1, 1, 1)}) = 3/54 pW,X(2, 2) = 33/54; P(W = 2, X = 2) = P(X = 2). But P(X = 2) = P(Z = 2) by symmetry, and from Question 3.11.7, this probability is 33/54. Then pW,X(2, 1) = 1 − 3/54 − 33/54 = 18/54 Finally, pW 1 (1) = (3/54)/(21/54) = 1/7; pW 1 (2) = (18/54)/(21/54) = 6/7; and pW 2 (2) = (33/54)/(33/54) = 1

3.11.9

p X x + y = n ( x) =

e− λ

λk

e− µ

P ( X = k , X + Y = n) P ( X = k , Y = n − k ) = P ( X + Y = n) P ( X + Y = n)

µ n−k

k

n−k

⎛ λ ⎞ ⎛ µ ⎞ n! (n − k )! = n k !(n − k )! ⎜⎝ λ + µ ⎠⎟ ⎝⎜ λ + µ ⎠⎟ (λ + µ ) e−( λ + µ ) n! but the right hand term is a binomial probability with parameters n and λ/(λ + µ).

=

k!

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Section 3.11: Conditional Densities

69

3.11.10 Let U be the number of errors made by Compositor A in the 100 pages. Then the pdf of U is Poisson with parameter λ = 200. Similarly, if V is the number of errors made by Compositor B, then the pdf for V is Poisson with µ = 300. From the previous question, P(U ≤ 259⎜U + V = 520) is binomial with parameter n = 520 and p = 200/(200 + 300) = 2/5. k 520 − k 259 ⎛ 520 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ The desired probability is ⎜ ⎟ ⎜⎝ ⎟⎠ ⎜⎝ 5 ⎟⎠ k =0 ⎝ k ⎠ 5

∑

P ( X > s + t and X > t ) P ( X > s + t ) = P( X > t ) P( X > t )

3.11.11 P(X > s + t| X > t) =

∫ (1/ λ ) ∫

(1/ λ )

=

∞

s +t ∞ t

3.11.12 (a) fX(x) =

e − x / λ dx e − x / λ dx

∫

∞

=

−(1/ λ ) e − x / λ −(1/ λ ) e

∞

s +t −x/λ ∞

=

(1/ λ )e − ( s +t ) / λ = e−s/λ = −t / λ (1/ λ )e

Also, P(X < 1, Y < 1) =

s

(1/ λ )e − x / λ dx = P(X > s)

t

2e − x e − y dy = 2e−2x, x > 0, so P(X < 1) =

x

∫

∞

1

x

0

0

∫∫

2e − ( x + y ) dydx =

∫

1 0

∫

1 0

2e−2 x dx = 1 − e−2 = 1 − 0.135 = 0.865 x

2e− x ⎡⎣ −e − y ⎤⎦ dx = 0

∫

1 0

(2e − x − 2e −2 x )dx

0.400 = 0.462 0.865 (b) P(Y < 1| X = 1) = 0, since the joint pdf is defined with y always larger than x.

= −2e − x + e −2 x

(c)

fY x ( y ) =

(d) E (Y | x) =

1

0

= 0.400. Then the conditional probability is

f X .Y ( x, y ) 2e − ( x + y ) = exe−y, x < y = −2 x f X ( x) 2e

∫

∞ x

ye x e− y dy = e x

y2 3.11.13 fX(x) = ( x + y )dy = xy + 0 2

∫

1

fY | x ( y ) =

3.11.14 fX(x) =

∫

f X ,Y ( x, y ) f X ( x)

1− x 0

=

2 dy = 2 y

f X ,Y ( x, y )

∫

∞

x

ye− y dy = e x [− e − y ( y + 1)]∞x = e x [e − x ( x + 1)] = x + 1

1

= x+ 0

1 ,0≤x≤1 2

x+ y ,0≤y≤1 1 x+ 2 1− x 0

= 2(1 − x), 0 ≤ x ≤ 1

2 1 ,0≤y≤1−x = f X ( x) 2(1 − x) 1 − x For each x, the conditional pdf does not depend on y, so it is a uniform pdf.

fY⏐x(y) =

=

1 ⎛ 2 y + 4x ⎞ 1 3.11.15 fX,Y(x, y) = fY⏐x(y)fX(x) = ⎜ (1 + 4 x) = (2 y + 4 x) ⎟ ⎝ 1 + 4x ⎠ 3 3 11 1 1 1 fY(y) = (2 y + 4 x)dx = (2 xy + 2 x 2 ) = (2 y + 2) , with 0 ≤ y ≤ 1 0 03 3 3

∫

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70

Chapter 3: Random Variables

2⎛ 3 ⎞ 3.11.16 (a) fX(x) = (2 x + 3 y )dy = ⎜ 2 xy + y 2 ⎟ 0 5 5⎝ 2 ⎠

∫

12

1

= 0

4 3 x + , with 0 ≤ x ≤ 1 5 5

2 (2 x + 3 y ) 4x + 6 y 5 ,0≤y≤1 (b) fY⏐x(y) = = 4 3 4x + 3 x+ 5 5 1 (c) f 1 ( y ) = (2 + 6 y ) Y 5 2

∫

P(1/4 ≤ Y ≤ 3/4) =

3.11.17 fY(y) =

∫

y 0

3/ 4 1 1/ 4

5

(2 + 6 y )dy = 10/20 = 1/2

2 dx = 2y

2 1 = ,0 1) P ( X > 1 / 2)

6 ⎛ 2 xy ⎞ 55 x + ⎟dydx = ⎜ 1/2 1 7 ⎝ 2⎠ 112 1 6 23 (2 x 2 + x)dx = We know fX from part (a) so the denominator is P(X > 1/2) = 1/2 7 28 55 23 55 = . The conditional probability requested is 112 28 92

First calculate the numerator: P(X > 1/2, Y > 1) =

1

∫ ∫

2

∫

Section 3.12: Moment-Generating Functions 3.12.1 Let X be a random variable with pX(k) = 1/n, for k = 0, 1, 2, …, n − 1, and 0 otherwise. n −1 n −1 1 1 n −1 t k 1 − ent MX(t) = E(etX) = etk p X (k ) = etk = (e ) = . n n k =0 n(1 − et ) k =0 k =0

∑

∑

(Recall that 1 + r + … + rn−1 =

∑

1− rn ). 1− r

3.12.2 fX(−3) = 6/10; fX(5) = 4/10. MX(t) = E(etX) = e−3t(6/10) + e5t(4/10) 3.12.3 For the given binomial random variable, 10

1 ⎛ 1 1 ⎞ E(etX) = MX(t) = ⎜1 − + et ⎟ . Set t = 3 to obtain E(e3X) = 10 (2 + e3 )10 ⎝ 3 3 ⎠ 3 k 1 ∞ ⎛ 3et ⎞ ⎛1⎞⎛3⎞ e ⎜ ⎟⎜ ⎟ = 3.12.4 MX(t) = ⎝4⎠⎝4⎠ 4 k = 0 ⎜⎝ 4 ⎟⎠ k =0 ∞

∑

=

1 4

tk

∑

k

1 1 = , 0 < et < 4/3 t 3e 4 − 3et 1− 4

3.12.5 (a) Normal with µ = 0 and σ2 = 12 (c) Binomial with n = 4 and p = ½

(b) Exponential with λ = 2 (d) Geometric with p = 0.3

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72

Chapter 3: Random Variables

3.12.6 MY(t) = E(etY) =

∫

1 0

ety y dy +

∫

2 1

ety (2 − y ) dy

1 ⎞ 1 2 2 ⎛1 1⎞ 2 ⎛1 = ⎜ y − 2 ⎟ ety + ety − ⎜ y − 2 ⎟ ety 1 ⎝t ⎝t t ⎠ 0 t t ⎠ 1 2 ⎛1 1 ⎞ ⎛ 1⎞ 2 ⎛2 1 ⎞ ⎛1 1 ⎞ = ⎜ − 2 ⎟ et − ⎜ − 2 ⎟ + e 2 t − e t − ⎜ − 2 ⎟ e 2 t + ⎜ − 2 ⎟ e t ⎝t t ⎠ ⎝ t ⎠ t ⎝t t ⎠ ⎝t t ⎠ t 1 1 2 1 = 2 + 2 e 2t − 2 et = 2 (et − 1)2 t t t t ∞

3.12.7 MX(t) = E(etX) =

∑ k =0

3.12.8 MY(t) = E(etY) =

∫

∞ 0

etk e − λ

λk k!

∞

t ( λ et ) k = e λ ( e −1) k! k =0

∑

= e− λ

ety ye − y dy =

∫

∞ 0

ye − y (1− t ) dy =

1 1− t

∫

∞ 0

y (1 − t )e − y (1− t ) dy

1 ⎛ 1 ⎞⎛ 1 ⎞ = ⎜ = , ⎟ ⎜ ⎟ ⎝ 1 − t ⎠ ⎝ 1 − t ⎠ (1 − t )2

since the integral is the mean of an exponential pdf with parameter (1 − t), which is

1 . 1− t

2 d t2 / 2 e = tet / 2 dt 2 2 2 d 2 M Y(2) (t ) = tet / 2 = t (tet / 2 ) + et / 2 = (t 2 + 1)et / 2 dt 2 2 2 d (3) M Y (t ) = (t 2 + 1)et / 2 = (t 2 + 1)tet / 2 + 2tet / 2 , and E(Y 3) = M Y(3) (0) = 0 dt

3.12.9 M Y(1) (t ) =

3.12.10 From Example 3.12.3, MY(t) =

λ (λ − t )

and M Y(1) (t ) =

Successive differentiation gives M Y(4) (t ) =

λ (λ − t )2

.

(4!) λ 24 (4!) λ = 4. . Then E(Y4) = M Y(4) (0) = 5 (λ − t ) λ5 λ

2 2 d at +b2t 2 / 2 e = (a + b 2 t )eat + b t / 2 , so M Y(1) (0) = a dt 2 2 2 2 (2) M Y (t ) = (a + b 2t )2 eat + b t / 2 + b 2 eat + b t / 2 , so M Y(2) (0) = a2 + b2. Then Var(Y) = (a2 + b2) − a2 = b2

3.12.11 M Y(1) (t ) =

3.12.12 Successive differentiation of MY(t) gives M Y(4) (t ) = α4k(k + 1)(k + 2)(k + 3)(1 − αt)−k−4. Thus, E(Y4) = M Y(4) (0) = α4k(k + 1)(k + 2)(k + 3) 3.12.13 The moment generating function of Y is that of a normal variable with mean µ = −1 and variance σ 2 = 8. Then E(Y2) = Var(Y) + µ2 = 8 + 1 = 9.

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Section 3.12: Moment-Generating Functions

73

d (1 − t / λ ) − r = (− r )(1 − t / λ )− r −1 (− λ ) = λ r (1 − t / λ ) − r −1 dt d M Y(2) (t ) = λ r (1 − t / λ ) − r −1 = ( − r − 1) λ r (1 − t / λ )− r − 2 (− λ ) = λ 2 r (r + 1)(1 − t / λ ) − r − 2 dt (r + k − 1)! Continuing in this manner yields M Y( k ) (t ) = λ r (1 − t / λ )− r − k . (r − 1)! ( r + k − 1)! Then E(Yk) = M Y( k ) (0) = λ r . (r − 1)!

3.12.14 M Y(1) (t ) =

3.12.15 MY(t) =

∫

b a

ety

b

1 1 dy = ety b−a (b − a )t

= a

1 (etb − e at ) for t ≠ 0 (b − a )t

1 ⎡ betb − ae at etb − e at ⎤ − M Y(1) (t ) = ⎢ ⎥ (b − a) ⎣ t t2 ⎦ ⎡ betb − ae at etb − e at ⎤ 1 − lim ⎢ E(Y) = lim M Y(1) (t ) = ⎥ . Applying L’Hospital’s rule gives t →0 (b − a ) t → 0 ⎣ t t2 ⎦

E(Y) =

1 ⎡ 2 b 2 − a 2 ⎤ ( a + b) 2 ( b − a ) − ⎢ ⎥= (b − a ) ⎣ 2 ⎦ 2

3.12.16 M Y(1) (t ) =

(1 − t 2 )2e2t − (−2t )e 2t (1 + t − t 2 )e2 t , so E(Y) = M Y(1) (0) = 2. = 2 2 2 2 2 (1 − t ) (1 − t )

M Y(2) (t ) =

2(1 − t 2 ) 2 [(1 − 2t )e 2t + 2(1 + t − t 2 )e 2t ] − 2(1 − t 2 )(−2t )2(1 + t − t 2 )e 2t (1 − t 2 ) 4

so M Y(2) (0) = 6. Thus Var(Y) = E(Y 2) − µ2 = 6 − 4 = 2. 3.12.17 Let Y =

1

λ

V, where fV(y) = ye−y, y ≥ 0. Question 3.12.8 establishes that MV(t) =

1 . By (1 − t ) 2

Theorem 3.12.3(a), MY(t) = MV(t/λ) = 1(1 − t / λ ) 2 . 3

⎛ ⎞ 1 1 3.12.18 M Y1 +Y2 +Y3 (t ) = M Y1 (t ) M Y2 (t ) M Y3 (t ) = ⎜ = 2 ⎟ (1 − t / λ )6 ⎝ (1 − t / λ ) ⎠

3.12.19 (a) Let X and Y be two Poisson variables with parameters λ and µ, respectively. Then MX(t) = e − λ + λ e and MY(t) = e − µ + µe . t

t

MX+Y(t) = MX(t)MY(t) = e − λ + λ e e − µ + µ e = e− ( λ + µ ) + ( λ + µ ) e . This last expression is that of a Poisson variable with parameter λ + µ, which is then the distribution of X + Y. t

t

t

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74

Chapter 3: Random Variables

(b) Let X and Y be two exponential variables, with parameters λ and µ, respectively. Then MX(t) =

λ

(λ − t )

µ

and MY(t) =

MX+Y(t) = MX(t)MY(t) =

λ

(µ − t)

.

µ

. (λ − t ) (µ − t) This last expression is not that of an exponential variable, and the distribution of X + Y is not exponential.

(c) Let X and Y be two normal variables, with parameters µ1, σ 12 and µ2, σ 22 respectively. Then MX(t) = e µ1t +σ 1 t

2 2

/2

and MY(t) = e µ2t +σ 2 t

2 2

µ1t +σ 12 t 2

µ2t +σ 22t 2

/2

.

( µ1 + µ2 ) t + (σ 12 +σ 22 ) t 2 / 2

MX+Y(t) = MX(t)MY(t) = e e =e . This last expression is that of a normal variable with parameters µ1 + µ2 and σ 12 and σ 22 , which is then the distribution of X + Y. /2

/2

3.12.20 From the moment-generating function of X, we know that it is binomial with n = 5 and p = 3/4. Then P(X ≤ 2) = (1/4)5 + 5(3/4)(1/4)4 + 10(3/4)2(1/4)3 = 0.104 n

3.12.21 Let S =

∑ i =1

n

Yi . Then Ms(t) =

∏M i =1

M Y (t ) = MS/n(t) = MS(t/n) = e µt + (σ

2

Yi

(

(t ) = e µt +σ

/ n )t 2 / 2

2 2

t /2

)

n

= e nµt + nσ

2 2

t /2

.

. Thus Y is normal with mean µ and variance σ 2 / n .

3.12.22 From the moment-generating function of W, we know that W = X + Y, where X is Poisson with parameter 3, and Y is binomial with parameters n = 4 and p = 1/3. Also, X and Y are independent. Then P(W ≤ 1) = pX(0)pY(0) + pX(0)pY(1) + pX(1)pY(0) = (e−3)(2/3)4 + (e−3)4(1/3)(2/3)3 + (3e−3)(2/3)4 = 0.059 3.12.23 (a) MW(t) = M3X(t) = MX(3t) = e − λ + λ e . This last term is not the moment-generating function of a Poisson random variable, so W is not Poisson. 3t

(b) MW(t) = M3X+1(t) = etMX(3t) = et e − λ + λe . This last term is not the moment-generating function of a Poisson random variable, so W is not Poisson. 3t

3.12.24 (a) MW(t) = M3Y(t) = MY(3t) = e µ (3t )+σ (3t ) / 2 = e(3 µ )t + 9σ t / 2 . This last term is the moment-generating function of a normal random variable with mean 3µ and variance 9σ 2 , which is then the distribution of W. 2

2

2 2

(b) MW(t) = M3Y+1(t) = etMY(3t) = et e µ (3t ) +σ (3t ) / 2 = e(3µ +1)t +9σ t / 2 . This last term is the moment-generating function of a normal random variable with mean 3µ + 1 and variance 9σ 2 , which is then the distribution of W. 2

2

2 2

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Chapter 4: Special Distributions Section 4.2: The Poisson Distribution 4.2.1

1 ; n = 6000. Let x = number of words misspelled. Using the 3250 0 6000 ⎛ 6000 ⎞ ⎛ 1 ⎞ ⎛ 3249 ⎞ exact binomial analysis, P(X = 0) = ⎜ = 0.158. For the Poisson ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎟⎠ ⎝ 3250 ⎠ ⎝ 3250 ⎠

p = P(word is misspelled) =

e−1.846 (1.846)0 ⎛ 1 ⎞ = 1.846, so P(X = 0) = = 0.158. The agreement approximation, λ = 6000 ⎜ ⎟ ⎝ 3250 ⎠ 0!

is not surprising because n is so large and p is so small (recall Example 4.2.1). 4.2.2

Let X = number of prescription errors. Then λ = np = 10 ⋅ 1 − P(X = 0) = 1 −

4.2.3

4.2.4

e −0.0313 (0.0313)0 = 0.031. 0!

1 1 , and λ = 500 ⋅ = 365 365 e−1.370 (1.370)0 e−1.370 (1.370)1 1.370, P(X ≤ 1) = P(X = 0) + P(X = 1) = = 0.602. + 0! 1! 1 (a) Let X = number of chromosome mutations. Given that n = 20,000 and p = 10,000 −2 3 (so λ = 2), P(X = 3) = e 2 /3! = 0.18. 1 (b) Listed in the table are values of P(X ≥ k) under the assumption that p = . If X is on 10,000 the order of 5 or 6, the credibility of that assumption becomes highly questionable.

Let X = number born on Poisson’s birthday. Since n = 500, p =

k 3 4 5 6 4.2.5

905 = 0.0313, and P(X ≥ 1) = 289, 411

P(X ≥ k) 0.3233 0.1429 0.0527 0.0166

Let X = number of items requiring a price check. If p = P(item requires price check) = 0.01 and n ⎛10 ⎞ = 10, a binomial analysis gives P(X ≥ 1) = 1 − P(X = 0) = 1 − ⎜ ⎟ (0.01)0 (0.99)10 = 0.10. Using ⎝0⎠ the Poisson approximation, λ = 10(0.01) = 0.1 and P(X ≥ 1) = 1 − P(X = 0) e −0.1 (0.1)0 = 1 − = 0.10. The exact model that applies here is the hypergeometric, rather than the 0! binomial, because p is a function of the previous items purchased. However, the variation in p will be essentially zero for the 10 items purchased, so the binomial and hypergeometric models in this case will be effectively the same.

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76

4.2.6

Chapter 4: Special Distributions

Let X = number of policy-holders who will die next year. Since n = 120, p =

1 , and 150

120 = 0.8, P(company will pay at least $150,000 in benefits) = P(X ≥ 3) = 150 2 e −0.8 (0.8) k 1 − P(X ≤ 2) = 1 − = 0.047. k! k =0

λ=

∑

4.2.7

1 , 200 1 1 e−0.6 (0.6) k (so λ = 120 ⋅ = 0.6), = P(X ≥ 2) = 1 − P(X ≤ 1) =1 − = 0.122 . k! 200 k =0

Let X = number of pieces of luggage lost. Given that n = 120, p =

∑

4.2.8

Let X = number of cancer cases. If n = 9500 and p =

1 9,500 , then λ = = 0.0095 1,000,000 1,000,000

e−0.0095 (0.0095) k = 0.00005. The fact that the latter is so k! k =0 small suggests that a lineman’s probability of contracting cancer is considerably higher than the one in a million value for p characteristic of the general population. 1

and P(X ≥ 2) = 1 − P(X ≤ 1) = 1 −

4.2.9

∑

Let X = number of solar systems with intelligent life and let p = P(solar system is inhabited). For n ⎛100,000,000,000 ⎞ 0 = 100,000,000,000, P(X ≥ 1) = 1 − P(X = 0) = 1 − ⎜ p ⋅ (1 − p)100,000,000,000. ⎟ 0 ⎝ ⎠ 100,000,000,000 −12 Solving 1 − (1 − p) = 0.50 gives p = 6.9 × 10 . Alternatively, it must be true that −λ 0 e λ 1− = 0.50, which implies that λ = −ln(0.50) = 0.69. But 0.69 = np = 1 × 1011 ⋅ p, 0! so p = 6.9 × 10−12. 109(0) + 65(1) + 22(2) + 3(3) + 1(4) = 0.61, so 200 e −0.61 (0.61) k the presumed Poisson model is pX(k) = , k = 0, 1, … Evaluating pX(k) for k = 0, 1, k! 2, 3, and 4+ shows excellent agreement between the observed proportions and the corresponding Poisson probabilities.

4.2.10 The average number of fatalities per corps-year =

No. of deaths, k 0 1 2 3 4+

Frequency 109 65 22 3 1 200

Proportion 0.545 0.325 0.110 0.015 0.005 1.000

pX(k) 0.5434 0.3314 0.1011 0.0206 0.0035 1.0000

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Section 4.2: The Poisson Distribution

77

4.2.11 The observed number of major changes = 0.44 (= x =

1 [237(0) + 90(1) + 22(2) + 7(3)]), so 356

e −0.44 (0.44)k , k = 0, 1, … Judging from the agreement k! evident in the accompanying table between the set of observed proportions and the values for pX(k), the hypothesis that X is a Poisson random variable is entirely credible.

the presumed Poisson model is pX(k) =

No. of changes, k 0 1 2 3+

Frequency 237 90 22 7 356

Proportion 0.666 0.253 0.062 0.020 1.000

pX(k) 0.6440 0.2834 0.0623 0.0102 1.0000

1 e −1.53 (1.53)k [9(0) + 13(1) + 10(2) + 5(3) + 2(4) + 1(5)] = 1.53, pX(k) = , k! 40 k = 0, 1, … Yes, the Poisson appears to be an adequate model, as indicated by the close agreement between the observed proportions and the values of pX(k).

4.2.12 Since x =

No. of bags lost, k 0 1 2 3 4 5+

4.2.13 The average of the data is

Frequency 9 13 10 5 2 1 40

Proportion 0.225 0.325 0.250 0.125 0.050 0.025 1.000

pX(k) 0.2165 0.3313 0.2534 0.1293 0.0494 0.0201 1.0000

1 [82(0) + 25(1) + 4(2) + 0(3) + 2(4)  0.363 . Then use the model 113

0.363k . Usual statistical practice suggests collapsing the low frequency categories, in this k! case, k = 2, 3, 4. The result is the following table.

e −0.363

No. of countires, k 0 1 2+

Frequency 82 25 6

pX(k) 0.696 0.252 0.052

Expected frequency 78.6 28.5 5.9

The level of agreement between the observed and expected frequencies suggests that the Poisson is a good model for these data.

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78

Chapter 4: Special Distributions

4.2.14 (a) The model pX(k) = e−2.157(2.157)k/k!, k = 0, 1, … fits the data fairly well (where x = 2.157), but there does appear to be a slight tendency for deaths to “cluster”— that is, the values 0, 5, 6, 7, 8, and 9 are all over-represented. No. of deaths, k 0 1 2 3 4 5 6 7 8 9 10+

Frequency 162 267 271 185 111 61 27 8 3 1 0

pX(k) 0.1157 0.2495 0.2691 0.1935 0.1043 0.0450 0.0162 0.0050 0.0013 0.0003 0.0001

Expected frequency 126.8 273.5 294.9 212.1 114.3 49.3 17.8 5.5 1.4 0.3 0.1

(b) Deaths may not be independent events in all cases, and the fatality rate may not be constant. 4.2.15 If the mites exhibit any sort of “contagion” effect, the independence assumption implicit in the 1 [55(0) + 20(1) + … + 1(7)] = 0.81, but pX(k) = Poisson model will be violated. Here, x = 100 e−0.81(0.81)k/k!, k = 0, 1, … does not adequately approximate the infestation distribution. No. of infestations, k 0 1 2 3 4 5 6 7+

Frequency 55 20 21 1 1 1 0 1

Proportion 0.55 0.20 0.21 0.01 0.01 0.01 0 0.01 1.00

pX(k) 0.4449 0.3603 0.1459 0.0394 0.0080 0.0013 0.0002 0.0000 1.0000

4.2.16 Let X = number of repairs needed during an eight-hour workday. Since E(X) = λ = 8 ⋅ e −1.6 (1.6) k = 0.783. k! k =0 2

1.6, P(expenses ≤ $100) = P(X ≤ 2) =

∑

4.2.17 Let X = number of transmission errors made in next half-minute. Since E(X) = λ = 4.5, 2 e−4.5 (4.5)k P(X > 2) = 1 − P(X ≤ 2) = 1 − = 0.826. k! k =0

∑

1 , then λ = 1.10. Therefore, P(X ≥ 2) = 1 − P(X ≤ 1) = 3 1 − e−1.10(1.10)0/0! − e−1.10(1.10)1/1! = 0.301.

4.2.18 If P(X = 0) = e−λλ0/0! = e−λ =

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1 = 5

Section 4.2: The Poisson Distribution

79

4.2.19 Let X = number of flaws in 40 sq. ft. Then E(X) = 4 and P(X ≥ 3) = 1 − P(X ≤ 2) = 2 e −4 4 k 1− = 0.762. k! k =0

∑

4.2.20 Let X = number of particles counted in next two minutes. Since the rate at which the particles are e −8.034 (8.034)3 ⎛ 482 ⎞ counted per minute is 4.017 ⎜ = , E(X) = 8.034 and P(X = 3) = = 0.028. ⎟ ⎝ 120 ⎠ 3! Now, suppose X = number of particles counted in one minute. Then P(3 particles are counted in next two minutes) = P(X = 3) ⋅ P(X = 0) + P(X = 2) ⋅ P(X = 1) + P(X = 1) ⋅ P(X = 2) + P(X = 0) ⋅ P(X = 3) = 0.028, where λ = 4.017. 4.2.21 (a) Let X = number of accidents in next five days. Then E(X) = 0.5 and P(X = 2) = e−0.5(0.5)2/2! = 0.076. (b) No. P(4 accidents occur during next two weeks) = P(X = 4) ⋅ P(X = 0) + P(X = 3) ⋅ P(X = 1) + P(X = 2) ⋅ P(X = 2) + P(X = 1) ⋅ P(X = 3) + P(X = 0) ⋅ P(X = 4). 4.2.22 If P(X = 1) = P(X = 2), then e−λλ1/1! = e−λλ2/2!, which implies that 2λ = λ2, or, equivalently, λ = 2. Therefore, P(X = 4) = e−224/4! = 0.09. ⎧ λ2 λ4 λ6 ⎫ e− λ λ 2 k = e − λ ⎨1 + + + + "⎬ = e−λ ⋅ cosh λ = 2! 4! 6! ⎩ ⎭ k = 0 (2 k )! ⎞ 1 −2 λ ⎟⎠ = 2 (1 + e ) . ∞

4.2.23 P(X is even) = ⎛ e λ + e− λ e ⎜⎝ 2 −λ

∞

4.2.24

f X +Y (w)=

∑

k =0 w

∑

pk (k ) pY ( w − k ) =

w

∑

e− λ

k =0

λk k!

e− µ

µ w− k ( w − k )!

1 w 1 w! λ k µ w− k = e− ( λ + µ ) ( λ + µ ) w w! k = 0 k !( w − k )! w! k =0 th The last expression is the w term of the Poisson pdf with parameter λ + µ. = e−( λ + µ )

1

∑ k !(w − k )!λ

k

µ w− k = e − ( λ + µ )

∑

∞

4.2.25 From Definition 3.11.1 and Theorem 3.7.1, P(X2 = k) =

⎛ x1 ⎞

∑ ⎜⎝ k ⎟⎠ p

k

(1 − p) x1 − k ⋅

x1 = k

− λ y+k ⎛y + k⎞ k e− λ ( λ p)k y e λ (1 ) ⋅ − ⋅ = p p k ⎟⎠ k! ( y + k )! y =0 ∞

Let y = x1 − k. Then P(X2 = k) =

∑ ⎜⎝

e− λ λ x1 . x1 ! ∞

[ λ (1 − p)] y = y! y =0

∑

e − λ ( λ p ) k λ (1− p ) e− λ p ( λ p ) k ⋅e = . k! k!

4.2.26 (a) Yes, because the Poisson assumptions are probably satisfied—crashes are independent events and the crash rate is likely to remain constant. 3 e−2.5 (2.5)k (b) Since λ = 2.5 crashes per year, P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − = 0.24. k! k =0

∑

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80

Chapter 4: Special Distributions

(c) Let Y = interval (in yrs.) between next two crashes. By Theorem 4.2.3, P(Y < 0.25) =

∫

0.25 0

2.5e −2.5 y dy = 1 − 0.535 = 0.465.

4.2.27 Let X = number of deaths in a week. Based on the daily death rate, E(X) = λ = 0.7. Let Y = interval (in weeks) between consecutive deaths. Then P(Y > 1) =

∫

∞ 1

0.7e −0.7 y dy = − e− u

∞ 0.7

= 0.50. 4.2.28 Given that fY(y) = 0.027e−0.027, P(Y1 + Y2 < 40) = e − u (−u − 1)

1.08 0

∫

40 0

(0.027) 2 ye −0.027 y dy =

∫

1.08 0

ue − u du =

= 1 − 0.706 = 0.29 (where u = 0.027y).

4.2.29 Let X = number of bulbs burning out in 1 hour. Then E(X) = λ = 0.11. Let Y = number of hours a bulb remains lit. Then P(Y < 75) =

∫

75 0

0.011e−0.011 y dy = − e − u

0.825 0

= 0.56. (where u = 0.011y).

Since n = 50 bulbs are initially online, the expected number that will fail to last at least 75 hours is 50 ⋅ P(Y < 75), or 28. 4.2.30 Assume that 29 long separations and 7 short separations are to be randomly arranged. In order for “bad things to come in fours” three of the short separations would have to occur at least once in the 30 spaces between and around the 29 long separations. For that to happen, either (1) 3 short separations have to occur in one space and the remaining 4 shorts in 4 other spaces (2) 3 short separations occur in one space, 2 short separations occur in another space, and 1 short separation occurs in each of two spaces or (3) 3 short separations occur in each of two spaces and the remaining short occurs in a third space. The combined probability of these three possibilities is ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ 30 ⎟ ⎜ 5 ⎟ + ⎜ 30 ⎟ 4! + ⎜ 30 ⎟ ⎜ 3⎟ ⎜⎝ 5 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 4 ⎟⎠ 2!1!1! ⎜⎝ 3 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎛ ⎞ ⎜ 36 ⎟ ⎜⎝ 29 ⎟⎠

= 0.126

Section 4.3: The Normal Distribution 4.3.1

(a) 0.5782

4.3.2

(a) (b) (c) (d) (e)

4.3.3

(a) Both are the same because of the symmetry of fZ(z).

(b)

0.8264

(c) 0.9306

(d)

0.9808 − 0.5000 = 0.4808 0.4562 − 0.2611 = 0.1951 1 − 0.1446 = 0.8554 = P(Y < 1.06) 0.0099 P(Z ≥ 4.61) < P(Z ≥ 3.9) = 1 − 1.0000 = 0.0000

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0.0000

Section 4.3: The Normal Distribution

81

(b) Since fZ(z) is decreasing for all z > 0,

4.3.4

(a)

∫

(b)

∫

1.24 0 ∞ −∞

e− z

2

6e − z

/2

2

dz = 2π

/2

∫

dz = 6 2π

1.24 0

∫

∞ −∞

∫

1 2 1 a− 2 a+

1 − z2 / 2 e dz is larger than 2π

∫

a +1 a

1 − z2 / 2 e dz . 2π

1 − z2 / 2 e dz = 2π (0.8925 − 0.5000) = 1.234 2π 1 − z2 / 2 e dz = 6 2π 2π

4.3.5

(a) −0.44

4.3.6

From Appendix Table A.1, z.25 = 0.67 and z.75 = −0.67, so Q = 0.67 − (−0.67) = 1.34.

4.3.7

Let X = number of decals purchased in November. Then X is binomial with n = 74,806 and p = 1/12. P(50X < 306,000) = P(X < 6120) = P(X ≤ 6119). Using the DeMoivre-Laplace approximation with continuity correction gives ⎛ 6119.5 − 74,806(1/12) ⎞ P(X ≤ 6119)  P ⎜ Z ≤ ⎟ = P(Z ≤ −1.51) = 0.0655 74,806(1/12)(11/12) ⎠ ⎝

4.3.8

Let X = number of usable cabinets in next shipment. Since np = 1600(0.80) = 1280 and np(1 − p) = 1600(0.80)(0.20) = 16, P(shipment causes no problems) = P(1260 ≤ X ≤ 1310) =

(b) 0.76

(c) 0.41

(d)

1.28

(e) 0.95

⎛ 1259.5 − 1280 X − 1280 1310.5 − 1280 ⎞ P⎜ ≤ ≤ ⎟⎠ = P(−1.28 ≤ Z ≤ 1.91) = 0.8716. ⎝ 16 16 16

4.3.9

Let X = number of voters challenger receives. Given that n = 400 and p = P(voter favors challenger) = 0.45, np = 180 and np(1 − p) = 99. (a) P(tie) = P(X = 200) = P(199.5 ≤ X ≤ 200.5) = ⎛ 199.5 − 180 X − 180 200.5 − 180 ⎞ ≤ ≤ P⎜ ⎟⎠ = P(1.96 ≤ Z ≤ 2.06) = 0.0053. ⎝ 99 99 99 (b) P(challenger wins) = P(X > 200) =P(X ≥ 200.5) = ⎛ X − 180 200.5 − 180 ⎞ ≥ P⎜ ⎟⎠ = P(Z ≥ 2.06) = 0.0197. ⎝ 99 99

4.3.10 (a) Let X = number of shots made in next 100 attempts. Since p = P(attempt is successful) = 0.70, P(75 ≤ X ≤ 80) = 80 ⎛100 ⎞ k 100 − k . ⎜⎝ k ⎟⎠ (0.70) (0.30) k = 75

∑

(b) With np = 100(0.70) = 70 and np(1 − p) = 100(0.70)(0.30) = 21, P(75 ≤ X ≤ 80) = P(74.5 ≤ ⎛ 74.5 − 70 X − 70 80.5 − 70 ⎞ X ≤ 80.5) = P ⎜ ≤ ≤ ⎟ = P(0.98 ≤ Z ≤ 2.29) = 0.1525. ⎝ 21 21 21 ⎠

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82

Chapter 4: Special Distributions

4.3.11 Let p = P(person dies by chance in the three months following birthmonth) =

1 . Given that n = 4

747, np = 186.75, and np(1 − p) = 140.06, P(X ≥ 344) = P(X ≥ 343.5) = ⎛ X − 186.75 343.5 − 186.75 ⎞ ≥ P⎜ ⎟ = P(Z ≥ 13.25) = 0.0000. The fact that the latter probability is ⎝ 140.06 140.06 ⎠ so small strongly discredits the hypothesis that people die randomly with respect to their birthdays. 4.3.12 Let X = number of correct guesses (out of n = 1500 attempts). Since five choices were available for each guess (recall Figure 4.3.4), p = P(correct answer) = 1/5, if ESP is not a ⎛ X − 1500(1/ 5) 325.5 − 300 ⎞ factor. Then P(X ≥ 326) = P(X ≥ 325.5) = P ⎜ ≥ ⎟ = P(Z ≥ 1.65) = 240 ⎠ ⎝ 1500(1/ 5)(4 / 5) 0.0495. Based on these results, there is certainly some evidence that ESP may be increasing the probability of a correct guess, but the magnitude of P(X ≥ 326) is not so small that it precludes the possibility that chance is the only operative factor. 4.3.13 No, the normal approximation is inappropriate because the values of n (= 10) and p (= 0.7) fail to p 0.7 =9 = 21. satisfy the condition n > 9 1− p 0.3 4.3.14 Let X = number of fans buying hot dogs. To be determined is the smallest value of c for which P(X >c) ≤ 0.20. Assume that no one eats more than one hot dog. Then X is a binomial random variable with n = 42,200 and p = P(fan buys hot dog) = 0.38. Since np = 16,036 and np (1 − p ) 1 ⎛ ⎞ c + 1 − − 16,036 ⎜ ⎟ 2 = 99.7, P(X > c) = 0.20 = P(X ≥ c + 1)  P ⎜ Z ≥ ⎟. 99.7 ⎜⎝ ⎠⎟ 1 ⎛ ⎞ c + 1 − − 16,036 ⎜ ⎟ 2 But P(Z ≥ 0.8416) = 0.20, so 0.8416 = ⎜ ⎟ , from which it follows that 99.7 ⎜⎝ ⎠⎟ c = 16,119. ⎛ −5.5 X − 100 5.5 ⎞ 4.3.15 P ( X − E ( X ) ≤ 5) = P(−5 ≤ X − 100 ≤ 5) = P ⎜ ≤ ≤ ⎟ ⎝ 50 50 50 ⎠ P(−0.78 ≤ Z ≤ 0.78) = 0.5646.

For binomial data, the central limit theorem and DeMoivre-Laplace approximations differ only if the continuity correction is used in the DeMoivre-Laplace approximation.

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Section 4.3: The Normal Distribution

83

4.3.16 Let Xi = face showing on ith die, i = 1, 2, …, 100, and let X = X1 + X2 + … + X100. Following the

( )

approach taken in Example 3.9.5 gives E(X) = 350. Also, Var(Xi) = E X i2 − [E(Xi)]2 = 2

1 2 35 3500 ⎛ 1⎞ , so Var(X) = (1 + 22 + 32 + 42 + 52 + 62 ) − ⎜ 3 ⎟ = . By the central limit theorem, ⎝ ⎠ 6 2 12 12 ⎛ X − 350 370.5 − 350 ⎞ then, P(X > 370) = P(X ≥ 371) = P(X ≥ 370.5) = P ⎜ ≥ ⎟  P(Z ≥ 1.20) ⎝ 3500 /12 3500 /12 ⎠ = 0.1151.

4.3.17 For the given X, E(X) = 5(18/38) + (−5)(20/38) = −10/38 = −0.263. Var(X) = 25(18/38) + (25)(20/38) − (−10/38)2 = 24.931, σ = 4.993. Then P(X1 + X2 + … + X100 > −50) ⎛ X + X 2 + ... + X 100 − 100(−0.263) −50 − 100(−0.263) ⎞ = P⎜ 1 > ⎟  P(Z > −0.47) 10(4.993) 100(4.993) ⎝ ⎠ = 1 − 0.3192 = 0.6808 4.3.18 If Xi is a Poisson random variable with parameter λi, then E(Xi) = Var(Xi) = λi . Let X = X1 + X2 + n

… + Xn be a sum of independent Poissons. Then E(X) = in Theorem 4.3.2 reduces to

X −λ

λ

∑λ

i

n

= Var(X). If λ =

i =1

∑ λ , the ratio i

i =1

.

4.3.19 Let X = number of chips ordered next week. Given that λ = E(X) = 50, P(company is unable to ⎛ X − 50 60.5 − 50 ⎞ fill orders) = P(X ≥ 61) = P(X ≥ 60.5) = P ⎜ ≥ ⎟ P(Z ≥ 1.48) = 0.0694. ⎝ 50 50 ⎠ 4.3.20 Let X = number of leukemia cases diagnosed among 3000 observers. If λ = E(X) = 3, 7 e−3 3k P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − = 1 − 0.9881 = 0.0119. Using the central limit theorem, k =0 k !

∑

⎛ X − 3 7.5 − 3 ⎞ 1 − P(X ≤ 7) = 1 − P(X ≤ 7.5) = 1 − P ⎜ ≤ ⎟  1 − P(Z ≤ 2.60) = 0.0047. The ⎝ 3 3 ⎠ approximation is not particularly good because λ is small. In general, if λ is less than 5, the normal approximation should not be used. Both analyses, though, suggest that the observer’s risk of contracting leukemia was increased because of their exposure to the test.

4.3.21 No, only 84% of drivers are likely to get at least 25,000 miles on the tires. If X denotes the mileage obtained on a set of Econo-Tires, P(X ≥ 25,000) = ⎛ X − 30,000 25,000 − 30,000 ⎞ P⎜ ≥ ⎟⎠ = P(Z ≥ −1.00) = 0.8413. ⎝ 5000 5000

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84

Chapter 4: Special Distributions

4.3.22 Let Y denote a child’s IQ. Then P(child needs special services) = P(Y < 80) + P(Y > 135) = ⎛ Y − 100 80 − 100 ⎞ ⎛ Y − 100 135 − 100 ⎞ P⎜ < > ⎟ + P ⎜⎝ ⎟⎠ = P(Z < −1.25) + P(Z > 2.19) = ⎝ 16 16 ⎠ 16 16 0.1056 + 0.0143 = 0.1199. It follows that 1400 × 0.1199 × $1750 = $293,755 should be added to Westbank’s special ed budget. 4.3.23 Let Y = donations collected tomorrow. Given that µ = $20,000 and σ = $5,000, ⎛ Y − $20,000 $30,000 − $20,000 ⎞ P(Y > $30,000) = P ⎜ > ⎟⎠ = P(Z > 2.00) = 0.0228. ⎝ $5,000 $5,000 4.3.24 Let Y = pregnancy duration (in days). Ten months and five days is equivalent to 310 days. The credibility of San Diego Reader’s claim hinges on the magnitude of P(Y ≥ 310)—the smaller that probability is, the less believable her explanation becomes. Given that µ = 266 and σ = 16, P(Y ⎛ Y − 266 310 − 266 ⎞ ≥ 310) = P ⎜ ≥ ⎟⎠ = P(Z ≥ 2.75) = 0.0030. While the latter does not rule out the ⎝ 16 16 possibility that San Diego Reader is telling the truth, pregnancies lasting 310 or more days are extremely unlikely. 4.3.25 (a) Let Y1 and Y2 denote the scores made by a random nondelinquent and delinquent, respectively. Then E(Y1) = 60 and Var(Y1) = 102; also, E(Y2) = 80 and Var(Y2) = 52. Since 75 is the cutoff between teenagers classified as delinquents or nondelinquents, 75 − 60 ⎞ ⎛ P(nondelinquent is misclassified as delinquent) = P(Y1 > 75) = P ⎜ Z > ⎟ = 0.0668. ⎝ 10 ⎠ 75 − 80 ⎞ ⎛ Similarly, P(delinquent is misclassified as nondelinquent) = P(Y2 < 75) = P ⎜ Z < ⎟= ⎝ 5 ⎠ 0.1587.

4.3.26 Let Y denote the cross-sectional area of a tube. Then p = P(tube does not fit properly) = ⎛ 12.0 − 12.5 Y − 12.5 13.0 − 12.5 ⎞ ≤ ≤ P(Y < 12.0) + P(Y >13.0) = 1 − P(12.0 ≤ Y ≤ 13.0) = 1 − P ⎜ ⎟⎠ = ⎝ 0.2 0.2 0.2 1 − P(−2.50 ≤ Z ≤ 2.50) = 1 − 0.9876 = 0.0124. Let X denote the number of tubes (out of 1000) that will not fit. Since X is a binomial random variable, E(X) = np = 1000(0.0124) = 12.4. 4.3.27 Let Y = freshman’s verbal SAT score. Given that µ = 565 and σ = 75, P(Y > 660) = ⎛ Y − 565 660 − 565 ⎞ P⎜ > ⎟⎠ = P(Z > 1.27) = 0.1020. It follows that the expected number doing ⎝ 75 75 better is 4250(0.1020), or 434. 4.3.28 Let A* and B* denote the lowest A and the lowest B, respectively. Since the top 20% of the grades will be A’s, P(Y < A*) = 0.80, where Y denotes a random student’s score. Equivalently, A * −70 ⎞ ⎛ P ⎜Z < ⎟ = 0.80. From Appendix Table A.1, though, P(Z < 0.84) = 0.7995  0.80. ⎝ 12 ⎠ A * −70 Therefore, 0.84 = , which implies that A* = 80. 12

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Section 4.3: The Normal Distribution

85

B * −70 ⎞ ⎛ Similarly, P(Y < B*) = 0.54 = P ⎜ Z < ⎟ . But P(Z < 0.10) = 0.5398  0.54, ⎝ 12 ⎠ B * −70 so 0.10 = , implying that B* = 71. 12

4.3.29

20 ⎞ ⎛ 20 − 40 Y − 40 60 − 40 ⎞ ⎛ −20 If P(20 ≤ Y ≤ 60) = 0.50, then P ⎜ ≤ ≤ ≤ Z ≤ ⎟ . But ⎟⎠ = 0.50 = P ⎜⎝ ⎝ σ σ σ σ σ ⎠ 20 P(−0.67 ≤ Z ≤ 0.67) = 0.4972  0.50, which implies that 0.67 = . The desired value for σ, then,

σ

is

20 , or 29.85. 0.67

103.5 + 144.5 = 124, 2 20.5 ⎞ ⎛ 103.5 − 124 Y − 124 144.5 − 124 ⎞ ⎛ −20.5 P(103.5 ≤ Y ≤ 144.5) = 0.80 = P ⎜ ≤ ≤ ≤Z ≤ . ⎟⎠ = P ⎜⎝ ⎝ σ σ σ σ σ ⎟⎠ 20.5 According to Appendix Table A.1, P(−1.28 ≤ Z ≤ 1.28)  0.80, so = 1.28, implying that σ

4.3.30 Let Y = a random 18-year-old woman’s weight. Since µ =

σ

= 16.0 lbs. 4.3.31 Let Y = analyzer reading for driver whose true blood alcohol concentration is 0.9. Then ⎛ Y − 0.9 0.08 − 0.09 ⎞ P(analyzer mistakenly shows driver to be sober) = P(Y < 0.08) = P ⎜ < ⎟ = ⎝ 0.004 0.004 ⎠ P(Z < −2.50) = 0.0062. The “0.075%” driver should ask to take the test twice. The “0.09%” driver has a greater chance of not being charged by taking the test only once. As, n the number of times the test taken, increases, the precision of the average reading increases. It is to the sober driver’s advantage to have a reading as precise as possible; the opposite is true for the drunk driver. 75 − 62.0 92 − 76.3 = 1.71; the normed score for Laura is = 7.6 10.8 1.45. So, even though Laura made 17 points higher on the test, the company would be committed to hiring Michael.

4.3.32 The normed score for Michael is

⎛ Y − 100 103 − 100 ⎞ = P(Z > 0.56) = 4.3.33 By the first corollary to Theorem 4.3.3, P (Y > 103) = P ⎜ > ⎝ 16 / 9 16 / 9 ⎟⎠ ⎛ Y − 100 103 − 100 ⎞ 0.2877. For any arbitrary Yi, P(Yi > 103) = P ⎜ i > ⎟⎠ = P(Z > 0.19) = 0.4247. ⎝ 16 16 Let X = number of Yi’s that exceed 103. Since X is a binomial random variable with n = 9 and ⎛9⎞ p = P(Yi > 103) = 0.4247, P(X = 3) = ⎜ ⎟ (0.4247)3 (0.5753)6 = 0.23. ⎝3⎠

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86

Chapter 4: Special Distributions

⎛ 1.9 − 2 2.1 − 2 ⎞ 4.3.34 If P(1.9 ≤ Y ≤ 2.1) ≥ 0.99, then P ⎜ ≤Z≤ ⎟ ≥ 0.99. But P(−2.58 ≤ Z ≤ 2.58) ⎝ 2/ n 2/ n ⎠ 2.1 − 2 , which implies that n = 2663.  0.99, so 2.58 = 2/ n

4.3.35 Let Yi = resistance of ith resistor, i = 1, 2, 3, and let Y = Y1 + Y2 + Y3 = circuit resistance. By the first corollary to Theorem 4.3.3, E(Y) = 6 + 6 + 6 = 18 and Var(Y) = (0.3)2 + (0.3)2 + (0.3)2 = ⎛ Y − 18 19 − 18 ⎞ 0.27. Therefore, P(Y > 19) = P ⎜ > ⎟ = P(Z > 1.92) = 0.0274. Suppose P(Y > 19) ⎝ 0.27 0.27 ⎠ ⎛ 19 − 18 ⎞ 19 − 18 = 0.005 = P ⎜ Z > , . From Appendix Table A.1, P(Z > 2.58)  0.005, so 2.58 = ⎟ ⎝ 3σ 2 ⎠ 3σ 2 which implies that the minimum “precision” of the manufacturing process would have to be σ = 0.22 ohms.

4.3.36 Let YP and YC denote a random piston diameter and cylinder diameter, respectively. Then P(pair needs to be reworked) = P (YP > YC ) = P (YP − YC > 0) ⎛ Y − Y − (40.5 − 41.5) 0 − (40.5 − 41.5) ⎞ = P⎜ P C > ⎟ = P(Z > 2.00) = 0.0228, or 2.28%. ⎜⎝ (0.3)2 + (0.4) 2 (0.3) 2 + (0.4)2 ⎟⎠ n n 2 2 2 2 2 ⎛t⎞ ⎛t ⎞ 4.3.37 M Y (t ) = M Y1 +...Yn ⎜ ⎟ = M Yi ⎜ ⎟ = e µt / n + σ t / 2 n = e µt + σ t / 2 n , but the latter is the ⎝ n ⎠ i =1 ⎝ n ⎠ i =1 moment-generating function for a normal random variable whose mean is µ and whose variance

∏

∏

n

is σ2/n. Similarly, if Y = a1Y1 + … + anYn, MY(t) =

∏M i =1

n

n

∏e

µi ai t + σ i2 ai2 t 2 / 2

(t ) =

n

∏M i =1

Yi

(ai t ) =

n

∑ ai µi t + ∑ ai2σ i2t 2 / 2

= e i=1

aiYi

i =1

. By inspection, Y has the moment-generating function of a

i =1

n

normal random variable for which E(Y) =

∑ i =1

ai µi and Var(Y) =

n

∑a σ 2 i

2 i

.

i =1

4.3.38 P (Y ≥ Y * ) = P(Y − Y * ≥ 0) , where E (Y − Y * ) = E (Y ) − E (Y * ) = 2 − 1 = 1 . 22 12 25 + = , because Y and Y * are independent. 9 4 36 ⎛Y − Y * −1 0 −1 ⎞ Therefore, P (Y ≥ Y * ) = P ⎜ ≥ = P ( Z ≥ −1.20) = 0.8849 25 / 36 ⎟⎠ ⎝ 25 / 36

Also, Var(Y − Y * ) = Var(Y ) − Var(Y * ) =

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Section 4.4: The Geometric Distribution

87

Section 4.4: The Geometric Distribution 4.4.1

Let p = P(return is audited in a given year) = 0.30 and let X = year of first audit. Then P(Jody escapes detection for at least 3 years) = P(X ≥ 4) = 1 − P(X ≤ 3) = 3

1−

∑ (0.70)

k −1

(0.30) = 0.343.

k =1

4.4.2

If X = attempt at which license is awarded and p = P(driver passes test on any given attempt) 1 1 = 10. = 0.10, then pX(k) = (0.90)k−1(0.10), k = 1, 2, …; E(X) = = p 0.10

4.4.3

No, the expected frequencies (= 5 ⋅ pX(k)) differ considerably from the observed frequencies, especially for small values of k. The observed number of 1’s, for example, is 4, while the expected number is 12.5. k −1

k

Obs. Freq.

1 2 3 4 5 6 7 8 9+

4 13 10 7 5 4 3 3 1 50

⎛3⎞ ⎛1⎞ pX(k) = ⎜ ⎟ ⎜ ⎟ ⎝4⎠ ⎝4⎠ 0.2500 0.1875 0.1406 0.1055 0.0791 0.0593 0.0445 0.0334 0.1001 1.0000

50 ⋅ pX(k) = Exp. freq. 12.5 9.4 7.0 5.3 4.0 3.0 2.2 1.7 5.0 50.0

1 1 = = 2. Barring any p 1 2 medical restrictions, it would not be unreasonable to model the appearance of a couple’s first girl (or boy) by the geometric probability function. The most appropriate value for p, though, would 1 1 not be exactly (although census figures indicate that it would be close to ). 2 2

4.4.4

If p = P(child is a girl) and X = birth order of first girl, then E(X) =

4.4.5

FX(t) = P(X ≤ t) = p

[t ]

∑ s=0

[t ]

(1 − p) s . But

∑ s =0

(1 − p) s =

1 − (1 − p )[t ] 1 − (1 − p )[t ] = , and the result 1 − (1 − p ) p

follows.

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88

4.4.6

Chapter 4: Special Distributions

Let X = roll on which sum of 4 appears for first time. Since p = P(sum = 4) = ⎛ 213 ⎞ ⎜⎝ ⎟ 216 ⎠

k −1

⋅

3 , k = 1, 2, … Using the expression for FX(k) given in Question 4.4.5, we can write 216

3 ⎞ ⎛ P(65 ≤ X ≤ 75) = FX(75) − FX(64) = 1 − ⎜1 − ⎝ 216 ⎠⎟ ⎛ 213 ⎞ ⎜⎝ ⎟ 216 ⎠

4.4.7

3 . pX(k) = 216

[75]

[64] 64 ⎛ ⎛ 3 ⎞ ⎞ ⎛ 213 ⎞ − ⎜1 − ⎜1 − ⎟ ⎟=⎜ ⎟ − ⎝ ⎝ 216 ⎠ ⎠ ⎝ 216 ⎠

75

= 0.058.

P(n ≤ Y ≤ n + 1) =

∫

n +1 n

λ e − λ y dy = (1 − e− λ y )

n +1

= e− λ n − e− λ ( n +1) = e − λ n (1 − e − λ )

n

Setting p = 1 − e−λ gives P(n ≤ Y ≤ n + 1) = (1 − p)np. 4.4.8

Let the random variable X* denote the number of trials preceding the first success. By inspection, ∞

pX*(t) = (1 − p)kp, k = 0, 1, 2, … Also, MX*(t) =

∑

etk ⋅ (1 − p )k p = p

t k

=p⋅

⎞ p ⎟⎠ = 1 − (1 − p)et . Let X denote the geometric random variable defined in Theorem

4.4.1. Then X* = X − 1, and M X * (t ) = e − t M X (t ) = e − t ⋅

4.4.9

∑[(1 − p)e ] k =0

k =0

⎛ 1 ⎜⎝ 1 − (1 − p )et

∞

pet p . = t 1 − (1 − p )e 1 − (1 − p )et

MX(t) = pet[1 − (1 − p)et]−1, so M X(1) (t ) = pet(−1)[1 − (1 − p)et]−2 ⋅ (−(1 − p)et)+ 1 [1 − (1 − p)et]−1pet. Setting t = 0 gives M X(1) (0) = E ( X ) = . Similarly, M X(2) (t) = p 2t t −3 t p(1 − p)e (−2)[1 − (1 − p)e ] ⋅ (−(1 − p)e ) + [1 − (1 − p)et]−2p(1 − p)e2t ⋅2 + [1 − (1 − p)et]−1pet + pet(−1)[1 − (1 − p)et]−2 ⋅ (−(1 − p)et) and M X(2) (0) = E(X 2) = 2

2− p . p2

2 − p ⎛ 1 ⎞ 1− p Therefore, Var( X ) = E ( X ) − [ E ( X )] = 2 − ⎜ ⎟ = 2 ⎝ p⎠ p p 2

2

4.4.10 No, because MX(t) = M X1 (t ) ⋅ M X 2 (t ) does not have the form of a geometric moment-generating function. 4.4.11 Let M X* (t ) = E (t X ) =

∞

∑ k =1

t k ⋅ (1 − p ) k −1 p =

p 1− p

∞

∑ k =1

[t (1 − p )]k =

p ∞ p = [t (1 − p )]k − 1 − p k =0 1− p

∑

⎤ p ⎡ 1 p pt = factorial moment-generating function for X. − = ⎢ ⎥ 1 − p ⎣ 1 − t (1 − p ) ⎦ 1 − p 1 − t (1 − p ) p −2 −1 Then M *(1) . X (t ) = pt(−1)[1 − t(1 − p)] (−(1 − p)) + [1 − t(1 − p)] p = [1 − t (1 − p)]2

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Section 4.5: The Negative Binomial Distribution

When t = 1, M X*(1) (1) = E(X) =

89

1 2 p (1 − p ) 2 − 2p . Also, M *(2) and M *(2) = X (t ) = X (1) = 3 p [1 − t (1 − p )] p2

E[X(X − 1)] = E(X2) − E(X). Therefore, Var(X) = E(X 2) − [E(X)]2 = =

2 − 2p 1 ⎛ 1 ⎞ + −⎜ ⎟ p ⎝ p⎠ p2

2

1− p . p2

Section 4.5: The Negative Binomial Distribution 4.5.1

Let X = number of houses needed to achieve fifth invitation. If p = P(saleswoman receives ⎛ k − 1⎞ invitation at a given house) = 0.30, pX(k) = ⎜ (0.30)4(0.70)k−1−4(0.30), k = 5, 6, … and ⎝ 4 ⎠⎟ ⎛ k − 1⎞ 5 k− 5 ⎟⎠ (0.30) (0.70) = 0.029. 4 k =5 7

P(X < 8) = P(5 ≤ X ≤ 7) =

∑ ⎜⎝

4.5.2

Let p = P(missile scores direct hit) = 0.30. Then P(target will be destroyed by seventh missile fired) = P(exactly three direct hits occur among first six missiles and seventh missile scores direct ⎛ 6⎞ hit) = ⎜ ⎟ (0.30)3 (0.70)3 (0.30) = 0.056. ⎝3 ⎠

4.5.3

Darryl might have actually done his homework, but there is reason to suspect that he did not. Let the random variable X denote the toss where a head appears for the second time. Then pX(k) = 2 k −2 ⎛ k − 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ , k = 2, 3, …, but that particular model fits the data almost perfectly, as the ⎝⎜ 1 ⎠⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ table shows. Agreement this good is often an indication that the data have been fabricated. k 2 3 4 5 6 7 8 9 10

4.5.4

pX(k) 1/4 2/8 3/16 4/32 5/64 6/128 7/256 8/512 9/1024

Obs. freq. 24 26 19 13 8 5 3 1 1

Exp. freq. 25 25 19 12 8 5 3 2 1

Let p = P(defective is produced by improperly adjusted machine) = 0.15. Let X = item at which ⎛ k − 1⎞ ⎛ k − 1⎞ machine is readjusted. Then pX(k) = ⎜ (0.15) 2 (0.85)k −1− 2 (0.15) = ⎜ (0.15)3 (0.85) k −3 , ⎟ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ k = 3, 4, … It follows that P(X ≥ 5) = 1 − P(X ≤ 4) 3 = 1 − [P(X = 3) + P(X = 4)] = 0.988 and E(X) = = 20. 0.15

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90

Chapter 4: Special Distributions

⎛ k − 1⎞ r r ∞ ⎛ k ⎞ r +1 r k −r k⎜ p (1 − p ) = p (1 − p )k − r = . ⎟ ⎜ ⎟ p k =r ⎝ r ⎠ p k =r ⎝ r − 1⎠ ∞

∑

∑

4.5.5

E(X) =

4.5.6

Let Y denote the number of trials to get the rth success, and let X denote the number of trials in excess of r to get the rth success. Then X = Y − r. Substituting into Theorem 4.5.1 gives ⎛ k + r − 1⎞ r ⎛ k + r − 1⎞ r pX(k) = ⎜ p (1 − p )k = ⎜ p (1 − p )k , k = 0, 1, 2, … ⎟ ⎟ ⎝ k ⎠ ⎝ r −1 ⎠

4.5.7

Here X = Y − r, where Y has the negative binomial distribution as described in Theorem 4.5.1. Using the properties (1), (2), and (3) given by the theorem, we can write E(X) = E(Y − r) r r (1 − p ) and Var(X) = Var(Y − r) = Var(Y) + Var(r) = E(Y) − E(r) = − r = p p r r t ⎡ ⎤ ⎡ ⎤ pe p r (1 − p) r (1 − p ) = −rt −rt ⎢ = . Also, MX(t) = MY−r(t) = e MY(t) = e 1 − (1 − p )et ⎥ ⎢1 − (1 − p )et ⎥ . +0= 2 2 ⎦ ⎣ ⎦ ⎣ p p 3

4.5.8

⎡ ⎤ (4 / 5)et For each Xi, M X i (t ) = ⎢ , i = 1, 2, 3. If X = X1 + X2 + X3, it follows that t ⎥ ⎣1 − (1 − 4 / 5)e ⎦ 3

MX(t) =

∏ i =1

9

9 k −9 ⎡ ⎤ ⎛ k − 1⎞ ⎛ 4 ⎞ ⎛ 1 ⎞ (4 / 5)et , which implies that pX(k) = ⎜ M X i (t ) = ⎢ ⎜ ⎟ ⎜ ⎟ , t ⎥ ⎝ 8 ⎠⎟ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎣1 − (1 − 4 / 5)e ⎦ 12

k = 9, 10, … Then P(10 ≤ X ≤ 12) =

∑p

X

(k ) = 0.66.

k =10

4.5.9

M X(1) (t )

⎡ ⎤ pet = r⎢ t ⎥ ⎣1 − (1 − p )e ⎦

r −1

[pet[1 − (1 − p)et]−2(1 − p)et + [1 − (1 − p)et]−1pet]. When t = 0,

⎡ p (1 − p ) p ⎤ r + ⎥= . M X(1) (0) = E(X) = r ⎢ 2 p⎦ p ⎣ p r

r*

k ⎡ ⎤i ⎡ ⎤ pet pet M X i (t ) = , where r* = 4.5.10 MX(t) = = ri . Also, ⎢ ⎥ ⎢ ⎥ t t ⎣1 − (1 − p )e ⎦ i =1 i =1 ⎣ 1 − (1 − p )e ⎦ i =1 k ri r * E(X) = E(X1 + X2 + … + Xk) = E(X1) + E(X2) + … + E(Xk) = and = p i =1 p k

k

∏

∏

∑

∑

k

Var(X) =

∑ i =1

Var( X i ) =

k

∑ i =1

ri (1 − p ) r * (1 − p ) . = p2 p2

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Section 4.6: The Gamma Distribution

91

Section 4.6: The Gamma Distribution 4.6.1

Let Yi = lifetime of ith gauge, i = 1, 2, 3. By assumption, fYi ( y ) = 0.001e−0.001y, y > 0. Define the random variable Y = Y1 + Y2 + Y3 to be the lifetime of the system. By Theorem 4.6.1, fY(y) = (0.001)3 2 −0.001 y y e , y > 0. 2

4.6.2

The time until the 24th breakdown is a gamma random variable with parameters r = 24 and λ = 3. The mean of this random variable is r/λ = 24/3 = 8 months.

4.6.3

If E(Y) =

r

λ

= 1.5 and Var(Y) =

0. Then P(1.0 ≤ Yi ≤ 2.5) =

∫

2.5 1.0

r

λ2

= 0.75, then r = 3 and λ = 2, which makes fY(y) = 4y2e−2y, y >

4 y 2 e−2 y dy = 0.55. Let X = number of Yi’s in the interval

(1.0, 2.5). Since X is a binomial random variable with n = 100 and p = 0.55, E(X) = np = 55. 4.6.4

4.6.5

f λY ( y ) =

1

λ

fY ( y / λ ) =

1 λr ⎛ y ⎞ λ Γ (r ) ⎜⎝ λ ⎟⎠

r −1

e− λ ( y / λ ) =

To find the maximum of the function fY ( y ) =

1 r −1 − y y e Γ (r )

λr Γ (r )

y r −1e− λ y , differentiate it with respect to y and

set it equal to 0; that is dfY ( y ) d λ r r −1 − λ y λr = y e = [(r − 1) y r − 2 e− λ y − λ y r −1e− λ y ] = 0 dy dy Γ (r ) Γ (r ) This implies

λr

y r − 2 e − λ y [(r − 1) − λ y ] = 0 , whose solution is ymode =

r −1

. Since the derivative Γ (r ) λ is positive for y < ymode , and negative for y > ymode , then there is a maximum. 4.6.6

∞ 2 1 z 2 e − z / 2 dz 2π −∞ 2 ∞ 2 − z2 / 2 2 ⎛ 1⎞ 2 ⎛1⎞ ⎛1⎞ z e dz = 1. Let y = z2. Then E ( Z 2 ) = Γ ⎜1 + ⎟ = ⎜ ⎟ Γ ⎜ ⎟ , which 0 ⎝ ⎠ 2 π π π ⎝2⎠ ⎝2⎠

Let Z be a standard normal random variable. Then E ( Z 2 ) = =

∫

∫

⎛1⎞ implies that Γ ⎜ ⎟ = π . ⎝2⎠

4.6.7

⎛ 7 ⎞ 5 ⎛ 5 ⎞ 5 3 ⎛ 3 ⎞ 5 3 1 ⎛ 1 ⎞ 15 ⎛ 1 ⎞ Γ⎜ ⎟ = Γ⎜ ⎟ = Γ = Γ = Γ by Theorem 4.6.2, part 2. ⎝ 2 ⎠ 2 ⎝ 2 ⎠ 2 2 ⎜⎝ 2 ⎟⎠ 2 2 2 ⎜⎝ 2 ⎟⎠ 8 ⎜⎝ 2 ⎟⎠ ⎛1⎞ Further, Γ ⎜ ⎟ = π by Question 4.6.6. ⎝2⎠

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92

4.6.8

Chapter 4: Special Distributions

m

E(Y ) = =

4.6.9

∫

∞

0

y ⋅

λr

m

(m + r − 1)! λ m (r − 1)!

∫

(r − 1)!

λ

∞

0

y

r −1 − λ y

e

m+ r

(m + r − 1)!

dy =

∫

∞

0

λr (r − 1)!

y m + r −1e − λ y dy =

y m + r −1e − λ y dy

(m + r − 1)! . λ m (r − 1)!

Write the gamma moment-generating function in the form MY(t) = (1 − t/λ)−r. Then M Y(1) (t ) = −r(1 − t/λ)−r−1(−1/λ) = (r/λ)(1 − t/λ)−r−1 and M Y(2) (t ) = (r/λ)(−r − 1)(1 − t/λ)r−2 ⋅ (−1/λ) r = (r/λ2)(r + 1)(1 − t/λ)−r−2. Therefore, E(Y) = M Y(1) (0) = and Var(Y) = M Y(2) (0) − ⎡⎣ M Y(1) (0)⎤⎦ 2

λ

=

r (r + 1)

λ

2

−

r

2

λ

2

=

r

λ2

.

d r (1 − t / λ )− r = r (1 − t / λ ) − r −1 (−1/ λ ) = (1 − t / λ )− r −1 and dt λ d r r r (r + 1) M Y(2) (t ) = (1 − t / λ ) − r −1 = (− r − 1)(1 − t / λ ) − r − 2 (−1/ λ ) = (1 − t / λ ) − r − 2 2 dt λ λ λ For an arbitrary integer m ≥ 2, we can generalize the above to see that r ( r + 1)...(r + m − 1) r (r + 1)...(r + m − 1) M Y( m ) (t ) = (1 − t / λ ) − r − m . Then E(Ym) = M Y( m ) (0) = . m m

4.6.10 MY(t) = (1 − t/λ)−r so M Y(1) (t ) =

λ

r (r + 1)...(r + m − 1)

λ

Γ ( r + m) But note that . The right hand side of the equation is equal to the = m λ Γ (r )λ m expression in Question 4.6.8 when r is an integer.

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5: Estimation Section 5.2: Estimating Parameters: The Method of Maximum Likelihood and Method of Moments 8

8

5.2.1

L(θ) =

∏θ

ki

8

∑ ki

8−

(1 − θ )1− ki = θ i=1 (1 − θ )

∑ ki i =1

= θ 5(1 − θ)3

i =1

dL(θ ) dL(θ ) = θ 53(1 − θ)2(−1) + 5θ 4(1 − θ)3 = θ 4(1 − θ)2(−8θ + 5) ⋅ = 0 implies θ e = 5/8 dθ dθ

5.2.2

L(p) = p(1 − p)(1 − p)p(1 − p) = p2(1 − p)3 2 3 5 8 1 ⎛1⎞ ⎛ 2⎞ ⎛1⎞ is greater than L(1/2) = ⎜ ⎟ = , so pe = 1/3. L(1/3) = ⎜ ⎟ ⎜ ⎟ = ⎝3⎠ ⎝ 3⎠ ⎝2⎠ 243 32 4

5.2.3

L(θ) =

∏ λe

− λ yi

=λ e 4

−λ

4

∑ yi

= λ4e−32.8λ⋅

i =1

i =1

dL( λ ) = λ4(−32.8)e−32.8λ + 4λ3e−32.8λ = λ3e−32.8λ(4 − 32.8λ) dλ dL( λ ) = 0 implies λe = 4/32.8 = 0.122 dλ n

n

5.2.4

L(θ) =

∏

θ 2ki e −θ

2

ki !

i =1

=

θ

2

∑ ki i =1

e − nθ

2

⋅

n

∏k ! i

i =1

⎛ n ⎞ ln L(θ) = ⎜ 2 ki ⎟ (ln θ) − nθ 2 + ln ⎝ i =1 ⎠

∑

n

∏k ! i

i =1

n

2

d ln L(θ ) = 0 implies dθ

3

5.2.5

L(θ) =

∏ i =1

yi3e− yi / θ = 6θ 4 3

ln L(θ) = ln

∏

yi3 −

i =1

∑

n

− 2nθ =

i =1

θ

⎛ ⎜ ⎝

1

θ

∑

2

ki

i =1

n

ki − 2nθ 2

θ

= 0 or θ e =

∑k

i

i =1

n

3

3

∏ i =1

⎞ − ∑ yi / θ yi3 ⎟ e i=1 ⎠ ⋅ 216θ 12

3

∑y

i

− ln 216 − 12 ln θ

i =1

3

d ln L(θ ) 1 = 2 dθ θ

3

∑ i =1

yi −

12

θ

∑ y − 12θ i

=

i =1

θ2

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94

Chapter 5: Estimation 3

d ln L(θ ) = 0 implies dθ 4

5.2.6

L(θ) =

∏2

θ

e

yi

i =1

−θ yi

∑ y − 12θ i

i =1

θ

2

8.8 − 12θ

θ2

−θ

θ4

=

=

4

∏

4

∑

yi

i =1

e

16

= 0 or θ e = 0.733

yi

i =1

4 ⎛ ln L(θ) = 4ln θ − ln ⎜16 ⎝ i =1

∏

d ln L(θ ) 4 4 = − y ⋅ dθ θ i =1 i d ln L(θ ) = 0 implies θ e = dθ

⎞ yi ⎟ − θ ⎠

4

∑

yi

i =1

∑

4

=

4

∑

yi

4 = 0.456 8.766

i =1

5

5.2.7

L(θ) =

∏θ y

θ −1

i

i =1

⎛ =θ ⎜ ⎝ 5

5

∏ i =1

θ −1

⎞ yi ⎟ ⎠

.

5

ln L(θ) = 5 ln θ + (θ − 1)

∑ ln y

i

i =1

5 +θ

5

∑

ln yi d ln L(θ ) 5 5 i =1 = + ln yi = dθ θ i =1 θ d ln L(θ ) 5 − 0.625θ = 0 implies = 0 or θ e = 8.00 θ dθ

∑

n

5.2.8

L( p ) =

n

∏ (1 − p)

ki −1

∑ ki − n

p = (1 − p ) i=1

pn

i =1

⎛ n ⎞ ln L( p ) = ⎜ ki − n ⎟ ln (1 − p ) + n ln p ⎝ i =1 ⎠

∑

n

∑

ki − n ln L( p) n ln L( p) i =1 and = 0 implies pe = =− + 1− p dp dp p

n n

∑k

i

i =1

n

For the data, n = 1011, and

∑k

i

= 1(678) + 2(227) + 3(56) + 4(28) + 5(8) + 6(14) = 1536, so

i =1

pe =

1011 = 0.658. The table gives the comparison of observed and expected frequencies. 1536

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Section 5.2: Estimating Parameters

95

No. of Occupants 1 2 3 4 5 6+

5.2.9

Observed Frequency 678 227 56 28 8 14

Expected Frequency 665.2 227.5 77.8 26.6 9.1 4.8

(a) From the Comment following Example 5.2.1, 1 n 1 λe = ki = [1(19) + 2(12) + 3(13) + 4(9)] = 2.00 . n i =1 59

∑

23 = 10.6 3! The full set of expected values is given in column of the following table.

(b) For example, the expected frequency for the k = 3 class is 59 ⋅ e−2

No. of No-hitters 0 1 2 3 4+

Observed Frequency 6 19 12 13 9

Expected Frequency 8.0 16.0 16.0 10.6 8.4

The last expected frequency has been chosen to make that column sum to n = 59. The techniques to be introduced in Chapter 10 would support the Poisson model. The difference between observed and expected frequencies is in part because the game of baseball changed significantly over the years from 1950 to 2008. Thus, the parameter λ would not be constant over this period. n

⎛1⎞ 5.2.10 (a) L(θ) = ⎜ ⎟ , if 0 ≤ y1, y2, …, yn ≤ θ, and 0 otherwise. Thus θ e = ymax, which for these ⎝θ ⎠ data is 14.2. n

⎛ 1 ⎞ , if θ1 ≤ y1, y2, …, yn ≤ θ2, and 0 otherwise. Thus θ1e = ymin and (b) L(θ) = ⎜ ⎝ θ 2 − θ1 ⎟⎠

θ 2e = ymax. For these data, θ1e = 1.8, θ 2e = 14.2. n

∏

64

yi 2 yi i =1 , if θ ≤ y1, y2, …, yn ≤ 1 and 0 otherwise. If θ > ymin, then L(θ) = 0. 5.2.11 L(θ) = = 2 (1 − θ 2 )6 i =1 1 − θ So θ e ≤ ymin. Also, to maximize L(θ), minimize the denominator, which in turn means maximize θ. Thus θ e ≥ ymin. We conclude that θ e = ymin, which for these data is 0.92. 6

∏

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96

Chapter 5: Estimation

⎛ n ⎞ n 2 yi ⎟ θ −2 n , if 0 ≤ y1, y2, …, yn ≤ θ, and 0 otherwise. To maximize L(θ) = 2 ⎜ ⎝ i =1 ⎠ i =1 maximize θ. Since each yi ≤ θ for 1 ≤ i ≤ n, the maximum value for θ under these constraints is the maximum of the yi, or θ e = ymax. n

5.2.12 L(θ) =

2 yi

∏θ

∏

⎛1⎞ θk ⎜ ⎟ 5.2.13 L(θ) = ⎝ yi ⎠ i =1 25

∏

θ +1

θ

25 25θ

=θ k

⎛ ⎜1/ ⎝

θ +1

⎞ yi ⎟ ⎠

25

∏ i =1

, yi ≥ k, 1 ≤ i ≤ 25

n

ln L(θ) = 25 ln θ + 25θ ln k − (θ + 1)

∑ ln y

1

i =1

d ln L(θ ) 25 = + 25ln k − dθ θ

n

∑ ln y , so θ

e

i

25

=

i =1

−25ln k +

n

∑ ln y

i

i =1

n

5.2.14 (a) L(α,β) =

∏ αβ y β

β

−1 −α yi

e

i

i =1

⎛ = α nβ n ⎜ ⎝

⎞ yi ⎟ ⎠

n

∏ i =1

⎛ In L(α,β) = n ln α + n ln β + ( β − 1) ln ⎜ ⎝ n ∂ ln L(α , β ) n = − yβ ∂α α i =1 i ∂ ln L(α , β ) Setting = 0 gives αe = ∂α

β −1

n

∏ i =1

−α

n

∑ yiβ i =1

e

⎞ yi ⎟ − α ⎠

n

∑ yβ i

i =1

∑

n n

∑ yβ i

i =1

(b) The other equation is

∂ ln L(α , β ) n = + ln ∂β β

n

n

∏ y − α β ∑ yβ i

i =1

i =1

−1

i

=0

∂ ln L(α , β ) = 0 provides the other equation. ∂α Solving the two simultaneously would be done by numerical methods.

Setting

n

5.2.15 Let θ = σ2, so L(θ) =

1

∏

2πθ

i =1

e

−

1 ( yi − µ )2 θ 2

n n 11 ln L(θ) = − ln 2π − ln θ − 2 2 2θ d ln L(θ ) n 1 1 =− + 2θ 2 θ 2 dθ

Setting

= 2π − n / 2θ − n / 2 e

−

11 2θ

n

∑ ( yi − µ )2 i =1

n

∑( y − µ)

2

i

i =1

n

1 ( yi − µ )2 = 2 i =1

∑

d ln L(θ ) 1 = 0 gives θ e = σ e2 = dθ n

− nθ +

n

∑( y − µ)

2

i

i =1

θ2

n

∑( y − µ)

2

i

i =1

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 5.2: Estimating Parameters

5.2.16

E (Y ) =

∫

θ

0

97

2 y3 y 2 dy = 2 3θ θ 2y

θ

0

2 2 3 = θ . Setting θ = y gives θ e = y = 75. The maximum 3 3 2

likelihood estimate is ymax = 92. 5.2.17 E(Y) =

θe =

∫

1

y (θ 2 + θ ) yθ −1 (1 − y )dy = (θ 2 + θ )

0

∫

1 0

yθ (1 − y ) dy =

θ θ +2

. Set

θ θ +2

= y , which yields

2y 1− y

5.2.18 For Y Poisson, E(Y) = λ. Then λe = y = 13/6. The maximum likelihood estimate is the same. 5.2.19 For Y exponential, E(Y) = 1/λ. Then 1/λ = y implies λe = 1/ y . 5.2.20 E(Y) = θ1 so θ1e = y . 2

E(Y ) =

θ1 +θ 2

∫θ

1 −θ 2

θ +θ

1 2 1 1 ⎡ y3 ⎤ 1 y dy = = θ12 + θ 22 ⎢ ⎥ 2θ 2 2θ 2 ⎣ 3 ⎦θ −θ 3 1 2

2

n

1 1 Substitute θ1e = y into the equation θ12 + θ 22 = 3 n

5.2.21 E(Y) = Setting

∫

∞ k

⎛1⎞ yθ k ⎜ ⎟ ⎝ yi ⎠

θ +1

θ

dy = θ k θ

∫

∞ k

y −θ dy =

∑ i =1

⎛1 yi2 to obtain θ 2e = 3 ⎜ ⎝n

n

∑y

2 i

i =1

⎞ − y2 ⎟ . ⎠

θk θ −1

θk = y gives θ e = y /( y − k ) θ −1

5.2.22 E(X) = 0 ⋅ θ0(1 − θ)1−0 + 1 ⋅ θ1(1 − θ)1−1 = θ. Then θ e = y , which for the given data is 2/5. 5.2.23 E(Y) = µ, so µe = y . E(Y2) = σ2 + µ2. Then substitute µe = y into the equation for E(Y2) to obtain σ e2 + y 2 =

1 n

n

∑y

2 i

or σ e2 =

i =1

1 n

n

∑y

2 i

i =1

5.2.24 From Theorem 4.5.1, E(X) = r/p. Var(X) = Then E(X 2) = Var(X) + E(X)2 = Set

− y2

r (1 − p) . p2

r (1 − p) + r 2 p2

r r (1 − p) + r 2 1 = x to obtain r = px , and substitute into the equation = p n p2 px (1 − p ) + ( px ) 1 = 2 n p 2

to obtain an equation in p:

n

∑x

2 i

.

i =1

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n

∑x

2 i

i =1

98

Chapter 5: Estimation

Equivalently, px − p 2 x + p 2 x 2 − p 2

Then re =

x2 1 x+ n

n

∑

xi2

1 n

n

∑x

= 0. Solving for p gives pe =

2 i

i =1

x 1 x+ n

.

n

∑

xi2

−x

2

i =1

. −x

2

i =1

5.2.25 From Chapter 4, E(X) = 1/p. Setting 1/p = x , gives pe =

1 . For the given data, pe = 0.479. x

The expected frequencies are: No. of clusters/song 1 2 3 4 5 6 7 8

Observed frequency 132 52 34 9 7 5 5 6

Expected frequency 119.8 62.4 32.5 16.9 8.8 4.6 2.4 2.6

The last expected frequency has been chosen to make that column sum to n = 59. 5.2.26 Var(Y ) = σˆ 2 implies E (Y 2 ) − E (Y )2 =

1 n

n

∑y

2 i

− y 2 . However by the first given equation,

i =1

E (Y ) = y . Removing these equal terms from the equation above gives the second equation of 2

2

Definition 5.2.3, or E (Y 2 ) =

1 n

n

∑y

2 i

.

i =1

Section 5.3: Interval Estimation 5.3.1

⎛ σ σ ⎞ The confidence interval is ⎜ y − zα / 2 , y + zα / 2 ⎟ ⎝ n n⎠ ⎛ 15 15 ⎞ = ⎜107.9 − 1.96 , 107.9 + 1.96 ⎟ = (103.7, 112.1). ⎝ 50 50 ⎠

5.3.2

⎛ σ σ ⎞ ⎛ 0.09 0.09 ⎞ The confidence interval is ⎜ y − zα / 2 = ⎜ 0.766 − 1.96 , y + zα / 2 ,0.766 + 1.96 ⎟ ⎟ ⎝ n n⎠ ⎝ 19 19 ⎠ = (0.726, 0.806). The value of 0.80 is believable.

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 5.3: Interval Estimation

5.3.3

5.3.4

99

⎛ σ σ ⎞ ⎛ 8.0 8.0 ⎞ = ⎜ 70.833 − 1.96 The confidence interval is ⎜ y − zα / 2 , y + zα / 2 , 70.833 + 1.96 ⎟ ⎟ ⎝ 6 6⎠ n n⎠ ⎝ = (64.432, 77.234). Since 80 does not fall within the confidence interval, that men and women metabolize methylmercury at the same rate is not believable. ⎛ σ σ ⎞ The confidence interval is ⎜ y − zα / 2 , y + zα / 2 ⎟ ⎝ n n⎠ ⎛ 40.7 40.7 ⎞ = ⎜188.4 − 1.96 ,188.4 + 1.96 ⎟ = (175.46, 201.34). Since 192 does fall in the ⎝ 38 38 ⎠ confidence interval, there is doubt the diet has an effect.

5.3.5

The length of the confidence interval is 2 zα / 2

σ n

=

2(1.96)(14.3) 56.056 56.056 = . For ≤ 3.06, n n n

2

⎛ 56.056 ⎞ = 335.58, so take n = 336. n≥ ⎜ ⎝ 3.06 ⎟⎠

5.3.6

(a) P(−1.64 < Z < 2.33) = 0.94, a 94% confidence level. (b) P(−∞ < Z < 2.58) = 0.995, a 99.5% confidence level. (c) P(−1.64 < Z < 0) = 0.45, a 45% confidence level.

5.3.7

The probability that the given interval will contain µ is P(−0.96 < Z < 1.06) = 0.6869. The probability of four or five such intervals is binomial with n = 5 and p = 0.6869, so the probability is 5(0.6869)4(0.3131) + (0.6869)5 = 0.501.

5.3.8

The given interval is symmetric about y .

5.3.9

The interval given is correctly calculated. However, the data do not appear to be normal, so claiming that it is a 95% confidence interval would not be correct.

⎛ 192 (192 / 540)(1 − 192 / 540) 192 (192 / 540)(1 − 192 / 540) ⎞ − 1.96 , + 1.96 5.3.10 ⎜ ⎟ = (0.316, 0.396) 540 540 540 ⎝ 540 ⎠

5.3.11 Let p be the probability that a viewer would watch less than a quarter of the advertisements during Super Bowl XXIX. The confidence interval for p is ⎛ 281 (281/1015)(1 − 281/1015) 281 (281/1015)(1 − 281/1015) ⎞ , + 1.64 ⎜ 1015 − 1.64 ⎟ 1015 1015 1015 ⎝ ⎠ = (0.254, 0.300) 5.3.12 Budweiser would use the sample proportion 0.54 alone as the estimate. Schlitz would construct the 95% confidence interval (0.36, 0.56) to claim that values < 0.50 are believable.

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100

Chapter 5: Estimation

5.3.13 In closest integer to 0.63(2253) is 1419. This gives the confidence interval ⎛ 1419 (1419 / 2253)(1 − 1419 / 2253) 1419 (1419 / 2253)(1 − 1419 / 2253) ⎞ , + 1.96 ⎜ 2253 − 1.96 ⎟ 2253 2253 2253 ⎝ ⎠ = (0.610, 0.650). Since 0.54 is not in the interval, the increase can be considered significant. 5.3.14

k (k / n)(1 − k / n) = 0.57 − 0.67 n n k (k / n)(1 − k / n) + 0.67 = 0.63 n n

Adding the two equations gives 2

k k = 1.20 or = 0.60 n n

k (0.60)(1 − 0.60) into the first equation above gives 0.60 − 0.67 n n = 0.57. Solving this equation for n gives n = 120.

Substituting the value for

5.3.15 2.58

p(1 − p) 1 (2.58) 2 ≤ 2.58 ≤ 0.01, so take n ≥ = 16,641 n 4n 4(0.01) 2

5.3.16 For Foley to win the election, he needed to win at least 8088 of the absentee votes, since 8088 > 8086 = 2174 + (14,000 − 8088). If X is the number of absentee votes for Foley, then it is binomial with n = 14,000 and p to be determined. ⎛ X − 14,000 p ⎛ 8088 − 14,000 p ⎞ 8088 − 14,000 p ⎞ P(X ≥ 8088) = P ⎜ ≥ ⎟ = P ⎜Z ≥ ⎟. 14,000 p(1 − p) ⎠ 14,000 p (1 − p ) ⎠ ⎝ 14,000 p (1 − p ) ⎝ 8088 − 14,000 p For this probability to be 0.20, = z.20 = 0.84. This last equation can be solved 14,000 p (1 − p ) by the quadratic formula or trial and error to obtain the approximate solution p = 0.5742. 5.3.17 Both intervals have confidence level approximately 50%. 5.3.18 g(p) = p − p2. g′(p) = 1 − 2p. Setting g′(p) = 0 gives p = 1/2. Also, g′′(p) = −2. Since the second derivative is negative at p = 1/2, a maximum occurs there. The maximum value of g(p) is g(1/2) = 1/4. 1.96 = 0.031. The number of in-favor responses was the closest integer 2 998 to 0.59(998) or 589. The 95% confidence interval is ⎛ 589 (589 / 998)(1 − 589 / 998) 589 (589 / 998)(1 − 589 / 998) ⎞ , + 1.96 ⎜ 998 − 1.96 ⎟ 998 998 998 ⎝ ⎠ = (0.559, 0.621)

5.3.19 The margin of error is

1.96 = 0.069. The sample proportion is 86/202 = 0.426. The 2 202 largest believable value is 0.426 + 0.069 = 0.495, so we should not accept the notion that the true proportion is as high as 50%.

5.3.20 From Definition 5.3.1, d =

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Section 5.3: Interval Estimation

101

5.3.21 If X is hypergeometric, then Var(X/n) =

p (1 − p ) N − n . n N −1

As before p(1 − p) ≤ 1/4. Thus, in Definition 5.3.1, substitute d =

1.96 2 n

N −n . N −1

5.3.22 (a) The 90% confidence interval is ⎛ 126 (126 / 350)(1 − 126 / 350) 126 (126 / 350)(1 − 126 / 350) ⎞ , + 1.64 ⎜ 350 − 1.64 ⎟ 350 350 350 ⎝ ⎠ = (0.318, 0.402) (126 / 350)(1 − 126 / 350) 3000 − 350 350 3000 − 1 = 0.039, which gives a confidence interval (0.321, 0.399)

(b) Use for the margin of error 1.64

5.3.23 If n is such that 0.06 = If n is such that 0.03 =

1.96 1.962 , then n is the smallest integer ≥ = 266.8 . Take n = 267. 4(0.06) 2 2 n 1.96 1.962 , then n is the smallest integer ≥ = 1067.1. Take n = 1068. 4(0.03) 2 2 n

5.3.24 For candidate A, the believable values for the probability of winning fall in the range (0.52 − 0.05, 0.52 + 0.05) = (0.47, 0.57). For candidate B, the believable values for the probability of winning fall in the range (0.48 − 0.05, 0.48 + 0.05) = (0.43, 0.53). Since 0.50 falls in both intervals, there is a sense in which the candidates can be considered tied. 5.3.25 Case 1: n is the smallest integer greater than 2 z.02 2.052 = 420.25, so take n = 421. = 4(0.05) 2 4(0.05)2 Case 2: n is the smallest integer greater than 2 z.04 1.752 = 478.5, so take n = 479. = 4(0.04) 2 4(0.04) 2 5.3.26 Take n to be the smallest integer ≥

2 z.005 p (1 − p ) 2.582 (0.40)(0.60) = 639.01, = (0.05)2 (0.05)2

so n = 640. 5.3.27 Take n to be the smallest integer ≥

2 z.10 1.282 = 1024. = 4(0.02) 2 4(0.02) 2

2 z.075 1.442 5.3.28 (a) Take n to be the smallest integer ≥ = 576. = 4(0.03)2 4(0.03)2

(b) Take n to be the smallest integer ≥

2 z.075 p (1 − p ) 1.442 (0.10)(0.90) = 207.36, so let n = 208. = (0.03) 2 (0.03) 2

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102

Chapter 5: Estimation

Section 5.4: Properties of Estimators 5.4.1

P(| θˆ − 3| > 1.0) = P( θˆ < 2) + P( θˆ > 4) = P (θˆ = 1.5) + P (θˆ = 4.5) = P ((1, 2)) + P ((4,5)) = 2 /10

5.4.2

For the uniform variable Y, FY ( y ) =

5.4.3

5.4.4

5.4.5

5.4.6

y

θ

θn

, 0 ≤ y ≤θ .

2.86 = 1 − 0.661 = 0.339 36

(b) For n = 3and θ = 3, P (| θˆ − 3 | < 0.2) = FYmax (3) − FYmax (2.8) = 1 −

2.83 = 1 − 0.813 = 0.187 33

⎛ X − 500(0.52) 250 − 500(0.52) ⎞ P(X < 250) = P ⎜ < ⎟ = P(Z < −0.90) = 0.1841 500(0.52)(0.48) ⎠ ⎝ 500(0.52)(0.48) ⎛ 19.0 − 20 21.0 − 20 ⎞ P(19.0 < Y < 21.0) = P ⎜ 2 λ ⎟ ≥ P (Y1 > 2 λ ) = e−2 λ . The proof now proceeds along the lines of Part (a). ⎝ i =1 ⎠ n

∑

5.7.4

(a) Let µn = E (θˆn ). E ⎡⎣(θˆn − θ )2 ⎤⎦ = E ⎣⎡ (θˆn − µn + µn − θ ) 2 ⎦⎤ = E ⎡⎣(θˆn − µn ) 2 + ( µn − θ ) 2 + 2(θˆn − µn )( µn − θ )⎤⎦ = E[(θˆ − µ ) 2 ] + E[( µ − θ ) 2 ] + 2( µ − θ ) E[(θˆ − µ )] n

n

n

2⎤

n

n

n

= E ⎡⎣(θˆn − µn ) ⎦ + ( µn − θ ) + 0 or E ⎡⎣(θˆn − θ )2 ⎤⎦ = E ⎡⎣ (θˆn − µn ) 2 ⎤⎦ + ( µn − θ ) 2 2

The left hand side of the equation tends to 0 by the squared-error consistency hypothesis. Since the two summands on the right hand side are non-negative, each of them must tend to zero also. Thus, lim (µn − θ)2 = 0, which implies lim (µn − θ) = 0, or lim µn = θ. n →∞

n →∞

n →∞

(b) By Part (a) lim µn − θ = 0. n →∞

( ) ( Var(θˆ ) by Chebyshev’s Inequality. ≥ε)≤ ε

For any ε > 0, lim P θˆn − θ ≥ ε = lim P (θˆn − µn ) − ( µn − θ ) ≥ ε n →∞

(

= lim P θˆn − µn n →∞

But

Var(θˆn )

ε2

=

n →∞

n

2

E[(θˆn − µn )2 ]

ε2

(

)

, and by Part (a), lim E ⎡⎣(θˆn − µn ) 2 ⎤⎦ = 0, n →∞

)

so lim P θˆn − θ ≥ ε = 0. Thus, θˆn is consistent. n →∞

5.7.5

E ⎣⎡(Ymax − θ ) 2 ⎤⎦ =

=

n

θn

∫

θ 0

∫

θ 0

n ⎛ y⎞ (y −θ ) ⎜ ⎟ θ ⎝θ ⎠ 2

( y n +1 − 2θ y n + θ 2 y n −1 )dy =

n −1

dy n ⎛ θ n + 2 2θ n + 2 θ n + 2 ⎞ ⎛ n 2n ⎞ − + = ⎜ − + 1⎟ θ 2 n ⎜ ⎟ n ⎠ ⎝ n + 2 n +1 ⎠ θ ⎝ n + 2 n +1

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Section 5.8: Bayesian Estimation

111

2n ⎛ n ⎞ Then lim E ⎡⎣(Ymax − θ ) 2 ⎤⎦ = lim ⎜ − + 1⎟ θ 2 = 0 and the estimator is squared error n →∞ n →∞ ⎝ n + 2 n +1 ⎠ consistent.

5.7.6

Because the symmetry of the pdf, the mean of the sample median is µ. Chebyshev’s inequality Var(Yn′+1 ) applies and P ( Yn′+1 − µ < ε ) > 1 − . 2

ε

(

)

1 =0, so lim P Yn' +1 − µ < ε = 1, and Yn+1 is consistent n →∞ n →∞ 8[ f ( µ ; µ )]2 n Y

Now, lim Var(Yn′+1 ) = lim n →∞

for µ.

Section 5.8: Bayesian Estimation 5.8.1

The numerator of gΘ(θ | X = k) is Γ (r + s ) r −1 Γ (r + s ) r pX(k|θ)fΘ(θ) = [(1 − θ ) k −1θ ] θ (1 − θ ) s −1 = θ (1 − θ ) s + k − 2 Γ ( r )Γ ( s ) Γ ( r )Γ ( s ) The term θ r (1 − θ ) s + k − 2 is the variable part of the beta distribution with parameters r + 1 and s + k -1, so that is the pdf of gΘ(θ| X = k).

5.8.2

The Bayes estimate is the mean of the posterior distribution, a beta pdf with a parameters k + 4 k+4 k+4 = . and n − k + 102. The mean of this pdf is k + 4 + n − k + 102 n + 106 k +4 n ⎛k ⎞ 106 ⎛ 4 ⎞ Note we can write = ⎜ ⎟+ ⎜ ⎟ n + 106 n + 106 ⎝ n ⎠ n + 106 ⎝ 106 ⎠

5.8.3

(a) Following the pattern of Example 5.8.2, we can see that the posterior distribution is a beta pdf with parameters k + 135 and n − k + 135. k + 135 k + 135 = , (b) The mean of the Bayes pdf given in part (a) is k + 135 + n − k + 135 n + 270 Note we can write

5.8.4 5.8.5

k + 135 n ⎛k ⎞ 270 ⎛ 135 ⎞ n ⎛k⎞ 270 ⎛ 1 ⎞ = ⎜⎝ ⎠⎟ + ⎜ ⎟= ⎜ ⎠⎟ + ⎜ ⎟ ⎝ ⎠ ⎝ n + 270 n + 270 n n + 270 270 n + 270 n n + 270 ⎝ 2 ⎠

k +1 n ⎛k ⎞ 2 ⎛1⎞ = ⎜⎝ ⎟⎠ + ⎜ ⎟ n+2 n+2 n n+ 2 ⎝2⎠

In each case the estimator is biased, since the mean of the estimator is a weighted average of the unbiased maximum likelihood estimator and a non-zero constant. However, in each case, the weighting on the maximum likelihood estimator tends to 1 as n tends to ∞, so these estimators are asymptotically unbiased.

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112

5.8.6

Chapter 5: Estimation

The numerator of gΘ(θ| X = k) is fY(y| θ)fΘ(θ) =

θr Γ (r )

y r −1e −θ y

µs Γ ( s)

θ s −1e − µθ =

µ s y r −1 Γ ( r )Γ ( s )

θ r + s −1e − ( y + µ )θ

We recognize the part involving θ as the variable part of the gamma distribution with parameters r + s and y + µ, so that is gΘ (θ| X = k). 5.8.7

Since the sum of gamma random variables is gamma, then W is gamma with parameters nr and λ. n

Then gΘ(θ| W = k) is a gamma pdf with parameters nr + s and

∑y +µ . i

i =1

5.8.8

The Bayes estimate is the mean of the posterior pdf, which in this case is

nr + s n

∑y +µ i

i =1

5.8.9

⎛ n ⎞ Γ (r + s ) k + r −1 pX(k| θ)fΘ(θ) = ⎜ ⎟ θ (1 − θ ) n − k + s −1 , ⎝ k ⎠ Γ ( r )Γ ( s ) ⎛ n ⎞ Γ (r + s ) so pX(k| θ) = ⎜ ⎟ ⎝ k ⎠ Γ ( r )Γ ( s )

1

∫θ 0

k + r −1

(1 − θ )n − k + s −1dθ ,

⎛ n ⎞ Γ ( r + s ) Γ ( k + r )Γ ( n − k + s ) n! (r + s − 1)! (k + r − 1)!(n − k + s − 1)! = ⎜ ⎟ = (n + r + s − 1)! Γ (n + r + s ) k !( n − k )! (r − 1)!( s − 1)! ⎝ k ⎠ Γ ( r )Γ ( s )

=

(k + r − 1)! ( n − k + s − 1)! n !(r + s − 1)! k !( r − 1)! (n − k )!( s − 1)! (n + r + s − 1)!

⎛ k + r − 1⎞ ⎛ n − k + s − 1⎞ = ⎜ ⎝ k ⎟⎠ ⎜⎝ n − k ⎟⎠

⎛ n + r + s − 1⎞ ⎜⎝ ⎟⎠ n

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 6: Hypothesis Testing Section 6.2: The Decision Rule 6.2.1

(a) Reject H0 if

y − 120 ≤ −1.41; z = −1.61; reject H0. 18 / 25

(b) Reject H0 if

y − 42.9 is either ≤ −2.58 or ≥ 2.58; z = 2.75; reject H0. 3.2 / 16

(c) Reject H0 if

y − 14.2 ≥ 1.13; z = 1.17; reject H0. 4.1/ 9

6.2.2

Let µ = true average IQ of students after drinking Brain-Blaster. To test H0: µ = 95 versus H1: µ ≠ 95 at the α = 0.06 level of significance, the null hypothesis should be rejected if y − 95 is either ≤ −1.88 or ≥ 1.88. Equivalently, H0 will be rejected if y is either z= 15 / 22 15 15 1) ≤ 95 − (1.88) = 89.0 or 2) ≥ 95 + (1.88) = 101.0. 22 22

6.2.3

(a) No, because the observed z could fall between the 0.05 and 0.01 cutoffs. (b) Yes. If the observed z exceeded the 0.01 cutoff, it would necessarily exceed the 0.05 cutoff.

6.2.4

Assuming there is no reason to suspect that the polymer would shorten a tire’s lifetime, the alternative hypothesis should be H1: µ > 32,500. At the α = 0.05 level, H0 should be rejected if 33,800 − 32,500 = 1.26, implying that the observed the test statistic exceeds z.05 = 1.64. But z = 4000 / 15 mileage increase is not statistically significant.

6.2.5

No, because two-sided cutoffs (for a given α) are further away from 0 than one-sided cutoffs. ⎛ 29.9 − 30 Y − 30 30.1 − 30 ⎞ By definition, α = P(29.9 ≤ Y ≤ 30.1| H0 is true) = P ⎜ = ≤ ≤ ⎝ 6.0 / 16 6.0 / 16 6.0 / 16 ⎠⎟ P(−0.07 ≤ Z ≤ 0.07) = 0.056. The interval (29.9, 30.1) is a poor choice for C because it rejects H0 for the y -values that are most compatible with H0 (that is, closest to µ0 = 30). Since the alternative is two-sided, H0 should be rejected if y is either 6.0 6.0 = 27.1 or 2) ≥ 30 + 1.91 ⋅ = 32.9. 1) ≤ 30 − 1.91 ⋅ 16 16

6.2.6

6.2.7

y − 12.6 is either ≤ −1.96 or ≥ 1.96. But y = 12.76 and 0.4 / 30 z = 2.19, suggesting that the machine should be readjusted. (b) The test assumes that the yi’s constitute a random sample from a normal distribution. Graphed, a histogram of the 30 yi’s shows a mostly bell-shaped pattern. There is no reason to suspect that the normality assumption is not being met.

(a) H0 should be rejected if

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114

Chapter 6: Hypothesis Testing

6.2.8

(a) Obs. z = −1.61, so P-value = P(Z ≤ −1.61) = 0.0537. (b) Obs. z = 2.75, so P-value = P(Z ≤ −2.75) + P(Z ≥ 2.75) = 0.0030 + 0.0030 = 0.0060. (c) Obs. z = 1.17, so P-value = P(Z ≥ 1.17) = 0.1210. Yes, all the P-values agree with the decisions reached in Question 6.2.1.

6.2.9

P-value = P(Z ≤ −0.92) + P(Z ≥ 0.92) = 0.3576; H0 would be rejected if α had been set at any value greater than or equal to 0.3576.

6.2.10 Let µ = true average blood pressure when taking statistics exams. Test H0: µ = 120 versus 125.2 − 120 H1: µ > 120. Given that σ = 12, n = 50 and y = 125.2, z = = 3.06. The 12 / 50 corresponding P-value is approximately 0.001 (= P(Z ≥ 3.06)), so H0 would be rejected for any usual choice of α. y − 145.75 is either ≤ −1.96 or ≥ 1.96. Here, y = 149.75 and 9.50 / 25 z = 2.10, so the difference between $145.75 and $149.75 is statistically significant.

6.2.11 H0 should be rejected if

Section 6.3: Testing Binomial Data—H0: p = po 6.3.1

(a) Given that the technique worked k = 24 times during the n = 52 occasions it was tried, 24 − 52(0.40) z= = 0.91. The latter is not larger than z.05 = 1.64, so H0: p = 0.40 would 52(0.40)(0.60) not be rejected at the α = 0.05 level. These data do not provide convincing evidence that transmitting predator sounds helps to reduce the number of whales in fishing waters. (b) P-value = P(Z ≥ 0.91) = 0.1814; H0 would be rejected for any α ≥ 0.1814.

6.3.2

Let p = P(A/HeJ mouse is right-pawed). Test H0: p = 0.67 versus H1: p ≠ 0.67. For α = 0.05, H0 should be rejected if z is either ≤ −1.96 or ≥ 1.96. Here, n = 35 and 18 − 35(0.67) = −1.96, implying that H0 k = number of right-pawed HeJ mice = 18, so z = 35(0.67)(0.33) should be rejected.

6.3.3

Let p = P(current supporter is male). Test H0: p = 0.65 versus H1: p < 0.65. Since n = 120 and 72 − 120(0.65) k = number of male supporters = 72, z = = −1.15, which is not less than or 120(0.65)(0.35) equal to −z.05 (= −1.64), so H0: p = 0.65 would not be rejected.

6.3.4

The null hypothesis would be rejected if z = happen, k ≥ 200(0.45) + 1.08 ⋅

k − 200(0.45) ≥ 1.08 (= z.14). For that to 200(0.45)(0.55)

200(0.45)(0.55) = 98.

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 6.4: Type I & Type II Errors

6.3.5

6.3.6

115

1 1 versus H1: p ≠ . Given that k = 26 and n = 60, 2 2 P-value = P(X ≤ 26) + P(X ≥ 34) = 0.3030. 1 1 versus H1: p < . Let p = P(person dies if month preceding birthmonth). Test H0: p = 12 12 16 − 348(1/12) Given that α = 0.05, H0 should be rejected if z ≤ −1.64. In this case, z = 348(1/12)(11/12) = −2.52, which suggests that people do not necessarily die randomly with respect to the month in which they were born. More specifically, there appears to be a tendency to “postpone” dying until the next birthday has passed.

Let p = P(Yi ≤ 0.69315). Test H0: p =

6.3.7

Reject H0 if k ≥ 4 gives α = 0.50; reject H0 if k ≥ 5 gives α = 0.23; reject H0 if k ≥ 6 gives α = 0.06; reject H0 if k ≥ 7 gives α = 0.01.

6.3.8

Let A1 be the event that “k ≥ 8” is the rejection region being used, and let A2 be the event that “k = 9” is the rejection region being used. Define B to be the event that H0: p = 0.6 is rejected. From Theorem 2.4.1, P(b) = P(B|A1)P(A1) + P(B|A2)P(A2). But P(B|A1) = 0.060466 + 0.010078 = 0.070544 and P(B|A2) = 0.010078. If p denotes the probability that A1 occurs, 0.05 = desired α = 0.070544 ⋅ p + 0.010078 ⋅ (1 − p), which implies that p = 0.66. It follows that the probability of rejecting H0 will be 0.05 if the “k ≥ 8” decision rule is used 66% of the time and the “k = 9” decision rule is used the remaining 34% of the time. 3 ⎛7 ⎞ k 7−k = 0.07 (a) α = P(reject H0|H0 is true) = P(X ≤ 3|p = 0.75) = ⎜⎝ k ⎟⎠ (0.75) (0.25) k =0 (b) p P(X ≤ 3|p) 0.75 0.07 0.65 0.20 0.55 0.39 0.45 0.61 0.35 0.80 0.25 0.93 0.15 0.99

6.3.9

∑

Section 6.4: Type I and Type II Errors 6.4.1

As described in Example 6.2.1, H0: µ = 494 is to be tested against H1: µ ≠ 494 using ±1.96 as y − 494 y − 494 the α = 0.05 cutoffs. That is, H0 is rejected if ≤ −1.96 or if ≥ 1.96. 124 / 86 124 / 86 Equivalently, the null hypothesis is rejected if y ≤ 467.8 or if y ≥ 520.2. Therefore, 1 − β = P(reject H0| µ = 500) = P (Y ≤ 467.8 µ = 500) + P(Y ≥ 520.2 µ = 500) ⎛ ⎛ 467.8 − 500 ⎞ 520.2 − 500 ⎞ = P ⎜Z ≤ ⎟⎠ + P ⎜⎝ Z ≥ ⎟ = P(Z ≤ −2.41) + P(Z ≥ 1.51) ⎝ 124 / 86 124 / 86 ⎠ = 0.0080 + 0.0655 = 0.0735.

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116

Chapter 6: Hypothesis Testing

y − 25.0 ≥ 1.28 (= z.10). Solving for y shows that the decision 2.4 / 30 2.4 rule can be re-expressed as “Reject H0 if y ≥ 25.0 + 1.28 ⋅ = 25.561.” 30

6.4.2

When α = 0.10, H0 is rejected if

6.4.3

The null hypothesis in Question 6.2.2 is rejected if y is either ≤ 89.0 or ≥ 101.0. Suppose µ = 90. Since σ = 15 and n = 22, 1 − β = P (Y ≤ 89.0) + P (Y ≥ 101.0) 89.0 − 90 ⎞ 101.0 − 90 ⎞ ⎛ ⎛ = P ⎜Z ≤ ⎟ + P ⎝⎜ Z ≥ ⎟ = P(Z ≤ −0.31) + P(Z ≥ 3.44) = 0.3783 + 0.0003 ⎝ ⎠ 15 / 22 15 / 22 ⎠ = 0.3786.

6.4.4

For n = 16, σ = 4, and α = 0.05, H0: µ = 60 should be rejected in favor of a two-sided H1 if either 4 4 y ≤ 60 − 1.96 ⋅ = 58.04 or y ≥ 60 + 1.96 ⋅ = 61.96. Then, for arbitrary µ, 1 − β = 16 16 P (Y ≤ 58.04 | µ ) + P (Y ≥ 61.96 | µ ) . Selected values of (µ, 1 − β) that would lie on the power curve are listed in the accompanying table.

µ 56 57 58 59 60 61 62 63 64 6.4.5

1−β 0.9793 0.8508 0.5160 0.1700 0.05 (=α) 0.1700 0.5160 0.8508 0.9793

y − 240 50 ≤ −2.33 or, equivalently, if y ≤ 240 − 2.33 ⋅ = 216.7. 50 / 25 25 Suppose µ = 220. Then β = P(accept H0|H1 is true) = P (Y > 216.7 µ = 220)

H0 should be rejected if z =

⎛ 216.7 − 220 ⎞ = P ⎜Z > ⎟ = P(Z > −0.33) = 0.6293. ⎝ 50 / 25 ⎠

6.4.6

(a) In order for α to be 0.07, P(60 − y * ≤ Y ≤ 60 + y * |µ = 60) = 0.07. Equivalently, ⎛ 60 − y * −60 Y − 60 60 + y * −60 ⎞ = P( −0.75 y * ≤ Z ≤ 0.75 y *) = 0.07. P⎜ ≤ ≤ ⎝ 8.0 / 36 8.0 / 36 8.0 / 36 ⎟⎠ But P(−0.09 ≤ Z ≤ 0.09) = 0.07, so 0.75 y * = 0.09, which implies that y * = 0.12.

(b) 1 − β = P(reject H0|H1 is true) = P(59.88 ≤ Y ≤ 60.12|µ = 62) ⎛ 59.88 − 62 60.12 − 62 ⎞ = P⎜ ≤Z≤ ⎟ = P(−1.59 ≤ Z ≤ −1.41) = 0.0793 − 0.0559 ⎝ 8.0 / 36 8.0 / 36 ⎠ = 0.0234.

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 6.4: Type I & Type II Errors

117

(c) For α = 0.07, ±zα/2 = ±1.81 and H0 should be rejected if y is either 8.0 8.0 1) ≤ 60 − 1.81 ⋅ = 57.59 or 2) ≥ 60 + 1.81 ⋅ = 62.41. Suppose µ = 62. Then 36 36 1 − β = P( Y ≤ 57.59|µ = 62) + P( Y ≥ 62.41|µ = 62) = P(Z ≤ −3.31) + P(Z ≥ 0.31) = 0.0005 + 0.3783 = 0.3788. 6.4.7

For α = 0.10, H0: µ = 200 should be rejected if y ≤ 200 − 1.28 ⋅

15.0 . Also, n

⎛ 200 − 1.28 ⋅ 15.0 / n − 197 ⎞ ⎛ ⎞ 15.0 µ = 197 ⎟ = 0.75, so P ⎜ 1 − β = P ⎜Y ≤ 200 − 1.28 ⋅ ⎟ = 0.75. ⎝ ⎠ n 15.0 / n ⎝ ⎠

But P(Z ≤ 0.67) = 0.75, implying that

200 − 1.28 ⋅ 15.0 / n − 197

15.0 / n smallest n satisfying the conditions placed on α and 1 − β is 95.

6.4.8

= 0.67. It follows that the

4 = 8.83 or 45

If n = 45, H0 will be rejected when y is either (1) ≤ 10 − 1.96 ⋅

4 = 11.17. When µ = 12, β = P(accept Ho|H1 is true) 45 ⎛ 8.83 − 12 11.17 − 12 ⎞ = P(8.83 ≤ Y ≤ 11.17|µ = 12) = P ⎜ ≤Z≤ ⎟ = P(−5.32 ≤ Z ≤ −1.39) = 0.0823. ⎝ 4 / 45 4 / 45 ⎠

(2) ≥ 10 + 1.96 ⋅

It follows that a sample of size n = 45 is sufficient to keep β smaller than 0.20 when µ = 12. 6.4.9

Since H1 is one-sided, H0 is rejected when y ≥ 30 + zα ⋅

9 . Also, 1 − β = power 16

⎛ ⎛ ⎞ 30 + zα ⋅ 9 / 16 − 34 ⎞ 9 = P ⎜ Y ≥ 30 + zα ⋅ µ = 34 ⎟ = 0.85. Therefore, 1 − β = P ⎜ Z ≥ ⎟ ⎝ ⎠ 16 9 / 16 ⎝ ⎠

= 0.85. But P(Z ≥ −1.04) = 0.85, so

30 + zα ⋅ 9 / 16 − 34 9 / 16

= −1.04, implying that zα = 0.74.

Therefore, α = 0.23. 6.4.10 (a) P(Type I error) = P(reject H0|H0 is true) = P(Y ≥ 3.20|λ = 1) =

∫

∞ 3.20

4⎞ ⎛ (b) P(Type II error) = P(accept H0|H1 is true) = P ⎜ Y < 3.20 λ = ⎟ = ⎝ 3⎠ = 0.91.

e − y dy = 0.04.

∫

3.20 0

3 −3 y / 4 e dy = 4

∫

2.4 0

e − u du

6.4.11 In this context, α is the proportion of incorrect decisions made on innocent suspects—that is, 9 , or 0.064. Similarly, β is the proportion of incorrect decisions made on guilty suspects— 140 15 here, , or 0.107. A Type I error (convicting an innocent defendant) would be considered 140 more serious than a Type II error (acquitting a guilty defendant). Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

118

Chapter 6: Hypothesis Testing

6.4.12 Let X = number of white chips in sample. Then α = P(X ≥ 2|urn has 5 white and 5 red) ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛10 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛10 ⎞ 1 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = . When the urn is 60% white, ⎝ 2 ⎠ ⎝1 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 0 ⎠ ⎝ 3 ⎠ 2 ⎛ 6⎞ ⎛ 4⎞

⎛10 ⎞ ⎛ 6 ⎞ ⎛ 4 ⎞

⎛10 ⎞

1

β = P(X ≤ 1|urn has 6 white and 4 red) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = . When the urn is ⎝ 0 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ 3 ⎛ 7 ⎞ ⎛ 3⎞ 70% white, β = P(X ≤ 1|urn has 7 white and 3 red) = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 3⎠

⎛10 ⎞ ⎛ 7 ⎞ ⎛ 3 ⎞ ⎜⎝ 3 ⎟⎠ + ⎜⎝1 ⎟⎠ ⎜⎝ 2 ⎟⎠

⎛10 ⎞ 11 ⎜⎝ 3 ⎟⎠ = 60 .

4

5 ⎛ y⎞ 6.4.13 For a uniform pdf, fYmax ( y ) = ⎜ ⎟ , 0 ≤ y ≤ θ when n = 5. θ ⎝θ ⎠

Therefore, α = P(reject H0|H0 is true) = P(Ymax ≥ k|θ = 2) =

∫

2 k

4

5⎛ y⎞ k5 dy = 1 − . For α to be ⎜ ⎟ 2⎝2⎠ 32

0.05, k = 1.98. 6.4.14 Level of significance = α = P(reject H0|H0 is true) = P(Y ≥ 0.90|fY(y) = 2y, 0 ≤ y ≤ 1) =

∫

1 0.90

2 ydy = 0.19.

⎛n⎞ 6.4.15 β = P(accept H0|H1 is true) = P(X ≤ n − 1|p) = 1 − P(X = n|p) = 1 − ⎜ ⎟ p n (1 − p)0 = 1 − pn. ⎝n⎠

When β = 0.05, p =

n

0.95 .

6.4.16 If H0 is true, X = X1 + X2 has a binomial distribution with n = 6 and p = ⎛ Therefore, α = P(reject H0|H0 is true) = P ⎜ X ≥ 5 p = ⎝

1 . 2

k 1 ⎞ 6 ⎛6 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎟= ⎜ ⎟ ⎜1 − ⎟ 2 ⎠ k = 5 ⎜⎝ k ⎟⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

∑

6− k

= 7 / 26 = 0.11.

⎛ 1 ⎞ 6.4.17 1 − β = P(reject H0|H1 is true) = P ⎜ Y ≤ θ ⎟ = ⎝ 2 ⎠

∫

1/2 0

(1 + θ ) yθ dy = yθ +1

1/ 2 0

⎛1⎞ =⎜ ⎟ ⎝2⎠

θ +1

e−6 6k = 0.062. k! k =0 2

6.4.18 (a) α = P(reject H0|H0 is true) = P(X ≤ 2| λ = 6) =

∑

e−4 4k k! k =0 2

(b) β = P(accept H0|H1 is true) = P(X ≥ 3| λ = 4) = 1 − P(X ≤ 2|λ = 4) = 1 −

∑

= 1 − 0.238 = 0.762 1⎞ ⎛ 6.4.19 P(Type II error) = β = P(accept H0|H1 is true) = P ⎜ X ≤ 3 p = ⎟ = ⎝ 2⎠

3

⎛ 1⎞ ⎜⎝1 − ⎟⎠ 2 k =1

∑

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

k −1

⋅

1 7 = 2 8

Section 6.5: A Notion of Optimality: The Generalized Likelihood Ratio

6.4.20 β = P(accept H0|H1 is true) = P(Y < ln 10|λ) =

∫

ln10 0

119

λ e − λ y dy = 1 − e−λln 10 = 1 − 10−λ

6.4.21 α = P(reject H0|H0 is true) = P(Y1 + Y2 ≤ k|θ = 2). When H0 is true, Y1 and Y2 are uniformly distributed over the square defined by 0 ≤ Y1 ≤ 2 and 0 ≤ Y2 ≤ 2, so the joint pdf of Y1 and Y2 is a 1⎛ 1 1⎞ plane parallel to the Y1Y2-axis at height ⎜ = fY1 ( y1 ) ⋅ fY2 ( y2 ) = ⋅ ⎟ . By geometry, α is the 4⎝ 2 2⎠ volume of the triangular wedge in the lower left-hand corner of the square over which Y1 and Y2 are defined. The hypotenuse of the triangle in the Y1Y2-plane has the equation 1 1 y1 + y2 = k. Therefore, α = area of triangle × height of wedge = ⋅ k ⋅ k ⋅ = k2/8. 2 4 For α to be 0.05, k = 0.4 = 0.63. 6.4.22 α = P(reject H0|H0 is true) = P(Y1Y2 ≤ k*|θ = 2). If θ = 2, the joint pdf of Y1 and Y2 is the 1 horizontal plane fY1 ,Y2 ( y1 , y2 ) = , 0 ≤ y1 ≤ 2, 0 ≤ y2 ≤ 2. Therefore, α = P(Y1Y2 ≤ k*|θ = 2) 4 2 k */ y 2 2 1 1 k* 1 k* k* k* ⎛k* ⎞ dy2 dy1 = dy1 = = 2⋅ ⋅ + + + ⎜ ln y1 = k */ 2 */ 2 0 */ 2 k k ⎠⎟ 2 4 4 4 4 y1 4 ⎝ 4

∫ ∫

∫

k* k* k* k* . By trial and error, k* = 0.087 makes α = 0.05. + ln 2 − ln 4 4 4 2

Section 6.5: A Notion of Optimality: The Generalized Likelihood Ratio n

6.5.1

L(ωˆ ) =

n

∏

(1 − p0 )

ki −1

∑ ki − n

p0 = p0n (1 − p0 ) i=1

= p0n (1 − p0 )k − n , where k =

i =1

n

∑k . i

From Case Study

i =1

n

n ˆ ) = ⎛ n ⎞ ⎛1 − n ⎞ 5.2.1, the maximum likelihood estimate for p is pe = . Therefore, L(Ω ⎜⎝ ⎟⎠ ⎜⎝ ⎟ k k⎠ k and the generalized likelihood ratio for testing H0: p = p0 versus ˆ ). H1: p ≠ p0 is the quotient L(ωˆ ) / L(Ω 10

6.5.2

Let y =

∑y . i

Then L(ωˆ ) =

i =1

10

Also, L(λ) =

10

∏λ e λ −

o yi

o

= λ010 e

− λ0

,

10

∑ yi i =1

= λ010 e − λ0 y .

i =1

∏ λe λ

− yi

= λ 10 e − λ y , so ln L(λ) = 10 ln λ − λy and

i =1

d ln L( λ ) 10 = − y . Setting the dλ λ

latter equal to 0 implies that the maximum likelihood estimate for λ is λe = 10

k −n

10 . Therefore, y

⎛ 10 ⎞

y − ˆ ) = ⎛ 10 ⎞ e ⎜⎝ y ⎠⎟ = (10/y)10e−10. The generalized likelihood ratio, then, is the quotient L (Ω ⎜⎝ y ⎟⎠

λ010 e− λ0 y /(10 / y )10 e −10 = ( λ0 e /10)10 y10 e− λ0 y . It follows that H0 should be rejected if λ = y10 e − λ0 y ≤ λ*, where λ* is chosen so that

∫

λ* 0

f λ (λ|H0 is true)dλ = 0.05.

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120

Chapter 6: Hypothesis Testing n

6.5.3

L(ωˆ ) =

n

∏ (1/

2π

1 − ( yi − µ0 )2 )e 2

= (2π ) − n / 2 e

−

∑

1 ( y − µ )2 2 i =1 i 0

. Since y is the maximum likelihood

i =1

n

ˆ ) = (2π )− n / 2 e estimate for µ (recall the first derivative taken in Example 5.2.4), L(Ω ˆ) = Here the generalized likelihood ratio reduces to λ = L(ωˆ ) / L(Ω

1 − (( y − µ0 ) /(1/ n ))2 e 2

−

∑

1 ( y − y )2 2 i=1 i

.

. The null

1 − (( y − µ0 ) /(1/ n ))2

hypothesis should be rejected if e 2 ≤ λ* or, equivalently, if ( y − µ0 ) /(1/ n ) > λ**, where values for λ** come from the standard normal pdf, fZ(z). 6.5.4

To test H0: µ = µ0 versus H1: µ = µ1, the “best” critical region would consist of all those samples n

for which

∏ ∑

( yi − µ0 ) 2 −

i =1 n

rejected if

∏

(1/ 2π )n e

1 − ( yi − µ1 )2 2

≤ k . Equivalently, H0 should be

n

∑(y − µ ) i

1

2

> 2lnk. Simplified, the latter becomes

i =1

∑ y > 2lnk + n ( µ i

i =1

n

/

i =1

i =1 n

2(µ1 − µ0)

1 − ( yi − µ0 )2 2

(1/ 2π ) n e

2 1

)

− µ02 . Consider the case where µ1 < µ0. Then µ1 − µ0 < 0, and

the decision rule reduces to rejecting H0 when y
0 for all u > 0. To verify that fU (u ) is a pdf requires proving that ∞ 1 ∞ ∞ 1 fU (u )du = 1. But fU (u )du = u n / 2 − 1e− u / 2 du = n/2 0 0 0 Γ (n / 2) 2

∫

=

∫

∫

1 Γ (n / 2)

∫

∞⎛u 0

⎞ ⎜⎝ ⎟⎠ 2

n/2 − 1

⎛n⎞ By definition, Γ ⎜ ⎟ = ⎝2⎠

7.3.2

Y −µ S/ n

e − u / 2 (du / 2) =

∫

∞ 0

1 ⎛n⎞ Γ⎜ ⎟ ⎝2⎠

∫

v n / 2 − 1e − v dv . Thus,

∞ 0

∫

v n / 2 − 1e− v dv , where v =

∞ 0

fU (u )dy =

u du and dv = . 2 2

1 ⎛n⎞ ⋅ Γ ⎜ ⎟ = 1. Γ (n / 2) ⎝ 2 ⎠

n 1 and for r and λ , respectively, in the moment-generating function for a gamma 2 2 pdf gives M χ 2 (t ) = (1 − 2t ) − n / 2 . Also, M χ(1)2 (t ) = (−n/2) (1 − 2t ) − n / 2 −1 (−2) = n (1 − 2t ) − n / 2 −1 and

Substituting

n

n

⎛ n ⎞ M χ(2)2 (t ) = ⎜ − − 1⎟ (n)(1 − 2t ) − n / 2 − 2 (−2) = (n2 + 2n) ⋅ (1 − 2t)−n/2 n ⎝ 2 ⎠

− 2

, so M χ(1)2 (0) = n and n

( )

M χ(2)2 (0) = n2 + 2n. Therefore, E ( χ n2 ) = n and Var χ n2 = n2 + 2n − n2 = 2n. n

7.3.3

If µ = 50 and σ = 10,

2

⎛ Yi − 50 ⎞ 2 ⎜⎝ ⎟⎠ should have a χ 3 distribution, implying that the numerical 10 i =1 3

∑

2 2 (=0.216) and χ .975,3 (= 9.348). Here, value of the sum is likely to be between, say, χ .025,3 2

2

2

2

⎛ Yi − 50 ⎞ ⎛ 65 − 50 ⎞ ⎛ 30 − 50 ⎞ ⎛ 55 − 50 ⎞ ⎟ +⎜ ⎟ +⎜ ⎟ = 6.50, so the data are not inconsistent ⎜ ⎟ = ⎜⎝ ⎝ ⎠ 10 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ i =1 3

∑

with the hypothesis that the Yi’s are normally distributed with µ = 50 and σ = 10. 7.3.4

Let Y =

(n − 1) S 2

Var(S2) = 7.3.5

7.3.6

σ2 2σ 4 n −1

(

)

. Then Var(Y) = Var χ n2−1 = 2(n − 1) =

(n − 1)2 Var( S 2 )

σ4

. If follows that

.

Since E(S2) = σ2, it follows from Chebyshev’s inequality that P(|S2 − σ2| < ε) > Var( S 2 ) 2σ 4 2 1− . But Var(S ) = → 0 as n → ∞ . Therefore, S2 is consistent for σ2. 2 − n 1 ε ⎛ Y − 200 ⎞ Y − 200 2 Let Y = χ 200 . Then = Z , in which case P ⎜ ≤ −0.25 ⎟ = 0.40 . Equivalently, ⎝ 400 ⎠ 400 2 Y ≤ 200 − 0.25 400 = 195, implying that χ .40,200 = 195.

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122

7.3.7 7.3.8

7.3.9

Chapter 7: Inferences Based on the Normal Distribution

(a) 0.983

(b) 0.132 (c) 9.00

V /7 ⎛ ⎞ < 3.29 ⎟ = P(2.51 < F7,9 < 3.29) = P(F7,9 < 3.29) − P(F7,9 ≤ 2.51) = 0.95 − 0.90 P ⎜ 2.51 < ⎝ ⎠ U /9 = 0.05. But P(3.29 < F7,9 < 4.20) = 0.975 − 0.95 = 0.025.

(a) 6.23 (b) 0.65 (c) 9 (d) 15 (e) 2.28

7.3.10 Since the samples are independent and S2/σ2 = χ n2−1 /(n − 1) , it follows that

S12 S22 / 2 = S12 / S22 2

σ σ has an Fn−1,n − 1 distribution. As n increases, the distributions of the unbiased estimators S12 and

S22 become increasingly concentrated around σ2 (recall Question 7.3.4), implying that F ratios converge to 1 as the two sample sizes get large. V /m , where U and V are independent χ 2 random variables with m and n degrees of U /n 1 U /n 1 = , which implies that has an F distribution with n and freedom, respectively. Then F V /m F m degrees of freedom.

7.3.11 F =

⎛ ⎞ 1 1⎞ ⎛1 7.3.12 If P(a ≤ Fm,n ≤ b) = q, then P ⎜ a ≤ ≤ b ⎟ = q = P ⎜ ≤ Fn , m ≤ ⎟ . From Appendix Table A.4, ⎝b Fn ,m a⎠ ⎝ ⎠ 1 P(0.052 ≤ F2,8 ≤ 4.46) = 0.90. Also, P(0.224 ≤ F8,2 ≤ 19.4) = 0.90. But = 0.224 and 4.46 1 = 19.23  19.4. 0.052 ⎛ t2 ⎞ 7.3.13 To show that fTn (t ) converges to fZ(t) requires proving that ⎜1 + ⎟ n⎠ ⎝

and

− ( n +1) / 2

⎛ n + 1⎞ Γ⎜ ⎝ 2 ⎠⎟ converges to 1/ 2π . To verify the first limit, write ⎛n⎞ nπ Γ ⎜ ⎟ ⎝2⎠

n / t2 ⎧⎪ ⎛ 1 ⎞ ⎫⎪ = ⎨ ⎜1 + ⎬ ⎝ n / t 2 ⎟⎠ ⎪ ⎩⎪ ⎭

approaches (e1 ) − t Γ (r ) =

2π ⎛ r ⎞ ⎜ ⎟ r ⎝e⎠

2

−t2 / 2

/2

⎛ t2 ⎞ ⋅ ⎜1 + ⎟ n⎠ ⎝

= e−t

2

/2

converges to e − t ⎛ t2 ⎞ ⎜⎝1 + n ⎟⎠

/2

− ( n +1) / 2

−1/ 2

. As n gets large, the last factor approaches 1 and the product

. Also, for large n, n! = 2π nn n e − n . Equivalently,

r

if r is large. An application of the latter equation shows that

⎛ n +1⎞ ⎛n⎞ Γ⎜ Γ ⎜ ⎟ converges to ⎝ 2 ⎠⎟ ⎝2⎠

2

n , which means that the constant in fTn (t ) converges to 2

1/ 2π , the constant in the standard normal pdf.

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Section 7.4: Drawing Inferences about

7.3.14

µ

123

∞

1 dx is the integral of the variable portion of a T1 pdf over the upper half of its range. 0 1 + x2 ⎛1⎞ Γ⎜ ⎟ π ∞ ∞ 1 1 ⎝2⎠ π Γ (1) 1 dt = , it follows that = . Since dx = ⋅ 2 0 0 1 2 2 Γ (1) 2 ⎛ ⎞ 1+ x π Γ ⎜ ⎟ (1 + t 2 ) ⎝2⎠

∫

∫

∫

7.3.15 Let T be a Student t random variable with n degrees of freedom. Then ∞ 1 E (T 2 k ) = C t 2 k dt , where C is the product of the constants appearing in the −∞ 2 ( n +1) / 2 ⎛ t ⎞ ⎜⎝1 + n ⎟⎠

∫

definition of the Student t pdf. The change of variable y = t / n results in the integral E (T 2 k ) = C *

∫

∞

−∞

y 2k

1

(1 + y )

2 ( n +1) / 2

dy for some constant C*.

∫

Because of the symmetry of the integrand, E (T 2 k ) is finite if the integral

y 2k

∞

0

(1 + y )

2 ( n +1) / 2

dy is

finite. But

∫

∞

0

(

y 2k

1+ y

)

2 ( n +1) / 2

dy
0, and αβ > 1, so the 2 2

the integral is finite.

Section 7.4: Drawing Inferences About µ (b) 0.80

(c) 0.85

(d) 0.99 − 0.15 = 0.84

(c) 1.7056

(d) 4.3027

7.4.1

(a) 0.15

7.4.2

(a) 2.508 (b) 1.079

7.4.3

Both differences represent intervals associated with 5% of the area under fTn (t ) . Because the pdf is closer to the horizontal axis the further t is away from 0, the difference t.05, n − t.10, n is the larger of the two.

7.4.4

Since

⎛ ⎞ Y − 27.6 Y − 27.6 is a Student t random variable with 8 df, P ⎜ −1.397 ≤ ≤ 1.397 ⎟ = 0.80 ⎝ ⎠ S/ 9 S/ 9

⎛ ⎞ Y − 27.6 and P ⎜ −1.8595 ≤ ≤ 1.8595 ⎟ = 0.90 (see Appendix Table A.2). ⎝ ⎠ S/ 9

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124

7.4.5

Chapter 7: Inferences Based on the Normal Distribution

⎛ Y − 15.0 ⎞ ⎛ ⎞ Y − 15.0 Y − 15.0 P⎜ ≥ k ⎟ = 0.05 implies that P ⎜ − k ≤ is a Student t ≤ k ⎟ = 0.95. But ⎝ ⎠ S / 11 S / 11 ⎝ S / 11 ⎠ random variable with 10 df. From Appendix Table A.2, P(−2.2281 ≤ T10 ≤ 2.2281) = 0.95, so k = 2.2281.

7.4.6

P(90.6 − k(S) ≤ Y ≤ 90.6 + k(S)) = 0.99 = ⎛ 90.6 − k ( S ) − 90.6 Y − 90.6 90.6 + k ( S ) − 90.6 ⎞ ⎛ k (S ) k (S ) ⎞ = P⎜ P⎜ ≤ ≤ ≤ T19 ≤ ⎟ = ⎟ ⎝ S / 20 ⎝ ⎠ S / 20 S / 20 S / 20 S / 20 ⎠ k (S ) 2.8609 ⋅ S P(−2.8609 ≤ T19 ≤ 2.8609), so = 2.8609, implying that k(S) = . S / 20 20

7.4.7

Since n = 20 and the confidence interval has level 90%, tα / 2,n −1 = t.05,19 = 1.7291. For these data 20

∑y

i

20

= 20.22 and

i =1

∑y

2 i

= 23.014. Then y = 20.22/20 = 1.011 and s =

i =1

20(23.014) − (20.22) 2 = 0.368. The confidence interval is 20(19) ⎛ s s ⎞ ⎛ 0.368 0.368 ⎞ , y + tα / 2,n −1 ,1.011 + 1.7291 ⎜⎝ y − tα / 2,n −1 ⎟⎠ = ⎜⎝1.011 − 1.7291 ⎟ = (0.869, 1.153). n n 20 20 ⎠

7.4.8

Given that n = 16 and the confidence level is 95%, tα / 2,n −1 = t.025,15 = 2.1314. 16

Here

∑y

i

= 24,256, so y =

i =1

1 (24,256) = 1516.0, and s is given to be 369.02. 16

⎛ 369.02 369.02 ⎞ The confidence interval is ⎜1516.0 − 2.1314 ,1516.0 + 2.1314 ⎟ ⎝ 16 16 ⎠ = ($1319.4, $1712.6).

7.4.9

(a) Let µ = true average age at which scientists make their greatest discoveries. 12

12

Since

∑y

i

i =1

= 425 and

∑y

2 i

y =1

= 15,627, y =

1 (425) = 35.4 and s = 12

12(15,627) − (425)2 12(11)

= 7.2. Also, tα/2,n−1 = t.025,11 = 2.2010, so the 95% confidence interval for µ is the range 7.2 7.2 ⎞ ⎛ , 35.4 + 2.2010 ⋅ ⎜⎝ 35.4 − 2.2010 ⋅ ⎟ , or (30.8 yrs, 40.0 yrs). 12 12 ⎠ (b) The graph of date versus age shows no obvious patterns or trends. The assumption that µ has remained constant over time is believable. 7.4.10 Given that n = 61 and the confidence level is 99%, tα / 2,n −1 = t.005,60 = 2.6603. Then y =

1 (6450) = 105.7, s = 60

61(684,900) − (6450)2 = 6.94 61(60)

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Section 7.4: Drawing Inferences about

µ

125

⎛ 6.94 6.94 ⎞ The 99% confidence interval for µ is ⎜105.7 − 2.6603 ⋅ ,105.7 + 2.6603 ⋅ ⎟ ⎝ 61 61 ⎠ = (103.3, 108.1).

7.4.11 For n = 24, tα / 2,n −1 = t.05,23 = 1.7139. For these data y = 4645/24 = 193.54 and s=

24(959, 265) − (4645)2 = 51.19. 24(23)

51.19 51.19 ⎞ ⎛ The confidence interval is ⎜193.54 − 1.7139 ,193.54 + 1.7139 ⎟ = (175.6, 211.4). ⎝ 24 24 ⎠ The medical and statistical definition of “normal” differ somewhat. There are people with medically norm platelet counts who appear in the population less than 10% of the time. ⎛ s s ⎞ 7.4.12 Given that n = 16, tα / 2,n −1 = t.025,15 = 2.1315, so ⎜ y − 2.1315 ⋅ , y + 2.1315 ⋅ ⎟ ⎝ 16 16 ⎠ s , implying that s = 4.88. Also, = (44.7, 49.9). Therefore, 49.9 − 44.7 = 5.2 = 2(2.1315) ⋅ 16 44.7 + 49.9 because the confidence interval is centered around the sample mean, y = = 47.3. 2

7.4.13 No, because the length of a confidence interval for µ is a function of s, as well as the confidence coefficient. If the sample standard deviation for the second sample were sufficiently small (relative to the sample standard deviation for the first sample), the 95% confidence interval would be shorter than the 90% confidence interval. 7.4.14 The range spanned by, say, a 99% confidence interval for µ would be a reasonable set of values for the company’s true average revenue. With n = 9, y = $59,540, and s = $6,680, the 99% ⎛ 6,860 6,860 ⎞ confidence interval for µ is ⎜ 59,540 − 3.554 ⋅ , 59,540 + 3.554 ⋅ ⎟ ⎝ 9 9 ⎠ = ($51,867, $67,213).

7.4.15 (a) 0.95 7.4.16 (a)

(b) 0.80

(c) 0.945 (d) 0.95

Given that n = 336 and the confidence level is 95%, we use the normal tables and take 1 336(10,518.84) − (1392.6)2 (1392.60) = 4.14, s = 336 336(335) = 3.764. Theorem 7.4.1 implies that the 95% confidence interval for µ is ⎛ 3.764 3.764 ⎞ , 4.14 + 1.96 ⋅ ⎜⎝ 4.14 − 1.96 ⋅ ⎟ = (3.74, 4.54). 336 336 ⎠ zα / 2 = z.025 = 1.96. Then y =

(b)

The normality assumption is egregiously violated for these data; note that the data’s histogram is sharply skewed. Theorem 7.4.1 is not appropriate.

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126

Chapter 7: Inferences Based on the Normal Distribution

7.4.17 Let µ = true average FEV1/VC ratio for exposed workers. Since

19

∑y

i

= 14.56 and

i =1

19(11.2904) − (14.56) 2 14.56 = 0.0859. To test = 0.766 and s = 19(18) 19 i =1 H0: µ = 0.80 versus H1: µ < 0.80 at the α = 0.05 level of significance, reject the null hypothesis if 0.766 − 0.80 t ≤ −t.05,18 = −1.7341. But t = = −1.71, so we fail to reject H0. 0.0859 / 19 19

∑

yi2

= 11.2904, y =

7.4.18 At the α = 0.05 level, H0: µ = 132.4 should be rejected in favor of H1:µ ≠ 132.4 143.8 − 132.4 y − 132.4 if ≥ t.025,83 = 1.9890. But t = = 17.4, making it clear that the skull 6.9 / 84 s / 84 differences between Etruscans and native Italians are too great to be ascribed to chance. A histogram of the 84 yi’s shows a distinctly bell-shaped pattern, so the normality assumption implicit in the t test appears to be satisfied. 7.4.19 Let µ = true average GMAT increase earned by students taking the review course. The hypotheses to be tested are H0: µ = 40 versus H1: µ < 40. Here,

15

∑y

i

= 556 and

i =1

556 15(20,966) − (556) 2 37.1 − 40 = 37.1, s = = 5.0, and t = = 15(14) 15 5.0 / 15 i =1 −2.25. Since −t.05,14 = −1.7613, H0 should be rejected at the α = 0.05 level of significance, suggesting that the MBAs ′R Us advertisement may be fraudulent. 15

∑

yi2 = 20,966, so y =

7.4.20 H0: µ = 0.618 should be rejected in favor of a two-sided H1 at the 0.01 level of significance if 0.6373 − 0.618 | t | ≥ t.005,33 = 2.7333 . Given that y = 0.6373 and s = 0.14139, the t statistic is 0.14139 / 34 = 0.80, so H0 is not rejected. These data do not rule out the possibility that national flags embrace the Golden Rectangle as an aesthetic standard. 7.4.21 Let u = true average pit depth associated with plastic coating. To test H0: µ = 0.0042 versus H1: µ < 0.0042 at the α = 0.05 level, we should reject the null hypothesis if t ≤ −t.05,9 = −1.8331. 0.0390 0.0039 − 0.0042 = 0.0039. Also, s = 0.00383, so t = = −2.48. Since For the 10 yi’s, y = 10 0.00383/ 10 H0 is rejected , these data support the claim that the plastic coating is an effective corrosion retardant. 7.4.22 The set of µ0’s for which H0: µ = µ0 would not be rejected at the α = 0.05 level of significance is the same as the 95% confidence interval for µ. 7.4.23 Because of the skewed shape of fY(y), and if the sample size was small, it would not be unusual for all the yi’s to lie close together near 0. When that happens, y will be less than µ, s will be considerably smaller than E(S), and the t ratio will be further to the left of 0 than fTn−1 (t ) would predict.

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Section 7.5: Drawing Inferences about σ

2

127

7.4.24 Both sets of ratios would have distributions similar to that of a Student t random variable with 2 df. Because of the shapes of the two fY(y)’s, though, both distributions would be skewed to the right, particularly so for fY(y) = 4y3 (recall the answer to Question 7.4.23). 7.4.25 As n increases, Student t pdfs converge to the standard normal, fZ(z) (see Question 7.3.13). 7.4.26 Only (c) would raise any serious concerns. There the sample size is small and the yi’s are showing a markedly skewed pattern. Those are the two conditions that particularly “stress” the robustness property of the t-ratio (recall Figure 7.4.6).

Section 7.5: Drawing Inferences About σ2 7.5.1

(a) 23.685 (b) 4.605

(c) 2.700

7.5.2

(a) 0.95

(b) 0.90

(c) 0.975 − 0.025 = 0.95 (d) 0.99

7.5.3

(a) 2.088

(b) 7.261

(c) 14.041 (d) 17.539

7.5.4

(a) 13

(b) 19

(c) 31

7.5.5

χ

7.5.6

⎛ S2 ⎞ ⎛ (n − 1) S 2 ⎞ < 2(n − 1) ⎟ = P χ n2−1 < 2(n − 1) . Values from the 0.95 column in a χ2 P ⎜ 2 < 2⎟ = P ⎜ 2 ⎝σ ⎠ ⎝ σ ⎠

(d) 17 3

2 .95,200

⎛ 2 2 ⎞ = 200 ⎜1 − + 1.64 = 233.9 9(200) ⎠⎟ ⎝ 9(200)

(

(

)

)

2 table show that for each n < 8, P χ n2−1 < 2(n − 1) < 0.95 . But for n = 9, χ .95,8 = 15.507, which

(

)

means that P χ 82 < 16 > 0.95.

7.5.7

⎛ (n − 1) S 2 ⎛ ⎞ (n − 1) S 2 (n − 1) S 2 ⎞ 2 2 1 P ⎜ χ α2 / 2,n −1 ≤ ≤ χ = − α = P ≤ σ ≤ 1−α / 2, n −1 ⎟ ⎜ 2 ⎟ , so σ2 χ α2 / 2,n −1 ⎠ ⎝ ⎠ ⎝ χ 1−α / 2,n −1 ⎛ (n − 1) s 2 (n − 1) s 2 ⎞ 2 , 2 ⎜ 2 ⎟ is a 100(1 − α)% confidence interval for σ . ⎝ χ 1−α / 2,n −1 χ α / 2,n −1 ⎠ Taking the square root of both sides gives a 100(1 − α)% confidence interval for σ .

7.5.8

18S 2 has a χ 2 distribution with 18 df, so for example 12.0 ⎛ ⎞ 18S 2 P ⎜ 8.231 ≤ ≤ 31.526 ⎟ 12.0 ⎝ ⎠

If n = 19 and σ 2 = 12.0,

= 0.95 = P(5.49 ≤ S 2 ≤ 21.02).

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128

Chapter 7: Inferences Based on the Normal Distribution 16

7.5.9

(a)

∑

yi = 1514, so y =

i =1

= 742.38. Since χ

1514 = 94.6. 16

2 .025,15

16

∑

yi2

= 154,398, so s 2 =

i =1

16(154,398) − (1514) 2 16(15)

2 = 6.262 and χ .975,15 = 27.488, a 95% confidence interval for σ is

⎛ 15(742.38) 15(742.38) ⎞ , , or (20.1, 42.2). ⎜ 27.488 6.262 ⎟⎠ ⎝ 2 2 (b) Given that χ .05,15 = 7.261 and χ .95,15 = 24.966, the two one-sided confidence intervals for σ

⎛ ⎛ 15(742.38) ⎞ 15(742.38) ⎞ = (0, 39.2) and , ∞ ⎟ = (21.1, ∞). are ⎜ 0, ⎜ 7.261 ⎟⎠ 24.966 ⎝ ⎝ ⎠ 16

7.5.10

∑

yi = 29.98, so y =

i =1

= 0.1921. Since χ

2 .025,9

29.98 = 2.998. 10

= 2.700 and χ

16

∑

yi2 = 91.609, so s 2 =

i =1 2 .975,9

10(91.609) − (29.98) 2 10(9)

= 19.023, a 95% confidence interval for σ is

⎛ 9(0.1921) 9(0.1921) ⎞ = (0.302, 0.800). ⎜ 19.023 , 2.700 ⎟⎠ ⎝ Since the standard deviation for 1-year CD rates of 0.262 falls below the interval, we have evidence that the variability of 5-year CD interest rates is higher.

7.5.11 Experimenters often prefer confidence intervals for σ (as opposed to σ2) because they are expressed in the same units as the data, which makes them easier to interpret. 7.5.12 (a) If

χ n2−1 − (n − 1) 2(n − 1)

= Z , then P (− zα / 2 ≤

χ n2−1 − (n − 1) 2(n − 1)

≤ zα/2)  1 − α

⎛ ⎞ (n − 1) S 2 = P ⎜ n − 1 − zα / 2 2(n − 1) ≤ ≤ n − 1 + zα / 2 2( n − 1) ⎟ 2 σ ⎝ ⎠ ⎛ ⎞ (n − 1) S 2 (n − 1) S 2 = P⎜ ≤σ 2 ≤ ⎟ , so n − 1 − zα / 2 2(n − 1) ⎠ ⎝ n − 1 + zα / 2 2(n − 1) ⎛ ⎞ (n − 1) s 2 (n − 1) s 2 , ⎜ ⎟ is an approximate 100(1 − α)% confidence ⎝ n − 1 + zα / 2 2(n − 1) n − 1 − zα / 2 2(n − 1) ⎠ ⎛ n − 1s n − 1s interval for σ 2 . Likewise, ⎜ , ⎜⎝ n − 1 + zα / 2 2(n − 1) n − 1 − zα / 2 2(n − 1)

⎞ ⎟ is an ⎟⎠

approximate 100(1 − α)% confidence interval for σ . (b) For the data in Table 7.5.1, n = 19 and s =

733.4 = 27.08, so the formula in Part a gives

⎛ 18(27.08) 18(27.08) ⎞ , ⎜ ⎟ = (21.1 million years, 46.0 million years) as the ⎜⎝ 18 + 1.96 36 18 − 1.96 36 ⎟⎠ approximate 95% confidence interval for σ .

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Section 7.5: Drawing Inferences about σ

2

129

⎛ (n − 1) s 2 (n − 1) s 2 ⎞ 261.90 2 2 7.5.13 If ⎜ 2 = 5.088. A trial and , 2 ⎟ = (51.47, 261.92), then χ .95, n −1 / χ .05,n −1 = 51.47 ⎝ χ .95,n −1 χ .05, n−1 ⎠ 16.919 2 2 / χ .05,9 = = 5.088, so n = 10, which error inspection of the χ 2 table shows that χ .95,9 3.325 9s 2 = 261.92. Therefore, s = 9.8. means that 3.325 n

1 2nY 7.5.14 (a) MY(t) = . Let X = = 1−θt θ

2

∑Y

i

i =1

θ

n

. Then MX(t) =

∏ i =1

⎛ 2t ⎞ ⎛ 1 ⎞ M Yi ⎜ ⎟ = ⎜ ⎝ θ ⎠ ⎝ 1 − 2t ⎟⎠

2n / 2

,

2 random variable. See Theorem 4.6.5. implying that X is a χ 2n

⎛ 2ny ⎛ ⎞ 2ny ⎞ 2nY , 2 (b) P ⎜ χ α2 / 2,2 n ≤ ≤ χ 12−α / 2,2 n ⎟ = 1 − α, so ⎜ 2 ⎟ is a θ ⎝ ⎠ ⎝ χ 1−α / 2,2 n χ α / 2,2 n ⎠ 100(1 − α)% confidence interval for θ.

7.5.15 Test H0: σ 2 = 30.42 versus H1: σ 2 < 30.42 . The test statistic in this case is (n − 1) s 2 18(733.4) 2 = 14.285. The critical value is χ α2 ,n −1 = χ .05,18 = 9.390. = χ2 = 2 2 σ0 30.4 Since the test statistic is not less than the critical value, we accept the null hypothesis, and we cannot assume that the potassium-argon method is more precise. 7.5.16 Test H 0 : σ 2 = 1 versus H1 : σ 2 > 1 . 30(19,195.7938) − (758.62) 2 = 0.425 30(29) 29(0.425) 2 The test statistic is χ 2 = = 12.325. The critical value is χ .95,29 = 42.557. 1 Since 12.325 < 42.557, we accept the null hypothesis and assume the machine is working properly.

The sample variance is

7.5.17 (a) Test H 0 : µ = 10.1 versus H1 : µ > 10.1 y − µ0 11.5 − 10.1 The test statistic is = 0.674 = s / n 10.17 / 24 The critical value is tα,n-1 = t0.05,23 = 1.7139. Since 0.674 < 1.7139, we accept the null hypothesis. We cannot ascribe the increase of the portfolio yield over the benchmark to the analyst’s system for choosing stocks. (b) Test H 0 : σ 2 = 15.67 versus H 0 : σ 2 < 15.67 23(10.17 2 ) 2 = 9.688. The critical value is χ .05,23 = 13.091. 15.67 2 Since the test statistic of 9.688 is less than the critical value of 13.091, we reject the null hypothesis. The analyst’s method of choosing stocks does seem to result in less volatility

The test statistic is χ 2 =

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Chapter 8: Types of Data: A Brief Overview Section 8.2: Classifying Data 8.2.1

Regression data

8.2.2

Two-sample data

8.2.3

One-sample data

8.2.4

Randomized block data

8.2.5

Regression data

8.2.6

Paired data

8.2.7

k-sample data

8.2.8

Regression data

8.2.9

One-sample data

8.2.10 Regression data 8.2.11 Regression data 8.2.12 One-sample data 8.2.13 Two-sample data 8.2.14 Regression data 8.2.15 k-sample data 8.2.16 Paired data 8.2.17 Categorical data 8.2.18 Paired data 8.2.19 Two-sample data 8.2.20 Categorical data 8.2.21 Paired data 8.2.22 k-sample data 8.2.23 Categorical data 8.2.24 Randomized block data 8.2.25 Categorical data

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132

Chapter 8: Types of Data: A Brief Overview

8.2.26 Two-sample data 8.2.27 Categorical data 8.2.28 Two-sample data 8.2.29 Paired data 8.2.30 k-sample data 8.2.31 Randomized block data 8.2.32 Two-sample data

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Chapter 9: Two-Sample Inferences Section 9.2: Testing H 0 : µ X = µY 9.2.1

s 2X = 13.79 and sY2 = 15.61 (n − 1) s X2 + (m − 1) sY2 11(13.79) + 8(15.61) = = 3.82 12 + 9 − 2 n+m−2 x−y 29.8 − 26.9 = 1.72 = t= s p 1/ n + 1/ m 3.82 1/12 + 1/ 9 sp =

Since t = 1.72 < t.01,19 = 2.539, accept H0. 9.2.2

For large samples, use the approximate z statistic z =

x−y s 2X n

+

sY2

−4.7 − (−1.6)

=

7.052 5.362 + 77 79

m

= −3.09

Since z = −3.09 < −1.64 = − z.05 , reject H0. 9.2.3

For large samples, use the approximate z statistic z =

x−y s 2X n

+

sY2

=

189.0 − 177.2

m

34.22 33.32 + 476 592

Since 5.67 > 1.64 = z.05 , reject H0. x−y

z=

=

491 − 498

9.2.4

= −1.76 s 2X sY2 1192 1292 + + 1126 5042 n m Since −1.76 < −1.64 = z.05 , reject H0.

9.2.5

z=

= −0.491 s 2X sY2 3.702 4.282 + + 93 28 n M Since − z.005 = −2.58 < −0.491 < 2.58 = z.005 , accept H0.

9.2.6

t=

x−y

sp

=

4.17 − 4.61

x−y 65.2 − 75.5 = = −1.68 1/ n + 1/ m 13.9 1/ 9 + 1/12

Since −t.05,19 = −1.7291 < t = −1.68, accept H0. 9.2.7

(n − 1) s X2 + (m − 1) sY2 3(266.92 ) + 3(224.32 ) = = 246.52 n+m−2 4+4−2 x−y 1133.0 − 1013.5 = = 0.69 1/ n + 1/ m 246.52 1/ 4 + 1/ 4

sp = s p = t=

sp

Since −t.025,6 = −2.4469 < t = 0.69 < t.025,6 = 2.4469, accept H0. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

= 5.67

134

9.2.8

Chapter 9: Two-Sample Inferences

5(15.12 ) + 8(8.12 ) = 11.317 6+9− 2 70.83 − 79.33 t= = −1.43 11.317 1/ 6 + 1/ 9

sp =

Since −t.005,13 = −3.0123 < t = −1.43 < t.005,13 = 3.0123, accept H0. 9.2.9

30(1.4692 ) + 56(1.3502 ) = 1.393 31 + 57 − 2 3.10 − 2.43 t= = 2.16 1.393 1/ 31 + 1/ 57

sp =

Since t = 2.16 > t.025,86=1.9880, reject H0. 9.2.10 Let H0: µX − 1 = µY and H1: µX − 1 < µY. 10(12) 2 + 10(162 ) = 14.9 10 + 10 − 2 (2.1 − 1) − 1.6 t= = −0.08 14.9 1/10 + 1/10

sp =

Since −t.05,18 = −1.7341 < −0.08 = t, accept H0. 9.2.11 (a) Reject H0 if | t | > t.005,15 = 2.9467, so we seek the smallest value of x − y such that t=

x−y s p 1/ n + 1/ m

=

x−y 15.3 1/ 6 + 1/11

> 2.9467, or x − y > (15.3)(0.5075)(2.9467)

= 22.880 (b) Reject H0 if t > t.05,19 = 1.7291, so we seek the smallest value of x − y such that x−y x−y = > 1.7291, or x − y > (214.9)(0.4494)(1.7291) t= s p 1/ n + 1/ m 214.9 1/13 + 1/ 8 =166.990 9.2.12 z =

x−y

σ / n +σ / m 2 X

2 Y

=

81.6 − 79.9 = 1.00 17.6 /10 + 22.9 / 20

The P-value is 2P(Z ≥ 1.00) = 2(1 − 0.8413) =2( 0.1587) = 0.3174 9.2.13 (a) Let X be the interstate route; Y, the town route. P(X > Y) = P(X − Y > 0). Var(X − Y) = Var(X) + Var(Y) = 62 + 52 = 61. ⎛ X − Y − (33 − 35) 2 ⎞ P(X − Y > 0) = P ⎜ > ⎟ = P(Z ≥ 0.26) = 1 − 0.6026 = 0.3974 ⎝ 61 61 ⎠ (b) Var( X − Y ) = Var( X ) + Var(Y ) = 62/10 + 52/10 = 61/10 ⎛ X − Y − (33 − 35) 2 ⎞ = P(Z > 0.81) = 1 − 0.7910 = 0.2090 P( X − Y ) > 0 = P ⎜ > ⎝ 61/10 61/10 ⎠⎟

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Section 9.2: Testing H 0 : µ X = µY

135

9.2.14 E ( X − Y ) = E ( X ) − E (Y ) = µX − µY Var( X − Y ) = Var( X ) + Var(Y ) = σ X2 / n + σ Y2 / m

Also, we know that X − Y is normal. The Z variable in Equation 9.2.1 is a Z transformation of a normal variable, and thus is standard normal. 9.2.15 It follows from Example 5.4.4 that E ( S X2 ) = E ( SY2 ) = σ 2 . E ( S p2 ) =

(n − 1) E ( S X2 ) + (m − 1) E ( SY2 ) (n − 1)σ 2 + (m − 1)σ 2 = =σ 2 n+m−2 n+m−2

9.2.16 Take ω = {(µX, µY): −∞ < µX = µY < ∞}. Since the X’s and Y’s are normal and independent, n+m m n m ⎡ 1 ⎛ n ⎞⎤ ⎛ 1 ⎞ 2 exp ( x ) ( yi − µ )2 ⎟ ⎥ , L(ω) = − − µ + f X ( xi ) fY ( yi ) = ⎜ ⎢ i ⎟ 2 ⎜ ⎝ 2πσ ⎠ ⎠ ⎥⎦ ⎢⎣ 2σ ⎝ i =1 i =1 i =1 i =1 where µ = µX = µY. Differentiating the expression with respect to µ and setting it equal to 0 yields

∏

∑

∏

n

∑

xi +

∑

m

∑y

nx + my . Substituting µˆ for µ in L(ω) n+m n+m gives the numerator of the generalized likelihood ratio. After algebraic simplification, we n+m ⎡ 1 ⎛ n 2 m 2 (nx + my ) 2 ⎞ ⎤ ⎛ 1 ⎞ exp yi − obtain L(ωˆ ) = ⎜ ⎢ − 2 ⎜ xi + ⎥. ⎝ 2πσ ⎟⎠ n + m ⎠⎟ ⎦⎥ i =1 ⎣⎢ 2σ ⎝ i =1

the maximum likelihood estimate µˆ =

i =1

i

i =1

∑

=

∑

The likelihood function unrestricted by the null hypothesis is n+m m ⎡ 1 ⎛ n ⎞⎤ ⎛ 1 ⎞ 2 L(Ω) = ⎜ exp − ( x − µ ) + ( yi − µY ) 2 ⎟ ⎥ ⎢ i X ⎟ 2 ⎜ ⎝ 2πσ ⎠ ⎠ ⎦⎥ i =1 ⎣⎢ 2σ ⎝ i =1

∑

Solving

∑

∂ ln L(Ω) ∂ ln L(Ω) = 0 and = 0 gives µˆ X = x and µˆY = y . ∂µ X ∂µY

ˆ ) , which simplifies to Substituting those values into L(Ω) gives L(Ω ˆ)=⎛ 1 ⎞ L (Ω ⎜⎝ ⎟ 2πσ ⎠

λ=

n+m

⎡ 1 ⎛ n exp ⎢ − 2 ⎜ xi2 − nx 2 + ⎣⎢ 2σ ⎝ i =1

⎡ 1 L(ωˆ ) = exp ⎢ − 2 ˆ) L (Ω ⎢⎣ 2σ

∑

m

∑y

2 i

i =1

⎞⎤ − my 2 ⎟ ⎥ ⎠ ⎦⎥

⎡ 1 ⎛ nm( x − y ) 2 ⎞ ⎤ ⎛ 2 ( nx + my )2 ⎞ ⎤ 2 nx + my − = exp ⎥ ⎢− 2 ⎜ ⎟⎥ . n + m ⎠⎟ ⎥⎦ ⎝⎜ ⎢⎣ 2σ ⎝ n + m ⎠ ⎥⎦

Rejecting H0 when 0 < λ < λ* is equivalent to ln λ < ln λ* or −2ln λ > −2lnλ* = λ**. ( x − y )2 x−y But −2ln λ = . Thus, we reject H0 when < − λ ** 1⎞ 1 1 2 ⎛1 σ ⎜ + ⎟ + σ ⎝n m⎠ n m x−y or > λ ** , and we recognize this as a Z test when σ2 is known. 1 1 + σ n m

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136

Chapter 9: Two-Sample Inferences

9.2.17 For the data given, x = 545.45, sX = 428, and y = 241.82, sY = 183. Then 545.45 − 241.82 x−y = 2.16 = t= 2 2 4282 /11 + 1832 /11 s X / n + sY / m s 2 4282 Let θˆ = X2 = = 5.47 . The degrees of freedom associated with this statistic is greatest sY 1832

integer in ν =

⎛ˆ n⎞ ⎜⎝θ + ⎟⎠ m

2

1 ˆ2 1 ⎛n⎞ θ + ⎜ ⎟ (n − 1) (m − 1) ⎝ m ⎠

2

=

11 ⎞ ⎛ ⎜⎝ 5.47 + ⎟⎠ 11

2

1 1 ⎛ 11 ⎞ (5.47) 2 + ⎜ ⎟ (11 − 1) (11 − 1) ⎝ 11 ⎠

2

= 13.5

Thus, the greatest integer is 13. Since t = 2.16 > t.05,13 = 1.7709, reject H0. 9.2.18 Decreasing the degrees of freedom also decreases the power of the test. 9.2.19 (a) The sample standard deviation for the first data set is approximately 3.15; for the second, 3.29. These values seem close enough to permit the use of Theorem 9.2.2. (b) Intuitively, the states with the comprehensive law should have fewer deaths. However, the average for these data is 8.1, which is larger than the average of 7.0 for the states with a more limited law. 9.2.20 For the data given, x = 29.43, s X2 = 93.073, and y = 35.73, sY2 = 234.946. Then x−y 29.43 − 35.73 = = −0.95 t= 2 2 93.073/ 9 + 234.946 / 7 s X / n + sY / m s2 93.073 Let θˆ = X2 = = 0.396 . The degrees of freedom associated with this statistic is greatest sY 234.946

integer in ν =

9⎞ ⎛ ⎜⎝ 0.396 + ⎟⎠ 7

2

1 1 ⎛9⎞ (0.396)2 + ⎜ ⎟ (9 − 1) (6 − 1) ⎝ 7 ⎠

2

= 8.08 , that is, 8

Since t = -0.95 > −t.01,8 = -2.896, do not reject H0.

Section 9.3: Testing H 0 : σ X2 = σ Y2 —The F Test 9.3.1

Since F.025,11,11 9.3.2

35.7604 = 0.308 . 115.9929 = 0.288 < 0.308 < 3.47 = F.975,11,11 , we can assume that the variances are equal.

From the case study, sX2 = 115.9929 and sY2 = 35.7604. The observed F =

0.2578 = 2.625. Since F.025,4,7 = 0.110 < 2.625 < 5.52 = F.975,4,7 , we can 0.0982 accept H0 that the variances are equal.

The observed F =

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Section 9.3: Testing H 0 : σ X2 = σ Y2 —The F Test

9.3.3

137

(a) The critical values are F.025,19,19 and F.975,19,19. These values are not tabulated, but in this case, we can approximate them by F.025,20,20 = 0.406 and F.975,20,20 = 2.46. The observed F = 2.41/3.52 = 0.685. Since 0.406 < 0.685 < 2.46, we can assume that the variances are equal. (b) Since t = 2.662 > t.025,38 = 2.0244, reject H0. 3.182 = 0.315. Since F.025,9,9 = 0.248 < 0.315 < 4.03 = F.975,9,9, we can accept 5.67 2 H0 that the variances are equal.

9.3.4

The observed F =

9.3.5

F = 0.202/0.372 = 0.292. Since F.025,9,9 = 0.248 < 0.292 < 4.03 = F.975,.9,9, accept H0.

9.3.6

The observed F = 398.75/274.52 = 1.453. Let α = 0.05. The critical values are F.025,13,11 and F.975,13,11. These values are not in Table A.4, so approximate them by F.025,12,11 = 0.301 and F.975,12,11 = 3.47. Since 0.301 < 1.453 < 3.47, accept H0 that the variances are equal. Theorem 9.2.2 is appropriate.

9.3.7 Let α = 0.05. F = 65.25/227.77 = 0.286. Since F.025,8,5 = 0.208 < 0.286 < 6.76 = F.975,8,5, accept H0. Thus, Theorem 9.2.2 is appropriate. 9.3.8

For these data, s X2 = 56.86 and sY2 = 66.5. The observed F = 66.5/56.86 = 1.170. Since F.025,8,8 = 0.226 < 1.170 < 4.43 = F.975, 8, 8, we can accept H0 that the variances are equal. Thus, Theorem 9.2.2 can be used, as it has the hypothesis that the variances are equal.

9.3.9

If σ X2 = σ Y2 = σ 2 , the maximum likelihood estimator for σ 2 is

σˆ 2 =

1 ⎛ n ( xi − x )2 + ⎜ n + m ⎝ i =1

∑

⎛ ⎞ Then L(ωˆ ) = ⎜ ⎝ 2πσˆ 2 ⎟⎠ 1

m

∑(y − y) i

i =1

( n+ m) / 2

−

e

2

⎞ ⎟. ⎠

n m ⎞ 1 ⎛ ( xi − x ) 2 + ( yi − y )2 ⎟ 2⎜ 2σˆ ⎝ i =1 ⎠ i =1

∑

∑

⎛ 1 ⎞ = ⎜ ⎝ 2πσˆ 2 ⎟⎠

(n+m) / 2

e−( n+m) / 2

If σ X2 ≠ σ Y2 the maximum likelihood estimators for σ X2 and σ Y2 are

σˆ X2 =

1 n

n

∑

( xi − x ) 2 and σˆY2 =

i =1

ˆ)=⎛ 1 ⎞ Then L(Ω ⎜ 2πσˆ 2 ⎟ ⎝ X ⎠ ⎛ 1 ⎞ = ⎜ 2 ⎟ ⎝ 2πσˆ X ⎠

n/2

e

The ratio λ =

−m / 2

n/2

−

e

1 m

m

∑( y − y) i

.

i =1

n ⎞ 1 ⎛ ( xi − x )2 ⎟ 2 ⎜ 2σˆ X ⎝ i=1 ⎠

⎛ 1 ⎞ ⎜ 2πσˆ 2 ⎟ ⎝ Y ⎠

2

∑

⎛ 1 ⎞ ⎜ 2πσˆ 2 ⎟ ⎝ Y ⎠

m/2

−

e

m ⎞ 1 ⎛ ( yi − y )2 ⎟ 2⎜ 2σˆY ⎝ i =1 ⎠

∑

m/2

e− n / 2

L(ωˆ ) (σˆ X2 ) n / 2 (σˆY2 )m / 2 equates to the expression given in the statement of the = ˆ) (σˆ 2 )( n + m ) / 2 L (Ω

question.

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138

Chapter 9: Two-Sample Inferences

9.3.10 Since µX and µY are known, the maximum likelihood estimator uses µX instead of x and µY instead of y . For the GLRT, λ is as in Question 9.3.9 with those substitutions.

Section 9.4: Binomial Data: Testing H 0 : p X = pY 9.4.1

9.4.2

9.4.3

9.4.4

9.4.5

x+ y 55 + 40 = = 0.2375 n + m 200 + 200 x y 55 40 − − n m 200 200 z= = = 1.76 pˆ (1 − pˆ ) pˆ (1 − pˆ ) 0.2375(0.7625) 0.2375(0.7625) + + n m 200 200 Since −1.96 < z = 1.76 < 1.96 = z.025, accept H0. pe =

x+ y 66 + 93 = = 0.188 n + m 423 + 423 x y 66 93 − − n m 423 423 z= = = −2.38 pˆ (1 − pˆ ) pˆ (1 − pˆ ) 0.188(0.812) 0.188(0.812) + + n m 423 423 For this experiment, H0: pX = pY and H1: pX < pY. Since z = −2.38 < −1.64 = −z.05, reject H0. pe =

24 + 27 = 0.836 29 + 32 24 27 − 29 32 z= = −0.17 0.836(0.164) 0.836(0.164) + 29 32 For this experiment, H0: pX = pY and H1: pX ≠ pY. Since −1.96 < z = −0.17 < 1.96 = z.025, accept H0 at the 0.05 level of significance.

Let α = 0.05. pe =

53 + 705 = 0.627 91 + 1117 53 705 − 91 1117 = −0.92 z= 0.627(0.373) 0.627(0.373) + 91 1117 Since −2.58 < z = −0.92 < 2.58 = z.005, accept H0 at the 0.01 level of significance. pe =

1033 + 344 = 0.590 1675 + 660 0.617 − 0.521 = 4.25 z= 0.590(0.410) 0.590(0.410) + 1675 660 Since z = 4.25 > 2.33 = z.01 , reject H0 at the 0.01 level of significance. pe =

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Section 9.4: Binomial Data: Testing H 0 : p X = pY

9.4.6

9.4.7

9.4.8

60 + 48 = 0.54 100 + 100 60 48 − 100 100 z= = 1.70 0.54(0.46) 0.54(0.46) + 100 100 The P value is P(Z ≤ −1.70) + P(Z ≥ 1.70) = 2(1 − 0.9554) = 0.0892. pe =

2915 + 3086 = 0.697 4134 + 4471 2915 3086 − 4134 4471 z= = 1.50 0.697(0.303) 0.697(0.303) + 4134 4471 Since −1.96 < z = 1.50 < 1.96 = z.025, accept H0 at the 0.05 level of significance. pe =

pe =

z=

9.4.9

175 + 100 = 0.358 609 + 160 175 100 − 609 160 = −7.93. Since z = −7.93 < −1.96 = −z.025, reject H0. 0.358(0.642) 0.358(0.642) + 609 160

78 + 50 = 0.256 300 + 200 78 50 − 300 200 z= = 0.25. In this situation, H1 is pX > pY. 0.256(0.744) 0.256(0.744) + 300 200 Since z = 0.25 < 1.64 = z.05, accept H0. The player is right. pe =

9.4.10 From Equation 9.4.1, [(55 + 60) /(160 + 192)](55+ 60) [1 − (55 + 60) /(160 + 192)](160 +192−55− 60) λ= (55 /160)55 [1 − (55 /160)]105 (60 /192)60 [1 − (60 /192)]132 =

115115 (237 237 )(160160 )(192192 ) . We calculate ln λ, which is −0.1935. 352352 (5555 )(105105 )(6060 )(132132 )

2 , accept H0. Then −2ln λ = 0.387. Since −2ln λ = 0.387 < 6.635 = χ .99,1

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139

140

Chapter 9: Two-Sample Inferences

Section 9.5: Confidence Intervals for the Two-Sample Problem 9.5.1

8(0.542 ) + 6(0.362 ) 14

The center of the confidence interval is x − y = 6.7 − 5.6 = 1.1. s p =

1 1 1 1 + = 1.7613(0.47) + = 0.42. The confidence interval n m 9 7 is (1.1 − 0.42, 1.1 + 0.42) = (0.68, 1.52). Since 0 is not in the interval, we can reject the null hypothesis that µX = µY.

= 0.47. The radius is tα / 2,n + m − 2 s p

9.5.2

The center of the confidence interval is x − y = 83.96 − 84.84 = −0.88. The radius is 1 1 1 1 + = 2.2281(11.2) + = 14.61. The confidence interval is (−0.88 − 14.61, 5 7 n m −0.88 + 14.61) = (−15.49, 13.73). Since the confidence interval contains 0, the data do not suggest that the dome makes a difference. tα / 2,n + m − 2 s p

9.5.3

In either case, the center of the confidence interval is x − y = 18.6 – 21.9 = −3.3. 11(115.9929) + 11(35.7604) = 8.71 22 1 1 1 1 The radius of the interval is t.005,22 s p + = 2.8188(8.71) + = 10.02 12 12 12 12 The confidence interval is (−3.3 − 10.02, −3.3 + 10.02) = (−13.32, 6.72). For the case of unequal variances, the radius of the interval is

For the assumption of equal variances, calculate s p =

s 2X sY2 115.9929 35.7604 + = 2.8982 + = 10.31 12 12 12 12 The confidence interval is (−3.3 − 10.31, −3.3 + 10.31) = (−13.61, 7.01). ⎛ ⎞ ⎜ ⎟ X − Y − ( µ X − µY ) Equation (9.5.1) is P ⎜ −tα / 2,n + m − 2 ≤ ≤ tα / 2,n + m − 2 ⎟ = 1 − α 1 1 ⎜ ⎟ Sp + ⎜⎝ ⎟⎠ n m t.005,17

9.5.4

⎛ 1 1 1 1⎞ so P ⎜ −tα / 2,n + m − 2 S p + ≤ X − Y − ( µ X − µY ) ≤ tα / 2,n + m − 2 S p + = 1 − α, or n m n m ⎠⎟ ⎝ ⎛ 1 1 1 1⎞ P ⎜ −( X − Y ) − tα / 2,n + m − 2 S p + ≤ − ( µ X − µY ) ≤ − ( X − Y ) + tα / 2, n + m − 2 S p + = 1 − α. n m n m ⎟⎠ ⎝

Multiplying the inequality above by −1 gives the inequality of the confidence interval of Theorem 9.5.1. 9.5.5

Begin with the statistic X − Y , which has E ( X − Y ) = µX − µY and Var( X − Y ) = σ X2 / n + σ Y2 / m . ⎛ ⎞ X − Y − ( µ X − µY ) Then P ⎜ − zα / 2 ≤ ≤ zα / 2 ⎟ = 1 − α, which implies ⎜⎝ ⎟⎠ σ X2 / n + σ Y2 / m

(

)

P − zα / 2 σ X2 / n + σ Y2 / m ≤ X − Y − ( µ X − µY ) ≤ zα / 2 σ X2 / n + σ Y2 / m = 1 − α.

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Section 9.5: Confidence Intervals for the Two-Sample Problem

Solving the inequality for µX − µY gives

(

141

)

P X − Y − zα / 2 σ X2 / n + σ Y2 / m ≤ µ X − µY ≤ X − Y + zα / 2 σ X2 / n + σ Y2 / m = 1 − α.

(

)

Thus the confidence interval is x − y − zα / 2 σ X2 / n + σ Y2 / m , x − y + zα / 2 σ X2 / n + σ Y2 / m .

9.5.6

The observed ratio is F =

s X2 0.0002103 = 2.20. The confidence interval is = sY2 0.0000955

⎛ s X2 ⎞ s X2 F , F0.975,9,7 ⎟ = (0.238(2.20), 4.82(2.20)) = (0.52, 10.60). Because the confidence 0.025,9,7 2 ⎜ s2 sY ⎝ Y ⎠ interval contains 1, it supports the assumption of Case Study 9.2.1 that the variances are equal.

9.5.7

9.5.8

⎛ s2 ⎞ ⎛ 137.4 s2 137.4 ⎞ (0.146), (5.29) ⎟ The confidence interval is ⎜ X2 F.025,5,7 , X2 F.975,5,7 ⎟ = ⎜ ⎠ 340.3 sY ⎝ sY ⎠ ⎝ 340.3 = (0.06, 2.14). Since the confidence interval contains 1, we can accept H0 that the variances are equal, and Theorem 9.2.1 applies.

Since

SY2 / σ Y2 has an F distribution with m − 1 and n − 1 degrees of freedom, S X2 / σ X2

⎛ ⎞ ⎛ S2 ⎞ S 2 /σ 2 σ 2 S2 P ⎜ Fα / 2, m −1,n −1 ≤ 2Y Y2 ≤ F1−α / 2,m −1, n−1 ⎟ = P ⎜ X2 Fα / 2,m −1,n −1 ≤ X2 ≤ X2 F1−α / 2,m−1,n −1 ⎟ = 1 − α . SX /σ X σ Y SY ⎝ ⎠ ⎝ SY ⎠ The inequality provides the confidence interval of Theorem 9.5.2.

9.5.9

The center of the confidence interval is

x y 126 111 − = − = 0.015. The radius is n m 782 758

y⎞ ⎛ x⎞⎛ x⎞ ⎛ y ⎞⎛ ⎛ 126 ⎞ ⎛ 126 ⎞ ⎛ 111 ⎞ ⎛ 111 ⎞ 1− 1− ⎜⎝ ⎟ ⎜1 − ⎟ ⎜ ⎟ ⎜1 − ⎟ ⎝⎜ n ⎠⎟ ⎝⎜ n ⎠⎟ ⎝⎜ m ⎠⎟ ⎝⎜ m ⎠⎟ 782 ⎠ ⎝ 782 ⎠ ⎝ 758 ⎠ ⎝ 758 ⎠ + = 1.96 + = 0.036. z.025 n m 782 758 The 95% confidence interval is (0.015 − 0.036, 0.015 + 0.036) = (−0.021, 0.051) Since 0 is in the confidence interval, one cannot conclude a significantly different frequency of headaches.

9.5.10 The center of the confidence interval is

x y 55 60 − = − = 0.031. The radius is n m 160 192

y⎞ 55 ⎞ ⎛ 60 ⎞ ⎛ 60 ⎞ ⎛ x⎞⎛ x⎞ ⎛ y ⎞⎛ ⎛ 55 ⎞ ⎛ ⎜⎝ ⎟⎠ ⎝⎜1 − ⎟⎠ ⎝⎜ ⎟⎠ ⎝⎜1 − ⎟ ⎜⎝ ⎠⎟ ⎝⎜1 − ⎠⎟ ⎝⎜ ⎠⎟ ⎝⎜1 − ⎠⎟ n n m m 160 160 192 192 ⎠ z.10 + = 1.28 + = 0.064. n m 160 192 The 80% confidence interval is (0.031 − 0.064, 0.031 + 0.064) = (−0.033, 0.095)

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142

Chapter 9: Two-Sample Inferences

9.5.11 The approximate normal distribution implies that ⎛ ⎞ X Y − − ( p X − pY ) ⎜ ⎟ n m P ⎜ − zα ≤ ≤ zα ⎟ = 1 − α ( X / n)(1 − X / n) (Y / m)(1 − Y / m) ⎜ ⎟ + ⎜⎝ ⎟⎠ n m ⎛ ( X / n)(1 − X / n) (Y / m)(1 − Y / m) X Y or P ⎜ − zα + ≤ − − ( p X − pY ) n m n m ⎝ ≤ zα

( X / n)(1 − X / n) (Y / m)(1 − Y / m) ⎞ + ⎟ = 1 − α which implies that n m ⎠

⎛ ⎛X Y ⎞ ( X / n)(1 − X / n) (Y / m)(1 − Y / m) + ≤ − ( p X − pY ) P ⎜ − ⎜ − ⎟ − zα ⎝ ⎠ n m n m ⎝ ⎛X Y ⎞ ≤ − ⎜ − ⎟ + zα ⎝ n m⎠

( X / n)(1 − X / n) (Y / m)(1 − Y / m) ⎞ + ⎟ = 1− α n m ⎠

Multiplying the inequality by −1 yields the confidence interval. 9.5.12 The center of the confidence interval is

x y 106 13 − = − = −0.083. The radius is n m 3522 115

106 ⎞ ⎛ 13 ⎞ ⎛ 13 ⎞ y⎞ ⎛ x⎞⎛ x⎞ ⎛ y ⎞⎛ ⎛ 106 ⎞ ⎛ ⎜⎝ ⎠⎟ ⎝⎜1 − ⎠⎟ ⎝⎜ ⎠⎟ ⎝⎜1 − ⎠⎟ ⎜⎝ ⎟⎠ ⎝⎜1 − ⎟⎠ ⎝⎜ ⎟⎠ ⎝⎜1 − ⎟ 3522 3522 115 115 ⎠ n n m m + = 1.96 + = 0.058 z.025 3522 115 n m The 95% confidence interval is (−0.083 − 0.058, −0.083 + 0.058) = (−0.141, −0.025) Since the confidence interval lies to the left of 0, there is statistical evidence that the suicide rate among women members of the American Chemical Society is higher.

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Chapter 10: Goodness-of-Fit Tests Section 10.2: The Multinomial Distribution 10.2.1 Let Xi = number of students with a score of i, i = 1, 2, 3, 4, 5. Then P(X1 = 0, X2 = 0, X3 = 1, 6! (0.116)0(0.325)0(0.236)1(0.211)2(0.112)3 = 0.000886. X4 = 2, X5 = 3) = 0!0!1!2!3! 10.2.2 Let X1 = number of round and yellow phenotypes, X2 = number of round and green phenotypes, 1

and so on. Then P(X1 = 1, X2 = 1, X3 = 1, X4 = 1) =

4! ⎛ 9 ⎞ ⎜ ⎟ 1!1!1!1! ⎝ 16 ⎠

1

⎛3⎞ ⎝⎜ 16 ⎠⎟

1

⎛3⎞ ⎝⎜ 16 ⎠⎟

1

⎛1⎞ = 0.0297. ⎝⎜ 16 ⎠⎟

10.2.3 Let Y denote a person’s blood pressure and let X1, X2, and X3 denote the number of individuals with blood pressures less than 140, between 140 and 160, and over 160, respectively. If µ = 124 140 − 124 ⎞ ⎛ and σ = 13.7, p1 = P(Y < 140) = P ⎜ Z < ⎟ = 0.8790, ⎝ 13.7 ⎠ 160 − 124 ⎞ ⎛ 140 − 124 p2 = P (140 ≤ Y ≤ 160) = P ⎜ ≤Z≤ ⎟ = 0.1167, and p3 = 1 − p1 − p2 = 0.0043. ⎝ 13.7 13.7 ⎠ 10 ! Then P(X1 = 6, X2 = 3, X3 = 1) = (0.8790)6 (0.1167)3 (0.0043)1 = 0.00265. 6!3!1!

10.2.4 Let Y denote a recruit’s IQ and let Xi denote the number of recruits in class i, i = 1, 2, 3. Then 90 − 100 ⎞ ⎛ p1 = P(class I) = P(Y < 90) = P ⎜ Z < ⎟ = 0.2643, p2 = P(class II) = P(90 ≤ Y ≤ 110) ⎝ 16 ⎠ 110 − 100 ⎞ ⎛ 90 − 100 ≤Z≤ = P⎜ ⎟⎠ = 0.4714, and p3 = P(class III) = P(Y > 110) = 1 − p1 − p2 ⎝ 16 16 7! = 0.2643. From Theorem 10.2.1, P(X1 = 2, X2 = 4, X3 = 1) = (0.2643)2 (0.4714) 4 (0.2643)1 2!4!1! = 0.0957.

10.2.5 Let Y denote the distance between the pipeline and the point of impact. Let X1 denote the number of missiles landing within 20 yards to the left of the pipeline, let X2 denote the number of missiles landing within 20 yards to the right of the pipeline, and let X3 denote the number of missiles for 5 which |y| > 20. By the symmetry of fY(y), p1 = P(−20 ≤ Y ≤ 0) = = P(0 ≤ Y ≤ 20) = p2, 18 5 5 8 − = . Therefore, P(X1 = 2, X2 = 4, X3 = 0) so p3 = P(|Y| > 20) = 1 − 18 18 18 2 4 0 6! ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛ 8 ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 0.00689. 2!4!0! ⎝ 18 ⎠ ⎝ 18 ⎠ ⎝ 18 ⎠

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144

Chapter 10: Goodness-of-Fit Tests

10.2.6 Let Xi, i = 1, 2, 3, 4, 5, denote the number of outs, singles, doubles, triples, and home runs, respectively, that the player makes in 5 at-bats. Then P(two-outs, two singles, one double) 5! = P(X1 = 2, X2 = 2, X3 = 1, X4 = 0, X5 = 0) = ⋅ (0.713)2(0.270)2(0.010)1(0.002)0(0.005)0 2!2!1!0!0! = 0.0111. 1/ 4 1/ 2 1⎞ 1 1⎞ 7 ⎛ ⎛1 10.2.7 (a) p1 = P ⎜ 0 ≤ Y < ⎟ = 3 y 2 dy = , p2 = P ⎜ ≤ Y < ⎟ = , 3 y 2 dy = 0 1/ 4 ⎝ ⎝4 4⎠ 64 2⎠ 64 3/ 4 1 3⎞ 19 37 ⎛1 ⎛3 ⎞ p3 = P ⎜ ≤ Y < ⎟ = , and p4 = P ⎜ ≤ Y ≤ 1⎟ = . 3 y 2 dy = 3 y 2 dy = 1/ 2 3/ 4 ⎝2 ⎝4 ⎠ 4⎠ 64 64 Then f X1 , X 2 , X 3 , X 4 (3,7,15,25) = P(X1 = 3, X2 = 7, X3 = 15, X4 = 25)

∫

∫

∫

∫

3

=

7

15

25

50! ⎛ 1 ⎞ ⎛ 7 ⎞ ⎛ 19 ⎞ ⎛ 37 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ . 3!7!15!25! ⎝ 64 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠

(b) By Theorem 10.2.2, X3 is a binomial random variable with parameters n (= 50) and ⎛ 19 ⎞ ⎛ 19 ⎞ ⎛ 45 ⎞ p3 ⎜ = ⎟ . Therefore, Var(X3) = np3(1 − p3) = 50 ⎜ ⎟ ⎜ ⎟ = 10.44. ⎝ 64 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ 10.2.8 M X1 , X 2 , X 3 (t1 , t2 , t3 ) = =

∑∑∑ e

t1k1 + t2 k2 + t3k3

⋅

n! ⋅ p1k1 p2k2 p3k3 k1 !k2 !k3 !

n! ∑∑∑ k !k !k ! ( p e ) ( p e ) ( p e ) 1

1

2

t1 k1

2

t2 k 2

3

t3 k3

, where the summation extends over all the values

3

of (k1, k2, k3) such that ki ≥ 0, i = 1, 2, 3 and k1 + k2 + k3 = n. Recall Newton’s binomial expansion. Applied here, it follows that the triple sum defining the moment-generating function

(

for (X1, X2, X3) can also be written p1et1 + p2 et2 + p3et3

(

10.2.9 Assume that M X1 , X 2 X 3 (t1 , t2 , t3 ) = p1et1 + p2 et2 + p3et3

(

= E (et1 X1 ) = p1et1 + p2 + p3

)

n

). n

) . Then M n

X1 , X 2 , X 3 (t1 ,0,0)

= (1 − p1 + p1et1 )n is the mgf for X1. But the latter has the form of

the mgf for a binomial random variable with parameters n and p1. 10.2.10 The log of the likelihood vector (k1, k2, …, kt) is log L = log p1k1 p2k2 ... ptkt = k1 log p1 + t

k2 log p2 +…+ kt log pt, where the pi’s are constrained by the condition that

∑p

i

= 1. Finding

i =1

the MLE for the pi’s can be accomplished using Lagrange multipliers. Differentiating t t ⎤ k ∂ ⎡ log L − λ pi with respect to each pi gives log L − λ pi ⎥ = i − λ , ⎢ ∂pi ⎣ i =1 i =1 ⎦ pi

∑

∑

i = 1, 2, …, t. But these derivatives equal 0 only if t

together with the fact that

∑p

i

i =1

ki = λ for all i. The latter equations, pi

= 1, imply that pˆ i =

ki , i = 1, 2, …, t. n

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Section 10.3: Goodness-of-Fit Tests: All Parameters Known

145

Section 10.3: Goodness-of-Fit Tests: All Parameters Known t

10.3.1

∑ i =1

( X i − npi ) 2 = npi

t

∑ i =1

( X i2 − 2npi X i + n 2 pi2 ) = npi

t

∑ i =1

t t X i2 − 2 X i + n pi = npi i =1 i =1

∑

∑

t

∑ i =1

X i2 −n. npi

⎛4⎞ ⎛6⎞ 10.3.2 If the hypergeometric model applies, π1 = P(0 whites are drawn) = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 2⎠ ⎛ 4⎞ ⎛6⎞

⎛10 ⎞

⎛10 ⎞ 15 ⎜⎝ 2 ⎟⎠ = 45 ,

24

π2 = P(1 white is drawn) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = , and π3 = P(2 whites are drawn) ⎝1 ⎠ ⎝1 ⎠ ⎝ 2 ⎠ 45 ⎛ 4⎞ ⎛6⎞ = ⎜ ⎟⎜ ⎟ ⎝ 2⎠ ⎝0⎠

⎛10 ⎞ 6 ⎜⎝ 2 ⎟⎠ = 45 . Let p1, p2, and p3 denote the actual probabilities of drawing 0, 1, and 2 15 24 6 white chips, respectively. To test H0: p1 = , p2 = , p3 = versus H1: at least one pi ≠ πi, 45 45 45 2 = 4.605. reject H0 if d ≥ χ 12−α ,k −1 = χ .90,2 (35 − 100(15 / 45))2 (55 − 100(24 / 45)) 2 (10 − 100(6 / 45)) 2 = 0.96, so Here, d = + + 100(15 / 45) 100(24 / 45) 100(6 / 45) H0 (and the hypergeometric model) would not be rejected.

10.3.3 If the sampling is presumed to be with replacement, the number of white chips selected would 0 2 ⎛2⎞ ⎛ 4 ⎞ ⎛ 6 ⎞ follow a binomial distribution. Specifically, π1 = P(0 whites are drawn) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 1

1

⎛2⎞ ⎛ 4 ⎞ ⎛ 6 ⎞ = 0.36, π2 = P(1 white is drawn) = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 0.48, and π3 = P(2 whites are drawn) ⎝1 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 2 0 ⎛2⎞ ⎛ 4 ⎞ ⎛ 6 ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 0.16. The form of the α = 0.10 decision rule is reject H0 if ⎝ 2 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 2 = 4.605. In this case, though, d ≥ χ 12−α ,k −1 = χ .90,2

(35 − 100(0.36)) 2 (55 − 100(0.48))2 (10 − 100(0.16))2 = 3.30. The null hypothesis d= + + 100(0.36) 100(0.48) 100(0.16) that the sampling occurred with replacement is not rejected.

10.3.4 If births occur randomly in time, then π1 = P(baby is born between midnight and 4 A.M.) =

1 6

5 . Let p1 and p2 denote the actual 6 probabilities of birth during those two time periods. The null hypothesis to be tested is 1 5 H0: p1 = , p2 = . At the α = 0.05 level of significance, H0 should be rejected if 6 6 2 d ≥ χ .95,1 = 3.841. Given that n = 2650 and that X1 = number of births between midnight and 4

and π2 = P(baby is born at a “convenient” time) = 1 − π1 =

(494 − 2650(1/ 6)) 2 (2156 − 2650(5 / 6)) 2 = 7.44. Since the + 2650(1/ 6) 2650(5 / 6) latter exceeds 3.841, we reject the hypothesis that births occur uniformly in all time periods.

A.M. = 494, it follows that d =

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146

Chapter 10: Goodness-of-Fit Tests

1 1 versus H1: p ≠ . Let 6 6 n = 2650 be the number of births and k = 494 is the number of babies born between midnight and 4 A.M. From Theorem 6.3.1, H0 should be rejected if z is either ≤ −1.96 or 494 − 2650(1/ 6) ≥ 1.96 = z.025 . Here z = = 2.73, so H0 is rejected. These two test procedures 2650(1/ 6)(5 / 6)

10.3.5 Let p = P(baby is born between midnight and 4 A.M.). Test H0: p =

2 are equivalent: If one rejects H0, so will the other. Notice that z.025 = (1.96)2 = 3.84 2 = χ .95,1 and (except for a small rounding error) z2 = (2.73)2 = 7.45 = χ2 = 7.44.

10.3.6 In the terminology of Theorem 10.3.1, X1 = 1383 = number of schizophrenics born in first quarter and X2 = number of schizophrenics born after the first quarter. By assumption, nπ1 = 1292.1 and nπ2 = 3846.9 (where n = 5139). The null hypothesis that birth month is unrelated to (1383 − 1292.1) 2 (3756 − 3846.9)2 2 schizophrenia is rejected if d ≥ χ .95,1 = 3.841. But d = + 1292.1 3846.9 = 8.54, so H0 is rejected, suggesting that month of birth may, indeed, be a factor in the incidence of schizophrenia. 10.3.7 Listed in the accompanying table are the observed and expected numbers of M&Ms of each color. Let p1 = true proportion of browns, p2 = true proportion of yellows, and so on. Color Brown Yellow Red Orange Blue Green

Observed Frequency 455 343 318 152 130 129 1527

pi 0.3 0.2 0.2 0.1 0.1 0.1

Expected Frequency 458.1 305.4 305.4 152.7 152.7 152.7 1527

To test H0: p1 = 0.30, p2 = 0.20, …, p6 = 0.10 versus H1: at least one pi ≠ πi, reject H0 if (455 − 458.1) 2 (129 − 152.7)2 2 d ≥ χ .95,5 = 11.070. But d = = 12.23, so H0 is rejected (these + ... + 458.1 152.7 particular observed frequencies are not consistent with the company’s intended probabilities). 10.3.8 Let the random variable X denote the length of a World Series. Then P(X = 4) = π1 4

1 ⎛1⎞ = P(AL wins in 4) + P(NL wins in 4) = 2 ⋅ P(AL wins in 4) = 2 ⎜ ⎟ = . Similarly, ⎝2⎠ 8 P(X = 5) = π2 = 2 ⋅ P(AL wins in 5) = 2 ⋅ P(AL wins exactly 3 of first 4 games) ⋅ P(AL wins 5th game) 3 1 ⎛4⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 1 = 2 ⋅ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅ = . Also, P(X = 6) = π3 = 2⋅ P(AL wins exactly 3 of first 5 games) ⋅ ⎝3 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 4 3

2

⎛5⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 5 P(AL wins 6th game) = 2 ⋅ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅ ⎜ ⎟ = , and P(X = 7) = π4 ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 16

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Section 10.3: Goodness-of-Fit Tests: All Parameters Known

147

5 . Listed in the table is the information necessary for 16 2 calculating the goodness-of-fit statistic d. The “Bernoulli model” is rejected if d ≥ χ .90,3 = 6.251.

= 1 − P(X = 4) − P(X = 5) − P(X = 6) =

For these data, d =

(9 − 6.25) 2 (11 − 12.50) 2 (8 − 15.625) 2 (22 − 15.625) 2 + + + = 7.71, so H0 is 6.25 12.50 15.625 15.625

rejected. Number of Games 4 5 6 7

Observed Frequency 9 11 8 22 50

Expected Frequency 6.25 12.5 15.625 15.625 50

10.3.9 Let pi = P(horse starting in post position i wins), i = 1, 2, …, 8. One relevant null hypothesis to 1 test would be that pi is not a function of i—that is, H0: p1 = p2 = … = p8 = versus H1: at least 8 1 2 one pi ≠ . If α = 0.05, H0 should be rejected if d ≥ χ .95,7 = 14.067. Each E(Xi) in this case is 8 1 (32 − 18.0)2 (21 − 18.0) 2 (11 − 18.0) 2 144 ⋅ = 18.0, so d = + + ... + = 18.22. Since 8 18.0 18.0 18.0 2 , we reject H0 (which is not surprising because faster horses are often awarded 18.22 ≥ χ .95,7 starting positions close to the rail). 10.3.10 Listed is the frequency distribution for the 70 yi’s using classes of width 10 starting at 220. If normality holds, each πi is an integral of the normal pdf having µ = 266 and σ = 16. Duration 220 ≤ y < 230 230 ≤ y < 240 240 ≤ y < 250 250 ≤ y < 260 260 ≤ y < 270 270 ≤ y < 280 280 ≤ y < 290 290 ≤ y < 300

Observed Frequency 1 5 10 16 23 7 6 2 70

πi 0.0122 0.0394 0.1071 0.1933 0.2467 0.2119 0.1226 0.0668

Expected Frequency 0.854 2.758 7.497 13.531 17.269 14.833 8.582 4.676 70

⎛ 230 − 266 Y − 266 240 − 266 ⎞ For example, π2 = P(230 ≤ Y < 240) = P ⎜ ≤ < ⎟⎠ ⎝ 16 16 16 = P(−2.25 ≤ Z < −1.63) = 0.0394. To account for all the area under fY(y), the intervals defining the first and last classes need to be extended to −∞ and +∞, respectively. That is, π1 = P(−∞ < Y < 230) and π8 = P(290 ≤ Y < ∞). Some of the expected frequencies (= 70 ⋅ πi) are too small (i.e., less than 5) for the χ 2 approximation to be fully adequate.

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148

Chapter 10: Goodness-of-Fit Tests

The first three classes need to be combined, giving 0.854 + 2.758 + 7.497 = 11.109. Also, the last two class should be combined to yield 8.582 + 2.758 = 13.258. With t = 5 final classes, then, the 2 = 7.779. Here, d = normality assumption is rejected if d ≥ χ .90,4 (16 − 11.109) 2 (8 − 13.258) 2 + ... + 11.109 13.258 = 10.73, so we would reject the null hypothesis that pregnancy durations are normally distributed.

10.3.11 Let the random variable Y denote the prison time served by someone convicted of grand theft auto. In the accompanying table is the frequency distribution for a sample of 50 yi’s, together 1 with expected frequencies based on the null hypothesis that fY(y) = y 2 , 0 ≤ y ≤ 3. For example, 9 11 E(X1) = 50 ⋅ π1 = 50 y 2 dy = 1.85. Combining the first two intervals (because E(X1) < 5) 09 1 2 yields k = 2 final classes, so H0: fY(y) = y 2 , 0 ≤ y ≤ 3 should be rejected if d ≥ χ .95,1 = 3.841. 9 (24 − 14.81) 2 (26 − 35.19)2 But d = + = 8.10, implying that the proposed quadratic pdf does not 14.81 35.19 provide a good model for describing prison time.

∫

Prison Time 0≤y