Collins Cambridge International AS & A Level Mathematics Probability and Statistics 1 Worked Solutions 9780008257767, 9780008482985

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Cambridge International AS & A Level Mathematics

Probability & Statistics 1 STUDENT’S BOOK: Worked solutions

Louise Ackroyd, Jonny Griffiths, Yimeng Gu Series Editor: Dr Adam Boddison

Pure Mathematics 1 International Students Book Title page.indd 1 57736_Pi_viii.indd 1 WS TITLE PAGE_Probability & Statistics 1.indd 1

14/11/17 10:46 pm 6/18/18 10:46 3:21 PM 31/07/18 AM

1

WORKED SOLUTIONS

Worked Solutions 1 Representation of data Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication of the answer.

Prerequisite knowledge

Exercise 1.1A

1

1

a The data is quantitative, as it is numerical, and discrete, as there are distinct individual values. (The original data could have been continuous, and then rounded to the nearest mark.)

b Monthly texts you send on your mobile is discrete, as you can count them.

b Median = 35, mode = 41 and range = 35 c

Grouped scores

c The number of burgers sold in a fast food restaurant is discrete, as you can count them.

Frequency

10–19

6

20–29

13

30–39

13

40–49

18

d The duration of a marathon is continuous, as you measure it.

A histogram with equal class widths is a sensible way to display the data.

Frequency

2 20

c An IGCSE grade in maths as a letter is qualitative.

10 5

19.5

29.5

39.5

d Examination scores in maths are quantitative and discrete.

49.5

Grouped scores

e Waist size is quantitative and continuous, unless it is given to the nearest cm/inch, in which case it is discrete.

The results show a modal group of 40–49. 2

3

e The age of a teacher in your school is continuous, as time is measured. But if someone is asked their age, they reply with a whole number, and that version of ‘age’ is discrete. a Gender is qualitative. b Height is quantitative, as it has a numerical meaning and is continuous.

15

0 9.5

a Daily rainfall in Penang is continuous, as you measure it.

Edam: 90 × 120 = 30 , Stilton: 102 × 120 = 34, 360 360 168 × 120 = 56 Cheddar: 360

f ‘Car owner or not’ is qualitative.

Score

1

2

3

4

5

6

Frequency

3

4

2

3

3

5

Mode = most common value = 6 Mean = sum of all the values divided by 20 = 74 = 3.7 20 Median = middle value when results are ordered = 4 Range = top result − bottom result = 6 − 1 = 5

3

g Weekly self-study time is quantitative and continuous unless measured to the nearest hour, in which case it is discrete. a 9 b 9 c 13.05 d The mean takes into account all the visits made, so it is the best measure to use.

1

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1 Representation of data

4 a Farah’s first week sums to 22.4 mm. 22.4 + 84.6 = 107, and 107 = 3.82 mm 28 b No effect, since 9.4 + 0.5 = 4.9 + 5.0. 5

Number of times stopped

Number of journeys

10 a Median = 39th value = 3 years Mode = 2 years



Mean = 276 ÷ 77 = 3.58 years

b The quartiles LQ, median and UQ divide the ordered data into four equal sets.

fx



1

3

3

2

5

10

3

11

33

4

21

84

5

22

110

6

17

102

7

14

98

8

7

56

100

496

Total



LQ = 19.5th value = 2, UQ = 58.5th value = 5, IQR = 3

c

1

2

3

4

5

6

7

8

9

10

Exercise 1.2A 1 Range = 41, Q1 = 20, Q3 = 44.5, so IQR = 24.5 2

∑ x = 1705,∑ x 2 = 537 615 so

( )

537 615 1705 2 − = 22 920 (4 s.f.), s.d.(x) = 151 (3 s.f.) 15 15 2 = 22 920 (4 s.f.), s.d.(x) = 151 (3 s.f.) 3 For example, −a, −a, −1, −1, −1, 0, 1, 1, 1, a, a Mean = 496 = 4.96 100 where a = 42.5 = 6.52 (3 s.f.) The best average to represent the data is the mean, IQR = 2, s.d. = 4 as it uses all the data, which contains no outliers. 4 Range = 20 – 5 = 15 6 a Mean = ∑ f × mid-point and so IQR = 14 – 8 = 6 ∑f Median = 100 + 1 = 50.5th value, so 5 2 537 615 1705 var(x) = − 15 15 Mode = 5

( )

∑ x = 191,∑ x 2 = 2275

6187.5 ÷ 320 = 19.335… = 19.3

b Since all the mid-points increase by a factor of 1.1, the mean increases by a factor of 1.1. c Since all the mid-points increase by a, the mean increases by a. d a = 0.1 × 19.3 = 1.93 7

var(x) =

Height, h (cm)

Frequency

fx

100 < h  120

5

550

120 < h  140

4

520

140 < h  160

12

1800

160 < h  180

13

2210

180 < h  200

8

1520

Total

42

6600

Estimate of the mean is correct: 6600 = 157.14 42 8 24 × 84.1 + 4 × 240.3 + 51 × 99.2 = 101.8 kg f × mid-point 79 9 i Answers to a, c and d are qualitative; answers to b, e and f are quantitative. ii  Answer to b is discrete; answers to e and f are continuous. iii Check students' answers

( )

2275 − 191 18 18

σ= s.d.(x)

2

= 3.71

There is an argument for the IQR, since this is unaffected by the high value of 20. On the other hand, the standard deviation does utilise all the data.

∑ xi

= 3985.1 = 332.091... = 332.1 cm 3 12 2 2 x 2 Variance = ∑ i − x = 1 323 540.07 − 3985.1 12 12 n

5 Mean =

12

(

)

= 10.1307… and so standard deviation = 3.18 cm3 6 Method 1 2

2

2

 1 − 31  +  4 − 31  +  5 − 31  +  6 − 31   5   5   5  5  

2

2

+  15 − 31  = 110.8  5 s.d.(x) σ = 110.8 = 4.71 (3 s.f.) 5 Method 2

∑ x = 31, ∑ x 2 = 303, s.d.(x) =

( )

303 − 31 5 5

2

= 4.71 (3 s.f.)

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Variance of combined samples WORKED SOLUTIONS

( ) = n(b + a ) + n(a + b ) − ( a + b ) 2n 2 =

7

∑b = 64.4, ∑ g = 68.6 ⇒ ∑b = 1030.4, ∑ g = 891.8



⇒ ∑ b, g = 1922.2 ⇒ b, g = 66.3 (3sf . .)

16

∑ b 2 − 64.42 ⇒ ∑ b 2 = 66 520.96, 16

∑ g 2 − 68.62 ⇒ ∑ g 2 = 61 295.78 9.1 = 13

⇒ ∑ b 2, g 2 = 127 816.74 ⇒ var(b, g ) =

(

127 816.74 1922.2 − 29 29

) = 14.1 (3 s.f.) 2

8 a This is FALSE. The variance could be 0.64, in which case the standard deviation would be 0.8. b This is TRUE. The range tells you the span of all the data, while the IQR tells you where the central 50% of the data lies. c This is FALSE. Given a set of data, we can subtract the same number from every member of the set to create a new set with a smaller mean but with an unchanged variance. d This is FALSE. The data set 1, 2, 3, 4, 5 has IQR = 3 and variance = 2. 9 L  et x be the number of sweets in a bag sampled on the first day, and y the number of sweets in a bag sampled on the second day.

a  ∑ x = 300,∑ y = 360,∑ x 2 = 3360,∑ y 2 = 4620







b In general ∑ x = na,∑ y = nb,∑ x 2 = n(b + a 2),











So mean(x, y) = 11, var(x, y) = 12

∑ y 2 = n(a + b 2)

2 So mean (x, y) = a + b , var(x, y) = a + b + (a − b) 2 4 2 2 ( a − b ) so var(x, y) – mean(x, y) = 0 4

− a+b 2

2

2

2



For the combined samples,



variance – mean =

a + b + (a − b)2 − a + b = (a − b)2  0 2 4 2 4  so variance mean. 10 a The range has the advantage that it is easy to calculate, but the disadvantage that it is affected significantly by extreme values. The company may have a few very highly paid people that could distort the picture. b The interquartile range is more sensible than the range as a measure of spread since it gives you the range for the central 50% of the data, so is unaffected by extreme values. However, in concentrating on the central 50%, it has the disadvantage of ignoring 50% of the data. c The variance includes all the data in its calculation, which is an advantage, but extreme values can still have a distorting impact. d You can say the same for the standard deviation as the variance, but one advantage of the s.d. over the variance is that its units are the same as for the data. 11 a She needs to score 74.

b  24, 34, 37, 39, 42, 54, a 20 She needs to score anything more than 54. 2

∑ x 2 − a 2 = b ⇒ ∑ x 2 = n(b + a 2)

9302 + a 2 −  230 + a  = 112 ⇒ a = 57 c   7  7 She needs to score 57.

∑ y 2 − b 2 = a ⇒ ∑ x 2 = n(a + b 2)



Then n

n

x + ∑ y na + nb a + b Combined mean = ∑ = = 2n 2n 2 Variance of combined samples



( ) = n(b + a ) + n(a + b ) − ( a + b ) 2n 2 =

∑x2 + ∑ y2 − 2n

2

a+b 2 2

1 

2 2 2 2 2 = b + a + a + b − a + ab + b 2 4 2 − 2ab − b 2 2 2 2 2 2 2 = b+ a + a+ b −a 4 2 b + a a − ab + b2 2 2 2 = + 4 4 2 = a + b + (a − b) 2 4

Let x represent data values from the first sample, and y data values from the second sample.

∑ x = na, ∑ y = nb

2n

2

2

13

10.2 =

∑x + ∑ y 2

2

2

2

(

9302 + a 2 −  230 + a  = 10 2 d   7  7 She needs to score 71.

) ⇒ a = 71 2

Exercise 1.3A 1 Various ideas such as height, weight, number of texts sent, number of music tracks owned, etc. Discrete variables are counted, while continuous data is measured.

2 2 2 2 = b + a + a + b − a + 2ab + b 2 4 3 2 2 2 2 2b −Cambridge a − 2ab −International b ©HarperCollinsPublishers AS & A Level Mathematics: Probability & Statistics 1 9780008257767 = 2b + 2a + 2a + 2018 4 2 2 = 2b + 2a + a − 2ab + b 4 4 57767_P001_013.indd 3 04/07/18 a + b (a − b)2

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1 REPRESENTATION OF DATA

2

b Median is similar but Molly’s is slightly higher, suggesting that she scores more. Larger range for Molly, suggesting that Jenson is more consistent overall. On the other hand, Jenson has a slightly larger IQR, suggesting that Molly’s middle 50% of scores are more consistent.

For a histogram, the frequency a bar represents is proportional to its area. Since 47 goes with an area of 8 cm2, one bean is represented by 8 cm2. 47 Since data is measured to the nearest cm, the true class intervals are: 5

2.5–8.5, 8.5–13.5, 13.5–25.5

b Q0: 36, Q1: 46, Q2: 56, Q3: 64, Q4: 73

These have widths of 6, 5, and 12 respectively.

(IQR = 18)

The width of 6 goes with the given width of 2 cm, so the other classes have widths 5 cm 3 and 4 cm. The 22 beans in the second class are represented by an area of 8 × 22 cm2 = 3.75, 47 and dividing by 5 gives a height of 2.25 cm. 3 5 a Width = cm 3 b Height = 2.25 cm

c

30

6

a We can treat the data as being continuous here. Grouping the data means a histogram is appropriate. b Because it includes both start and end values. c

True distance (km)

Number of people

40

50

60

70

80

Test mark Price of printer ($)

Frequency

Class width

Frequency density

50 < x  70

24

20

1.20

70 < x  100

31

30

1.03

100 < x  130

42

30

1.40

Frequency density

130 < x  160

16

30

0.53

160 < x  200

18

40

0.45

200 < x  300

4

100

0.04

39.5–49.5

67

67 = 6.70 10

49.5–59.5

124

124 = 12.40 10

59.5–64.5

4023

4023 = 804.6 5

64.5–69.5

2981

2981 = 596.2 5

69.5–84.5

89

89 = 5.93 15

84.5–149.5

75

75 = 1.15 65

1.4

Frequency density

3

a Mode = 60

1.2 1.0 0.8 0.6 0.4 0.2 50

100

150

200

250

300

Printer cost ($) 4

a Jenson Q0: 3, Q1: 12, Q2: 31, Q3: 55, Q4: 66 Molly

The histogram has a hump to the left and a tail to the right. Most of the printers cost less than $150, but the small number of more expensive printers means the mean price will be above the median price.

Q0: 12, Q1: 17, Q2: 34, Q3: 57, Q4: 98

Jenson Molly

7

a i True ii True iii False. The IQR for B is smaller than that for A.

0

10 20 30 40 50 60 70 80 90 100

iv False. Q3 − Q2 is larger for B.

Number of runs

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1

WORKED SOLUTIONS

8

b There is a larger range for Class B than for Class A. The IQR for B is smaller than that for A, however, so the middle 50% in class B is more tightly grouped. The medians for the two student groups are close, suggesting that attainment is roughly equal for the two groups. a

Exercise 1.4A 1

Number in family

Number of families

Cumulative frequency

1

15

15

2

20

35

3

22

57

4

23

80

Class Frequency width density

Mass of raisin (g)

Frequency

0 < x  0.5

4

0.5

8

5

11

91

0.5 < x  1.0

8

0.5

16

6

4

95

1.0 < x  2.0

14

1

14

2.0 < x  3.0

6

1

6

3.0 < x  3.5

4

0.5

8

Median = 48th value, and is approximately 2.6 from the cumulative frequency diagram.

Cumulative frequency

Frequency density

15

10

5

0.5

1

1.5

2

2.5

3

65 88776420 955333310 55420 630

70 60

×

50 40

×

30 20

0

× Median = 2.6 1

2

3

After 0 1 2 3 4 5

5589 123566899 0004467789 089 Key: 4 2 3 = 24 marks before, 0 23 marks after

The median mark for the group has risen from 23 to 30; the lesson seems to have had an effect (although there could be other reasons for the increase).

4

5

6

Number in family

b The 0.5–1.0 class Before

×

×

80

10

3.5

Mass of raisin (g)

9

×

90

20

2

Kick distance, l (m)

Upper bound

Cumulative frequency

5  l < 10

10

5

10  l < 20

20

58

20  l < 30

30

87

30  l < 50

50

102

50  l < 70

70

113

70  l < 100

100

120

10 You have Q1 − 1.5 × (Q3 − Q1) = 3, Q3 + 1.5 × (Q3−Q1) = 83 Rearranging gives 5Q1 − 3Q3 = 6, 5Q3 − 3Q1 = 166 Solving these simultaneously gives Q1 = 33, Q3 = 53

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1 REPRESENTATION OF DATA

a

b a = 15, b = 85, central 70% of results lie in (31,72). 120

c Mean =

110

The mean and median are similar, so the distribution is fairly symmetrical.

Cumulative frequency

100 90 80

4

70

a Mode = 60 b Q1: 46, Q2: 56, Q3: 64

60

(IQR = 18)

50

c

40

50

30

45

Cumulative frequency

20

P40 = 18

10 0

10 20 30 40 50 60 70 80 90 100 110 120

Kick distance, l (m) b 40 × 120 = 48th value ≈ 18 100 Kick distance, l (m)

c

Midpoint

Frequency

fx

5

37.5

10  l < 20

15

53

795

20  l < 30

25

29

725

30  l < 50

40

15

600

50  l < 70

60

11

660

70  l < 100

85

7

595

120

3412.5

Total Mean ≈ 3412.5 ≈ 28.4 120

35 30 25 20 15 10 5 30 35 40 45 50 55 60 65 70 75 80

Test mark

d P90 − P10 = 69 – 41 = 28 marks. This measure is useful because it omits extreme values. 5

Number of goals

0

1

3

4

5

6

7

8

Frequency

7

9 15 9

6

7

4

4

2

Cumulative frequency

7 16 31 40 46 53 57 61 63

a Median ≈ 48

Cumulative frequency

850

500

50 40 30 20 10

1 150 0

2

60

1000

Cumulative frequency

40

0

7.5

5  l < 10

3

∑ fx = 50 855 = 50.9 ∑ f 1000

50

100

2

3

4

5

Number of goals

6

7

8

The median value is slightly more than 2 goals per game.

Number of aphids

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1 

WORKED SOLUTIONS

Cumulative frequency

18.0

15

18.5

42

19.0

60

20.0

72

25.0

87

30.0

91

90

Cumulative frequency

6 a Upper class values

70 60 50 40 30 20 10

b

25

Cumulative frequency

100

30



80



70 60 50



40



30

Median = 18.6cm

20

18

20

22

24

26

28

30

× mid-point 1773.45 = = 19.5 (3 s.f.) 91 ∑f

Thus the mean is greater than the median, since the bulk of the results are to the left of the distribution, with a long tail to the right. b 140 minutes





c Q3 − Q1 = 80 − 46 = 34 minutes d Approximately 42 minutes e Cannot tell using just the box plot. 1.6 hours is 1 h 36 minutes. Using the cumulative frequency curve, we can draw up from 96 minutes to the curve, and then read off approximately 175 on the vertical axis.

Time (seconds) 26 27 28 29 30 31 32 33+

1

3 15 35 23 4

1



The median is the value of x at which y = 320: 300 − 80 = 5.5 40 25 × 600 = 150, 75 × 600 = 450 , so 100 100 450 = 22.25. 150 − 80 = 1.75, Q3 = Q1 = 20 40 Thus the interquartile range is 20.5.

Exercise 1.5A 1 a  ∑ x = 56, ∑ x 2 = 560 , mean = 8, b

f Box plot is more compact, but only shows five values (the quartiles), so the cumulative frequency curve shows more information.

Cumulative frequency

P85 – P15 is approximately 29.7 – 27.6 = 2.1 seconds.

standard deviation = 4

This suggests that about 25 sessions out of the 200 last longer than 96 minutes.

1

85 × 82 = 69.7 100

A cumulative frequency diagram can never be decreasing, which rules out b, c and d. A frequency can never be negative, which rules out e. 10 The maximum value for y is 600, and the minimum value is zero. The curve is never decreasing and starts with a y-value of zero, so it can be a complete cumulative frequency curve.

7 a 63 minutes

Frequency

15 × 82 = 12.3, 100



c Median ≈ 18.6 cm

∑f

You need P15 and P85, and

9 Only a and f could be parts of a genuine cumulative frequency diagram.

Length (cm)

d Mean =

35

Time (seconds)

90

10

8

80

0

4 19 54 77 81 82 82

∑ x = 350,∑ x 2 = 25 500, mean = 70, standard deviation = 14.1 (3 s.f.)

c

∑ x = 91, ∑ x 2 = 1275 , mean = 13,

standard deviation = 3.63 2 a  ∑ x = 50, ∑ x 2 = 298 , mean = 5, standard deviation = 2.19 b

∑ x = 115.8, ∑ x 2 = 1981.58 , mean = 16.5, standard deviation = 3.07

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1 Representation of data

3 a  ∑ fx = 194,∑ fx 2 = 1216 , mean =



s.d.(x) =

175 042 2641 8 a Mean x = 40 = 66.025 = 66.0 (3 s.f.),var(x) = 40

∑ fx = 6.06 ∑f

175 042 − 66.0252 = 16.749... ⇒ s.d.(x) = 4.09 (3 s.f.) x = 2641 = 66.025 = 66.0 (3 s.f.),var(x) = 40 40

∑ fx 2 −  ∑ fx  ∑ f  ∑ f 

2

− 66.0252 = 16.749... ⇒ s.d.(x) = 4.09 (3 s.f.)

= 1.12

b The smallest possible variance is 0, which would mean

b Both median and mode = 6.

4

All three averages are similar, which is due to the symmetrical nature of the distribution. However, the median or mode is a more useful value because it is an actual shoe size.

∑ x = 27, ∑ x = 245, n = 13

Mean =

∑x n

x ∑ 2 must be a whole number, so the smallest it can be is 174 373.

∑ fm = 770 + 1750 + 2092.5 + 3335 + 3487.5 + 2040 + 1260

∑ x = 27 = 2.08 2

 ∑x  −  =  n 

245 −  27  13  13 

2



= 3.81

200

200 = 73.675 = 73.7 (3 s.f.)

∑ fm 2 − 73.6752 = 1110 737.5 − 73.6752 = 125.682... 200

200

⇒ standard deviation ⇒ σ = 11.2 (3 s.f.) b Since the apple has the mean mass, the mean of the new batch of apples will be unchanged.

x 5 Mean = ∑ = 102 = 6.8 n 15 2

Since there is now less variation in the masses of the apples, the standard deviation of the masses of the new batch will be slightly reduced.

1181  102  Variance = 15 −  15  = 32.5 (3 s.f.) 2

Standard deviation = 1181 −  102  = 5.70 (3 s.f.) 15  15  − = x − 10, s.d.(w) = s.d.(x), so 6 a If w = x – 10, then w

10 ∑(v − 127.5) ≈ 19 × 0 + 24 × 10 + 35 × 20 + 41 × 30

+ 27 × 40 + 9 × 50 = 3700

x = 10 + 12 = 10.6, s.d.(x) = 144 − 0.6 2 = 2.62 20 20

∑(v − 127.5)2 ≈ 19 × 02 + 24 × 102 + 35 × 202 + 41 × 302

∑(x − 8) = 10.6 − 8 ⇒ ∑(x − 8) = 52,

+ 27 × 40 2 + 9 × 50 2 = 119 000

b

20

6.84 =

∑(x − 8)2 − 2.62 ⇒ ∑(x − 8)2 = 272 (3 s.f.) 20

c y = 48 + a = 10.6 ⇒ a = 9 30

40



9 a Mean

n 13 Standard deviation = 2

∑ x 2 − 66.0252 = 0 ⇒ ∑ x 2 = 174 372.025

( )

2

var(y) = 314 − 48 = 7.91 (3 s.f.) (So in fact we  30 30 do not need to know the value of a here).

7 a t = 20, s.d.(t ) = 4.60, c = 20, s.d.(c) = 1.41 b The country route, as there is a smaller deviation so it is easier to predict how long it will take, and is no slower on this evidence.

Mean ≈

3700 + 127.5 = 151 (3 s.f.) 155

Variance ≈

2

119 000  3700  − = 198 (3 s.f.)  155  155

11 If y = x – 25, then y = x − 25, s.d.(y) = s.d.(x). y = 124 = 6.2, s.d.(y) = 11.8. This implies 20 x = 31.2, s.d.(x) = 11.8. 12 If y = x – 70, then y = x − 70, s.d.(y) = s.d.(x). y = −315 = − 6.3, s.d.(y) = 3.41. This implies 50 x = 63.7, s.d.(x) = 3.41 (3 s. f.)

2

c

2106 + p 2  100 + p  − = 20, which becomes 6  6  2 5p – 200p + 1916 = 0 This solves to give p = 16 or 24 (to the nearest integer).

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Variance =

∑ fx 2 −  ∑ fx  ∑ f  ∑ f 

2

(

)

WORKED 6 094 779 705 SOLUTIONS 1 263 541 = − 1069 1069 = 4 304 297.79

Exam-style questions 1 a

14 15 16 17

5 0 0 0

≈ $2075 (to the nearest dollar) 6 a The mid-points are 3 and 25.5 respectively. b 7 students are represented by an area of 10  cm2, so 1 student is represented by an area of 10  cm2. The 7 true class width of the 6–10 class is 5 hours, and the true class width of the 31–49 class is 19, so the width of the 31–49 class on the histogram will be 19 = 7.6 cm. The height of the 31–49 class will be 2.5

Key: 14 5 = 1.45 m b

1.5

1.6

10 7 19 2.5

1.7

Height (m)

7 a

b IQR = 139, Q1 − 1.5 × IQR = −87.5, Q3 + 1.5 × IQR = 468.5 All the 11 values are inside the range (−87.5, 468.5), so there are no outliers in the data.

4 2 2 6 2

13 14 15 16 17 18

4 Girls 23789 256 667 239 Key: 3 14 2 = boys 143, girls 142

7 8 9 6 7 9

15

22

21

18

Frequency density

2

3.75

5.5

3

1.2

80 70 60 50 40 30 20 10

∑ x = 2392

0

for current boys. For boys with

5

10 15 20 25 30 35 40 45

t (minutes)

new boy added ∑ x = 160 × 60 = 2560.



Thus new boy’s score is 168.

b Approximately 86 − 53 = 33 people

4 100 + 2(7) = 114 cm, and 100 − 2(7) = 86 cm.

10

90

IQR for the girls = 167 − 147 = 20 c

Frequency

100

Girls 3 5 6 3

∑ fx = 602 = 15.05 hours ∑ f 40

5–10 10–14 14–18 18–25 25–40

t

Cumulative frequency

b 13 14 15 16 17 18

9 322 6322 952 63 44

= 0.188 cm.

c Estimate of the mean =

2 a Median = Q2 = 187, Q1 = 121, Q3 = 260

3 a Boys

1 

s.d. = 4304297.79 = 2074.68

5 6 6 8 8 1 1 2 3 5 6 1 1 2 4 0

1.4

2

So if the rainfall in the city is outside the range (86, 114) the year can be taken as exceptional.

5 a Mean profit =

∑ fx = 1 263 541 1069 ∑f

= $1182

(to the nearest dollar) b Variance = =

∑ fx 2 −  ∑ fx  ∑ f  ∑ f 

2

(

6 094 779 705 1 263 541 − 1069 1069

= 4 304 297.79

c Mean =

∑ fx = 1643.5 = 19.1 (3 s.f.) 86 ∑f

d Standard deviation ≈

∑ fx 2 − x 2 = ∑f

37 074.25  1643.5  −  86  86

2

  = 8.12 minutes (3 s.f.) e Q1 ≈ 13

)

2



Median = Q2 ≈ 18



Q3 ≈ 23, P60 = 18 minutes

s.d. = 4304297.79 = 2074.68 ≈ $2075 (to the nearest dollar)

9

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1 REPRESENTATION OF DATA

8

a The median is the 123rd result, which lies in the 170–180 class.

10 a

Cumulative frequency

90

b

Frequency density

8 7 6 5 4

70 60

2

40 30 20

0

×

×

12 14 16 18 20 22 24 26

c Mean ≈

Mean ≈ 20.8 cm

∑ fx = 42445 = 173 sec (3 s.f.) 246 ∑f

Standard deviation ≈ 1.96 cm (3 s.f.)

a The class widths are 200, 40, 40, 70, 150, 300, so the frequency densities are 0.03, 0.5, 0.75, 0.343, 0.08, 0.0267. The area of the 200–240 class is 14, so the area per household is 0.7, so the area for the 280–350 class is 16.8 cm2. The width of the 280–350 class is 2 × 70 = 40 3.5 cm, so its height must be 4.8 cm.

×

90

c Mean is best as no obvious outliers, and it uses all available data. 11 a Mean = 146 +160 = $164.87 30 2 Variance = 1024 − 146 ⇒ standard deviation = $3.23 30 30

b

∑ x 2 − 320∑ x + 30 × 1602 ⇒ ∑ x 2 = 815 744

60

×

50

c Mean = 4946 = 164.9 (4 s.f.) 30 Variance = 815 744 − 164.8666 2 30 ⇒ standard deviation = $3.23 (3 s.f.)

40 30

×

20

× 0

∑ x = 146 + 160 × 30 = 4946,∑ x 2 = 30

Alternative method: multiplying out, 1024 =

70

10

2

2   ∑x   × 2 ∑ x = 146 + 160 × 30 = 4946,∑ x = 30  var(x) +  30   = 815 744  

×

80

( )

( ) ⇒ standard deviation = $3.23

1024 − 146 30 30

b 100

∑ fx = 2077,∑ fx 2 = 43524.5

b

Length of song (sec)

Cumulative frequency

Median ≈ 20.8 ×

Foot length, l (cm)

130 140 150 160 170 180 190 200 210 220

×

Median ≈ $270

100 200 300 400 500 600 700 800 900 1000

Monthly shopping bill ($) c

×

50

10

3

×

80

1

9

×

100

The upper quartile is the 185th result, which also lies in the 170–180 class.

∑ fx = 30 660,∑ fx 2 = 10 984900 30660 = $307 (3 s.f.) 100 Standard deviation ≈ Mean ≈

12 a Mass, m (grams)

30  m< 40

40  m< 50

50  m< 60

60  m< 80

80  m< 120

Frequency

13

37

56

8

6

Cumulative frequency

13

50

106

114

120

10 984 900 − 306.6 2 = $126 (3 s.f.) 100

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1

WORKED SOLUTIONS

So here, the standard deviation is more than three-quarters of the IQR.

Cumulative frequency

120

Note: if the distribution was more symmetrical and less skewed, the rule would work better.

90

13 a Mean = 4930 = 290 kg 17 290 + 2(10.5) = 311 kg b New mean = 4930 + 292 = 290.1 kg 18 2.5 × 50 14 a Width = = 12.5 cm 10 b Height = 4 × 34 = 9.7 cm 14

60

30

c 90

120

90

Mass, m (g) i Median ≈ 51g b

ii LQ ≈ 44 g, UQ ≈ 57 g

30  m< 40

40  m< 50

50  m< 60

60  m< 80

80  m< 120

Frequency

13

37

56

8

6

Class width

10

10

10

20

40

Mass, m (grams)

Frequency density

1.3

3.7

5.6

0.4

0.15

Cumulative frequency

60

30

Q3 ≈ 38

80

×

70 60

×

50 40 30

Q1 ≈ 21

20

×

10

Median ≈ 29

×

0

20

40

60

80

100

Time (milliseconds)

6

52nd value = median ≈ 29

5

Frequency density

×

×

100

26th value = Q1 ≈ 21 78th value = Q3 ≈ 38

4

IQR ≈ 17 d Median is best due to the extreme values in the data set. 15 a The given bar has an area of 9 cm2, so each plant is represented by an area of 9 cm2. That means 29

3

2

1

60

30

90

Mass, m (g) c For this data, IQR ≈13 g. ∑ fm ≈ 53 Mean ≈ 120 s.d. ≈

∑ fm 2 − 532 = 13.7 (3 s.f.) 120

120

the area of the 25  m < 35 bar is 9 × 4 cm2. 29 The width of the 25  m < 35 bar must be double the width of the 5  m < 10 bar, at 3 cm. That means the height of the 25  m < 35 bar is 9 × 4 = 0.414 cm (3 s.f.). 3 × 29 b Mean =

∑ fx = 10.7 kg ∑f

Standard deviation =

∑ fx 2 − x 2 = 6.84 kg ∑f 11

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1 Representation of data

16 a Mean =

∑ fx = ∑f

18 a Mean =



23 × 105 + 33 × 64 + 43 × 42 + 53 × 30 + 64 × 21 + 80 × 10 272





= 37.011... = 37 years, no months s.d. =

∑ fx 2 − x 2 = ∑f

(

437185 10 067 − 272 272

)

2

= 15 years 5 months (nearest month)

b The modal class is the one with the highest frequency density. This is the class containing 18–27 year olds. c

2

773 071 329  310 467  − = 15 626.11... 125  125 



Variance =



s.d. = 125 s (2 minutes 5 seconds)

b

∑(t − 2500) = −2033,∑(t − 2500)2 = 1 986 329

19 a Australia 853 98210 976641 885430 8611 542

= 15.4 (3 s.f.)

310 467 = 2484 s (41 minutes 24 seconds) 125

48 49 50 51 52 53

England 0148 233679 1147888 246699 034 Key: 0 48 3 = 6 48.0 seconds Australia, 48.3 seconds England

Age 18–27 28–37 38–47 48–57 58–69 70–89 Freq

105

64

42

30

21

10

CF

105

169

211

241

262

272

b Australia England

300 48.0

50.0

51.0

52.0

53.0

54.0

Time (seconds)

250

Cumulative frequency

49.0

c Both diagrams show the data well. Box -andwhisker plots show that overall the English cricketers have slightly quicker times. Although Australia has a smaller range, the interquartile range of the two teams is similar. The stem-andleaf diagram has the advantage that the original data is preserved.

200

150

100

50

20 For distribution 1, (lowest value, Q1, Q2, Q3, highest value) = (8, 10, 14, 22, 30)

0

For distribution 2, (lowest value, Q1, Q2, Q3, highest value) = (8, 14, 22, 26, 30) 15

30

45

60

75

90

Age (years)



 edian ≈ 33, LQ ≈ 24, UQ ≈ 46, IQR ≈ 22 (all M in years)

d T  he modal class would be the most meaningful measure to choose here, although it would help to know what percentage of all drivers are in each class to make a fully-informed choice. 17 IQR = 6

∑ x = 55 + a,∑ x 2 = 385 + a 2 ⇒

0

5

10 15 20 25 30

The values for distribution 2 are significantly higher, as shown by the differing values for the median. The measures of spread, however, are the same here; the variation in each set of data would seem to be the same.

2



2 s.d. = 385 + a −  55 + a  = 6  11  11

This gives 10a2 – 110a – 3146 = 0, and so a = 24

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WORKED SOLUTIONS

1 

Mathematics in life and work 1 Drug Patients L

2

Q

L × Q

C (£ 1000)

CE Continue?

A

15

1

0.9

0.9

10

11.1

no

B

31

2

0.7

1.4

12

8.6

no

C

64

3

0.7

2.1

15

7.1

yes

D

42

4

0.6

2.4

8

3.3

yes

E

34

5

0.4

2.0

17

8.5

no

F

17

6

0.3

1.8

3

1.7

yes

∑ fx = 2447,∑ fx 2 = 33 031 Mean = 2447 = 12.1 (3 s.f.) 203 Standard deviation = 2

33 031  2447  − = 4.17 (3 s.f.)  203  203 3 Median = 102nd value = 3 years Q1 = 51st value = 3 years Q3 = 153rd value = 5 years 4 CE is the largest for drug A. 5 NICE would wish to discontinue drugs A, B and E, and continue with drugs C, D and F .

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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS

2 Probability, permutations and combinations Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication of the answer.

Prerequisite knowledge 1

b P(both yellow) = 16 = 0.16 100 c P(one blue and one yellow) = 0.24 + 0.24 = 0.48

a

x

Exercise 2.1A

K

B

1

59

24

30

2

3

?

Alternatively: There are 2! ways to arrange the two S books. There are 6! ways to arrange the remaining 6 books. Finally there are 7 possible positions for the S books in relation to the N books, so 7 × 6! × 2! = 10 080 ways in total.

a 1 3 b 4 15

3

c 11 15 a P(K and J) = 0.2

So the number of ways where the two S books are not together is 40 320 − 10 080 = 30 240.

b P(K and H) = 0 c P(H′) = 1 − 0.35 = 0.65

 2  6   2  6  c     +     = 55  1   3   2  2

d P(K′ and J′ and H′) = 1 − (0.25 + 0.2 + 0.1 + 0.35) = 0.1 4

4

e P(K or H) = 0.25 + 0.2 + 0.35 = 0.8 a

6 10 4 10

Blue

4 10

24 Yellow 100 24 Blue 100

Blue

6 10

The group must consist of two from one nation and one from each of the others. So the total number of possible choices is:  5  4  3  2 × 4 × 3 × 4 + 2 ×  2 × 5 × 3 × 4 +  2 × 5 × 4 × 4

36 100

6 10

= 1440

Yellow 4 10

14

b 8 × 7 × 6 = 336 5! × 21! = 6.13 × 1021 (5! arrangements for the vowels, and then a following 21! arrangements for the remaining letters.) a 8! = 40 320 b The number of ways where the two S books are together = 7! + 7! = 10 080.

b 120 − 59 − 30 − 24 = 7 members 2

a 8! = 40 320

Yellow 16 100

5

a The number of ways of arranging all 12 books is 12! = 479 001 600. The number of ways of arranging the eight Plato books is 8! = 40 320, while the number of ways of arranging the four Socrates books is 4! = 24. So the probability of having Plato-Socrates is 8!4! while the probability of Socrates-Plato 12!

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WORKED SOLUTIONS

2 

m(m − 1) n(n − 1) = 9 n(n − 1)(n − 2) m ⇒ (m − ) is 4!8! , and so the probability of having all 5 − 2) 1 m ) n(n − 1) 6= 9 n(n − 1)(n 12! 2 2(m − 110 m ⇒ 5(m − 1) 2 2 10 6 − 2)   the Plato books andmall (mthe − 1)Socrates n(n − 1) =books 9 n(n − 1)(n − 2) m =⇒3(5n(m − 1) = 3(n − 2) 2 2 10 6 together is 2 × 8!4! = 2 . n − 2) 12! = 3(495 So the smallest values are m – 1 = 3, n – 2 = 5, or b We now need the probability of Socratesm = 4, n = 7. Plato-Socrates, where the arrangement is 10 a 3 × 8! = 120 960     b  3 × 3 × 6! = 6480 0–8–4 or 1–8–3 or 2–8–2 or 3–8–1 or 4–8–0. c 5 × 4 × 6! = 14 400 The probabilities here are:



8! 4! + 4 × 8! × 3! + 4 × 3 × 8! × 2! 12! 12! 12! × × × ! × 1 ! 4 ! × 8! 4 3 2 8 + + 12! 12! = 5 × 8! × 4! = 1 12! 99

6 a 6! = 60 2!3! b 5! × 2 = 20 3!2! c 4! = 4 3! 7 a  Regarding the Es and Ls and Ss as distinct letters, there are 8! arrangements of the letters.

In these, the Ls can be arranged in 2! ways, as can the Es and the Ss. Thus the total number of distinct arrangements is 8! = 7! = 5040. 2!2!2!

b Count instead the number of arrangements where the two Ls DO appear together.

This is (regarding LL as one letter) 7! = 1260. 2!2!



So the number of distinct arrangements where the two Ls do not appear together is



5040 – 1260 = 3780.

8 a K(H)Q(H) appears in 7! arrangements, while Q(H)K(H) appears in another 7!, so there are 2 × 7! ways for this to happen altogether = 10 080 arrangements. b K(S) and K(C ) are together in 10 080 arrangements, so they are not together in 8! – 10 080 arrangements = 30 240 arrangements. c If K(H) and Q(H) are together and K(S) and K(C ) are together, then there are (regarding K(H)Q(H) as one card and K(S)K(C) as another) 6! × 2! × 2! = 2880 arrangements.  m  n  m  n 9 You have     = 0.9      1   3  2   2

Exercise 2.2A

1 a  If there are 200 students in total, 46 must be in Nis. The probability is therefore 46 = 0.23. 200 b Probability of ‘Qom’ house = 0.31.

Probability of ‘Aba’ house = 0.26.



 robability of not ‘Qom’ house or ‘Aba’ P house is therefore 1 − (0.31 + 0.26) = 0.43.

2 a Half of the tickets are even numbers so 1 or 0.5. 2 b Square ticket numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. There are 10 square ticket numbers.

Probability of selecting a square ticket 10 number = = 0.1. 100 Therefore the probability of not selecting a square ticket number = 1 − 0.1 = 0.9.

c Ten numbers end in 5, ten start with 5, and one number does both, so 19 numbers feature the digit 5. The probability is 0.19. 3 20 = 3x + 8 12 = 3x  x = 4 4 Probability of selecting a black ball = = 0.2. 20 Therefore, the probability of not selecting a black ball = 0.8. 4 a

6! = 3 8! 28

( ) 3!

b P(green and yellow together) 7! × 2 = 1 . = 3! 8 ! 4 3! 3 P(not together) = . 4 6! 3 c 8! = 28 3!

( )

 10   5 a  4  = 2  30 261  4 

 n n! Using   = r !(n − r )! r

15

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2 Probability, permutations and combinations

Thus P(team of five contains the goalkeeper,

 20  4  = 323 b  30 1827  4 

c

two defenders and two attackers) = 100 = 50 . 462 231

 10  13  7  2  × 13 × 7 +  2  × 10 × 7 +  2 × 10 × 13

 24 9 a   × 4 × 3 = 127 512  4  14  13  10  1   2   1  10 920 65 b = = = 0.0856 (3 s.f.) 127 512 759  24 4 3 × ×  4 

 30  4  =

12 285 13 = 27405 29

 10  22  2   2  × 2 20790 15 = = = 0.163 (3 s.f.) c 127512 92  24 × 4 × 3  4 

6 Probability of Alan getting a job (up to age 24) ≈ 1 − 0.132 = 0.868. The above figure is a crude estimate. Is Alan better or worse qualified than the average unemployed Australian? Does he spend more or less time than the average seeking work? Are the job opportunities for Australians the same as for those from overseas? Is Alan willing to compromise on levels of pay? There are a large number of unknowns. It could be that since Alan is choosing to move to Australia, he may have specific work in mind that increases his chances of finding a job. 7 a



Heads

Tails

Spades

SH

ST

Hearts

♥H

♥T

Diamonds

DH

DT

Clubs

CH

CT

b Red suits = hearts and diamonds, so the probability is 2 = 1 8 4 c H, T

T, H

= 1 – P (spade/diamond/club or ten) = 1 –  3 + 1  = 3  4 52  13 c P((B or (T or H))) = 1 – P(B or (T or H )) = 1 – P(B or T or H ) =1  – P(T or H ) (since if B happens, H and T happen)  3 1 =1– +   52 4  9 = 13

Exercise 2.3A

All eight outcomes are equally likely.

H, H

3 4 b P((H or T )) = 1 – P(H or T )

10 a P(H) =

1

x

0.3

T, T

Spades

SHH SHT STH STT

Hearts

♥HH ♥HT ♥TH ♥TT

Diamonds

DHH DHT DTH DTT

Clubs

CHH CHT CTH CTT

All 16 outcomes are equally likely. d 2 = 1 16 8

B

A

0.2

0.1

0.4 a 0.6 b 0.1

8 Total number of choices for the team of five is 11

C 5 = 462 . Total number of teams with the goalkeeper, two defenders and two attackers = 5C 2 × 5C 2 = 100 .

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2 

WORKED SOLUTIONS

2

x

5

x

Pianists

B

A

42

guitar

violin 0.15 0.15

0.35

36

10

68

0.35 a 0.7

P(piano but not violin and not guitar) = 42 = 7 156 26

b 0.5 c 0.65 3

6

x

x L

J

S

K

0.15

x y

z

0.25

0.25

0.35

w

P( J or L but not both) = 0.4.

x + y = y + z = 3y, x + y + z = 0.75, so 5y = 0.75, y = 0.15, x = z = 0.3

7

x

Gardeners

a 0.15 b 0.45

hoe

c 0.55 d 0.25

x

e 0.3 4

fork

x

10

7

24 x

14

x

9 spade

B

A

3x + 64 = 100, so x = 12.

0.1 0.4

0.2

P(spade) = 14 + 12 + 7 + 12 = 0.45 100 8 The probabilities must add up to 1, so 4x + 4y = 1, x + y = 0.25.

0.3 a 0.4 b 0.1



P(B) = 0.3 implies 3y + x = 0.3. Subtracting the two equations, 2y = 0.05, so y = 0.025, x = 0.225 Thus P(C) = 3x + y = 0.7.

c 0.2

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2 Probability, permutations and combinations

9 Say the number of members doing all three Saystrokes the number is a. of members doing all three strokes is a. x crawl

Exercise 2.4A 1

x B

A

butterfly

30–(39–a)

24–a 33–(43–a) a 15–a 19–a 30–(34–a)

1 4

1 5

? 11 20

backstroke So 7 = 3a–23 and so a = 10

a 0 1 b 5

x crawl

butterfly 1

14 10

5

11 c 20

0

2 A and B are mutually exclusive if and only if P(A ∪ B) = P(A) + P(B).

9

6

1 Let A be ‘the sum on the dice is 4’, so P(A) = . 12 Let B be ‘the scores are equal’, so P(B) = 1 . 6 8 2 P(A ∪ B) = = , which does not equal 1 + 1 = 1 . 36 9 12 6 4 Thus A and B are not mutually exclusive. 1 3 B 2 B 5 1 9 Y 2

5



backstroke a P (all three strokes) = 10 = 1 50 5 5 b P (none of these) = =1 50 10 10 x A

B ar

ar

3

4 9

ar 2 a 5 8

a



a + ar + ar2 + ar3 = 1 ⇒ a =

 

8 15

Total

Wears a hat

5 12

1 4

2 3

Does not wear a hat

1 12

1 4

1 3

1 2

1 2

1

18

Y

d 4 × 5 + 5 × 1 = 5 9 8 9 2 9

Does not wear a scarf

1 P ( S ∩ H ′ ) 12 1 = = b 4 1 P ( H ′) 3

3 8

c 4 × 3 = 1 9 8 6

Wears a scarf

Total

B

b 1 2

Thus P(A) = 1 and P(B) = 1 3 5 11 a

Y

5 8

4 a R 4 7 2 7

B

1 7 G

2 3

H

8 21

1 3

T

4 21

1 2

H

1 7

1 2 1 2 1 2

T H T

1 7 1 14 1 14

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WORKED SOLUTIONS

b P(red and heads) =

8 21

2 7 1 4 4 a 7 × 13 + 2 × 13 = 13

1 7



P(blue and heads) =



P(green and heads) =



P(heads) =

b P(at least one blue)= 1 – (P(RR) + P(GG) + P(RG) + P(GR))

1 14

= 1 −

8 1 1 25 + + = = 0.595 21 7 14 42

5

B 0.55

W

0.6

B

0.1

W

0.9 0.5

B W

0.5

B

0.2

W

S

FR 0.75

2 3 1 6 4 36 × − × − = 7 13 2 13 13 91

c P(both the same) = P(RR) + P(GG) + P(BB) = 2 × 3 + 1 × 6 + 3 × 2 = 30 7 13 2 13 14 13 91

P (different) = 1 – 30 = 61 91 91

8 a x bus

L

0.25

0.4

0.4 L

0.45 0.6

train 45

6

18

S

0.8 B a 0.6 × 0.45 × 0.4 + 0.4 × 0.25 × 0.5 = 0.158

75

b 0.6 × 0.45 × 0.6 + 0.6 × 0.55 × 0.9 + 0.4 × 0.25 × 0.5 + 0.4 × 0.75 × 0.8 = 0.749 c 0.4 × 0.25 × 0.5 + 0.4 × 0.75 × 0.8 = 0.29 6 a i 0.65 × 0.4 × 0.25 = 0.065



iii 0.1 × 0.45 × 0.55 + 0.1 × 0.45 × 0.45 + 0.1 × 0.55 × 0.55 + 0.45 × 0.55 × 0.9 = 0.298 (3 s.f.)

6 = 0.0417 144 24 51 P(B) × P(T ) = × = 0.0590 144 144 P(B ∩ T ) =

ii 0.2 × 0.85 × 0.75 = 0.128

( ) ( )

These are not equal, so B and T are not independent.

b x

b i P(both walk) = 0.9 × 0.25 P(both cycle) = 0.55 × 0.7

bus

train

P(same method ) = 0.61

45

6+x

18 – x

ii P(different methods) = 0.39 2 13

7 B

4 13

3 14 7 14

3 13 G 4 13

4 14

2 

3 13 R 3 13

B

75

7 13 G R B 6 13 G

P(B ∩ T ) = P(B) × P(T ),so

( )

(6 + x) (51 + x) 24 = × , 144 144 144 which solves to give x = 3

9 a A and B are mutually exclusive. b x C

R B 7 13 G

A 3

6

2 4

1

5 B

R

19

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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS

Probabilities of 1, 2, 3, 4, 5, 6 are kp, p, p, p, p, p so p(5 + k) = 1, and p = 1 . 5+k P(B)P(C) = P(B∩C)⇒ k+2 3 2 = ⇒k=4 5+k5+k k+5

5

x F

E 1 10

10 Since A and B are mutually exclusive, P(A∩B) = 0. If they are also independent, then P(A∩B) = P(A)P(B). Thus P(A)P(B) = 0, and so either P(A) = 0 or P(B) = 0 or both.

2

P(E ∩ F ) P(F ) = 0.15 ÷ 0.5

a P(E | F ) =

P(spade) = 1 , P(jack) = 1 , P(spade and jack) = 1 52 4 13 1 52 1 4

= 0.3

= 1 13 Or, since the events spade and jack are clearly independent, P(jack | spade) = P(jack). 6 P(W) = 14 , P(W ∩ H) = 31 31 P(jack | spade) =

P (H ∩ W ) P (W ) 6 14 = ÷ 31 31 3 = 7 1 1 1 P(H, T or T, H) = , P(H, H) = , P(T, T) = 2 4 4

b P(E′ ∩ F ) = P(F ) − P(E ∩ F ) = 0.5 − 0.15 = 0.35 c P(E′ ∩ F′) = 1 − P(E ∪ F ) = 1 − 0.6

P(H | W) =

3

P(T , T ) a P(T, T | T) = P(at least one T ) =

1 4 3 4

1 = 3

= 4

1 2 3 4

= 0.4 6

a

5 6

Red

6 12

=2 3

Yellow

5 11

M 0.1 0.4

P(M P) P(P) = 0.4 ÷ 0.5 = 0.8 P(M ∩ P ') 0.2 2 d P(M | P′) = = = 0.5 5 P(P ')

c P(M | P) =

Yellow

Red

Red Yellow

2 6 4 6 2 6

Yellow

Red 3 6 3 6

0.2

Yellow Red Yellow

 1 5 5  1 6 4 b P(B) =  × ×  +  × ×   2 11 6   2 11 6  +  1 × 6 × 4  +  1 × 5 × 3   2 11 6   2 11 6 

0.3 a P(M ∪ P) = 0.7 2 b P(P | M) = P(P ∩ M ) = 0.4 = 0.6 3 P(M )

6 11

Red

1 6 4 6

Yellow

6 11

x P

Red

5 11 6 12

b P(H, T | T) = P(H ,T or T , H ) P(at least one T )

7 20

2 5

Exercise 2.5A 1

3 20

=

2 3

c P(A ∩ B) = P(R, R, R) + P(Y, Y, R) = 25 + 5 132 44 =

10 33

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WORKED SOLUTIONS

d (R, Y, Y) and (Y, R, Y) are the only disallowed combinations. So P(A ∪ B) = 1 – 1 − 1 11 11 =

b P(G) = 0.1 × 0.3 + 0.9 × 0.5 = 0.48

9

9 11

7

a

25 132 10 33

( )

Red

7 9

Blue

)(

=

b 0

e P( A | B ′ ) =

1 1 10 a 4 = P(R | S) + P(S | R) P(S) P(R) + P(R ∩ S) P(S ∩ R) P(S) + P(R) = P(R ∩ S) 0.8 = P(R ∩ S) =

Heads

4 5

(

a 1

P( A ∩ B ′ ) P( A ∩ B ′ ) = P(B ′ ) 1 − P(B) P( A) − P( A ∩ B) = 1 3 = 3(0.5 − 0.2) = 0.9

=5 8

Tails

2 5 1 5

2 2 7 4 b P(tails) = 9 × 5 + 9 × 5

P(W ∩ G ) 0.03 = = 0.0625 P(G ) 0.48

Heads

3 5 2 9

c P(W | G ) =

c P(A ∩ B) = P(B | A)P(A) = 0.2 P( A ∩ B) 0.2 2 d P(B) = P( A | B) = 0.3 = 3

25 5 10 e P(R, R, R) + P(Y, Y, Y) = 132 + 44 = 33 P(all red | same colour) =

Tails

)

Thus P(R ∩ S) = 0.2. 2 5 1 8 b 4= + = ⇒ P(S | R) = 3 3P(S | R) P(S | R) 3P(S | R)

32 45

c P(heads) = 13 45

P(R ) =

P(blue | heads) =

P(blue ∩ heads) P(heads)

=

7 13 ÷ 45 45

=

7 13

1

(

)(

a

0.3 0.1

)

( 3060 ) = 30 ÷ 48 = 0.625 48 ( 160 )

 3   4   5  3   4   5  a   ×  ×  +  ×  ×   2  2   3  2  3   2  3  4   5  +   ×   ×   = 360  3  2   2

G

b 10! × 3! = 2177 2800

W 0.7 0.5

0.9

a 1 − 56 = 0.65 160 b 24 + 32 = 0.35 160 c

2

53 = 81

P(R ∩ S) 0.2 = = 0.3, so P(S) = 0.5 2 P(S | R) 3

Exam-style questions

d Probability of same colour = P(red and red) + P(blue and blue) 2 2 7 7 = × + × 9 9 9 9

8

2

G' G

3

a P(Uche does not) = 1 − 0.62 = 0.38 P(Eli does not) = 1 − 0.17 = 0.83 P(Ellis does not) = 1 − 0.68 = 0.32 0.38 × 0.83 × 0.32 = 0.101 (3 s.f.)

W'

b P(at least one) = 1 – P(none) 0.5

G'

= 1 – 0.100 928 = 0.899 (3 s.f.)

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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS

c P(two hit) = 0.62 × 0.17 × 0.32 + 0.62 × 0.83 × 0.68 + 0.38 × 0.17 × 0.68 = 0.428 (3 s.f.) 0.428 P(two hit | at least one hits) = = 0.476 (3 s.f.) 0.899 4

6

a 1 3 R

x

a

1 3

climbing 26

2 4

35

1 4

40

35 11 19

16

1 4 2 3 G

Therefore the probability of a student doing none of the activities is 35 ÷ 185 = 0.189 (3 s.f.). Therefore the probability of a student doing bowling only is 19 ÷ 185 = 0.103 (3 s.f.).

( ) ( )

R

a

Time taken  1 min

Time taken > 1 min

Total

Question right

1 8

13 24

2 3

Question wrong

1 8

5 24

1 3

Total

1 4

3 4

1

2x

2 3 B 1 3

x 2 3 G

13 24 2 3

()

= 13 16

1 3

1 3

1–3x

=

1 3

1 3

35 185 104 185

P(slowly and right) P(right)

1 12

R

1 6

B

1 12

c

= 0.336 (3 s.f.)

b P(slowly | right ) =

1 6 1 6

1 1 1 b P(B ∩ G) = P(B G) + P(G B) = 12 + 12 = 6

c 19 out of 185 do bowling only.

d P(no activities | not bowling) =

1 6

G

1 3

b 35 out of 185 do none of the activities.

B

R

B

table tennis

1 6

G

2 3

3

bowling

5

1 3

R

1 3

R

1–3x 3

B

1–3x 3

1–3x 3 R 4x 3 G

G

2x 3

R

2x 3

B

x 3

(1 − 3x) + 2x = 1 ⇒ 5 − 15x + 10x = 3 3 3 5 2 So x = 5

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WORKED SOLUTIONS

7

a p = 15, q = 9, r = 7, s = 1 b P(art | music ) = P(both art and music) P(music) =

9 32 16 32

( )

= 9 16

c P(art) = 24 = 3 ≠ 9 , so these are not 32 4 16 independent events. d 8

7 15 × = 0.106 (3 s.f.) 32 31

 6   3  2  6   3   2  6   3  2 a       +       +       = 144  2  1   1   1   2  1   1   1   2 b 2 × 9! = 725 760 c 6 × 5 × 7! = 151 200

9

a 47 + 62 – 100 = 9 boys both cycle and swim. 91 = 0.91 = P (student does just one sport). b 100 P(cycling only) c P(cycling | one sport only) = P(one sport only) = 0.38 ÷ 0.91 = 0.418 (3 s.f.) d Because some boys do both sports. Alternative answer: P(C ∪ S) not equal to P(C) + P(S)

11 The game will be fair if the probability of spinner 1 beating spinner 2 is equal to spinner 2 beating spinner 1. The same applies between spinner 1 and spinner 3, and between spinner 2 and spinner 3. Probability of spinner 1 beating spinner 2: =

( 63 × 15 ) + ( 62 × 35 ) + ( 16 × 45 ) = 1330

Probability of spinner 2 beating spinner 1: =

( 15 × 63 ) + ( 15 × 63 ) + ( 15 × 65 ) + ( 15 × 1) = 1730

in 1 versus 3, P(1 win) = 16 , P(3 win) = 14 , 30 30 14 in 2 versus 3 P(2 win) = 25 These probabilities are unequal; therefore the game is unfair. 12 a

0.02

Misshapen (M)

0.98

Not misshapen (M′)

0.03

Misshapen (M)

0.97

Not misshapen (M′)

0.05

Misshapen (M)

0.95

Not misshapen (M′)

J 0.25 0.45

K

0.3 L

e P(C & S) = 9 ÷ 100 = 0.09 ≠ 0.47 × 0.62 10 a Treat the jacks as a block, so you have the queens, the kings and the jack block, which is 9 things to arrange. There are 4! ways to arrange the jacks within the jack block.

b P(M′ ∩ J) = 0.25 × 0.98

Total number of arrangements is 12!, so 9! 4! = 0.0182 (3 s.f.). P(4 jacks together) = 12! b Ways to arrange the four kings = 4!, and the other cards = 8!

= 0.0335

So P(2K at one end, 2K at the other) 4! × 8! = 12! = 0.00202 (3 s.f.) c 3 ×  4  4  4 = 288  2  1   1  d

288 = 0.582 (3 s.f.)  12  4 

2

= 0.245 c P(M) = P(M ∩ J ) + P(M ∩ K) + P(M ∩ L) = 0.005 + 0.0135 + 0.015 P(K '∩ M ) P(M ) [ P ( M ∩ J ) + P(M ∩ L)] = P(M )

d P(K ′ | M) =

= 0.02 ÷ 0.0335 = 0.597 (3 s.f.) 5 of the numbers from 2 to 10 are even, so the 13 a 9 answer is 5 . 9 b Two odd numbers add to give an even number, but multiply to give an odd number, so S and T are not mutually exclusive. c P(S) = P(two cards are both even) + P(two cards are both odd) = 20 × 19 + 16 × 15 = 31 36 35 36 35 63

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2 Probability, permutations and combinations



17 a

P(T) = P(two cards are both odd)



16 15 4 × = = 36 35 21





0.35

You could argue instead that P(S | T) = 1 ≠ P(S). 14 a 9 × 8 × 7 × 6 × 5 = 15 120

0.4

b Digits that add up to 9 or 18. (1, 2, 6), (1, 3, 5), (2, 3, 4), (9, 8, 1), (9, 7, 2), (9, 6, 3), (9, 5, 4), (8, 7, 3), (8, 6, 4) are the possible triplets. Each produce 6 numbers, so 54 in total. 54 = 3 9 × 8 × 7 = 504, so P(divisible by 9) = 504 28 c Each group has 3 members. Number of ways:  3   3  3 =       × 3 = 81  2  1   1 



15 a 0.8 − 0.36 = 0.44

T'

0.75

T

0.25

T'

0.25

T

0.75

T'

P

0.25 F

c P(F | T ) =

R = Red

G = Green

c (0.44 × 0.56) + (0.36 × 0.66) + (0.2 × 0.25) = 0.534 d 1 − 0.534 = 0.466      (0.466)5 = 0.022 (3 d.p.)

16 Let us assume that there are 100 students at the school (probabilities are not affected by school size); x B

G

1

2

3

4

5

6

1

1

2

3

4

5

6

2

2

4

6

8

10

12

3

3

6

9

12

15

18

4

4

8

12

16

20

24

5

5

10

15

20

25

30

6

6

12

18

24

30

36

5

6

    b

R = Red 1

C

1 G = Green

31%

22%

P(F ∩ T ) 0.252 = = 0.103(3 s.f.) P(T ) 0.6075

18 a

b 1 − 0.44 = 0.56

26%

0.3

b P(T ) = 0.35 × 0.7 + 0.4 × 0.75 + 0.25 × 0.25 = 0.608

d The first three cards he picks must add up to a 1 multiple of 3, so the answer is . 3



T

S

4 4 31 ≠ × , so S and T are 21 21 63 not independent. P(S ∩ T) = P(T) =



0.7

21%

2

3

yes yes

5 6

a 0.26

4 yes

yes

3 4

2

yes

yes yes yes

b 0.21 c P(B ∩ C ) = 0.26 = 0.456(3 s.f.) P(C ) 0.57 d P(G)P(C ) = 52 57 ≠ 31 = P(G ∩ C ) , so not 100 100 100 independent.



P(RG is square) = 8 = 2 36 9

5 P(even ∩ square) 36 5 = = P(square) 8 8 36 5 3 d P(odd | square) = 1 − = 8 8

c P(even | square) =

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WORKED SOLUTIONS

2 

 5  24 19 a     = 2760  3  2  b

 5  24  5  24  1  4  +  2  3  = 73 370

20 a 12! = 6 × 11! 2! 10! b 8 × 4 × 2! = 16 × 10! c 8 × 8! d

2 1485

21 a Red even

Red odd

Total

Green even

1 3

1 6

1 2

Green odd

1 4

1 4

1 2

Total

7 12

5 12

1

b P(product even | sum even) =



P(green even or red even | both odd or both even) P(both even) = P(both odd or both even) =

(

1 4 3 = 7 1+1 3 4

)

Mathematics in life and work 1 Gearbox

Exhaust

Expected cost

0.9

Okay 0.855 $0

$0

0.95

0.1

Fail

$47.50

0.05

0.9

Okay 0.045 $3000

$135

0.1

Fail

$17.50

Okay 0.095 $500

Fail 0.005 $3500

2 $200 3 The premium should be $220 per year.

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3 Discrete random variables

3 Discrete random variables Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication of the answer.

Prerequisite knowledge

2

1 a {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 3 b 8 4 1 = c 8 2 2 a

a 1 2 3

1 1 1 + = 6 6 3

4

1 ×1 ×2= 2 36 6 6 33 3 Mean = = 1.1 30 Variance = 77 – 1.12 = 1.36 (3 s.f.) 30 b

5 6 8

4 a 6! = 6 × 5 × 4 … = 720 b 10C4 = 10! = 210 (6!4!)

10

Exercise 3.1A

12

9

1 a

15

x

2

3

4

5

6

7

8

9 10 11 12

P(X = x)

1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36

16 18 20

b

24 0.180

25

0.160

P (X = x)

0.140

30

0.120 0.100

36

0.080 0.060 0.040 0.020 0.000 2



3

4

5

6

7

8

9 10 11 12

The distribution is unimodal and symmetrical.

26

x

P(A = a) 1 36 1 18 1 18 1 12 1 18 1 9 1 18 1 36 1 18 1 9 1 18 1 36 1 18 1 18 1 18 1 36 1 18 1 36

3 a 0.1 + 0.2 + a + 0.25 + 0.3 = 1 a = 1 – (0.1 + 0.2 + 0.25 + 0.3) = 0.15 So P(Y = 1) = 0.15 b P(Y  0) = P(Y = −1) + P(Y = 0) = 0.1 + 0.2 = 0.3

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3 

WORKED SOLUTIONS

c P(Y > 1) = P(Y = 2) + P(Y = 3)



or

= 0.25 + 0.3

P(X  1) = P(X = 1) + P(X = 2) + P(X = 3)

= 0.55 4 a 2

3

4

5

6

7

8

1 16

1 8

3 16

1 4

3 16

1 8

1 16

x P(X = x)

Therefore P(X is a prime number) =



1 +1+1+1 = 9 or 0.5625 16 8 4 8 16

2

3

4

5

6

P(X = x)

1k

2k

3k

4k

5k

6k

P(X = x)

1 21

2 21

3 =1 21 7

4 21

5 21

6 =2 21 7

k + 2k + 3k + 4k + 5k + 6k = 1

21k = 1 k = 1 (substitute into the table) 21 b P(2  X < 4) = P(X = 2) + P(X = 3)

= 1 − P(X = 2) =1− 1 6 5 = 6

8



There are three ways of having 1 blue: G,G,B G,B,G B,G,G



There are three ways of having 2 blues: B,B,G B,G,B G,B,B



There is one way of having 3 blues: B,B,B 1

P(X = x)

8 27

2

()

4 9

b P(X  1) = 1 − P(X = 0)

=1− 8 27 19 = 27

1

2

3

4

P(R = r)

0.1r

0.1r

0.1r

k

P(R = r)

0.1

0.2

0.3

k

r

6 a There is one way of having 0 blues: G,G,G

( 23 )

2 1 + 21 7

= 2 + 4 +2=4 21 21 7 7

2 d P(2  X < 11) = 3

1 2 3× × 3 3

=

5 = 21 c P(even number) = P(X = 2) + P(X = 4) + P(X = 6)

c P(X  3) = 1 − P(X < 3)

P(X = x)

∑p = 1



=1+1+1 6 6 6 1 = or 0.5 2

3

19 27

1

b P(X < 7) = P(X = 2) + P(X = 3) + P(X = 5)

0

=

x

5 a  The outcome for each value of x is equal over six possible outcomes. 1 Therefore k = . 6

x

4 2 1 + + 9 9 27

7 a

b Possible prime numbers are: 2, 3, 5, 7

=

2

()

1 3× 3

2 9



9  The probability of all the possible outcomes add up to 1, therefore: 0.4 + 0.25 + 4k + 2k + k = 1 7k = 0.35

3 2

2 × 3

The value of k = 1 – (0.1 + 0.2 + 0.3) = 1 – 0.6 = 0.4

() 1 3

3

1 27

k = 0.05

The probability that at least two under graduates are late is: P(X  2) = 1 – P(X  1) = 1 – (0.4 + 0.25) = 0.35 10 a The probability of all the possible outcomes add up to 1, therefore: 0 + 3r + 9r + 4r + 3r + r = 1 20r = 1

r = 0.05

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3 Discrete random variables

b 0

1

2

3

4

5

P(C2 = c)

0.25k

1.25k

2.25k

3.25k

3.5k

4.5k

P(C2 = c)

1 1 1 × = 4 15 60

1 5 1 × = 4 15 12

3 9 1 × = 4 15 20

13 1 13 × = 4 15 60

7 7 1 × = 2 15 30

3 9 1 × = 2 15 10

c

0.25k + 1.25k + 2.25k + 3.25k + 3.5k + 4.5k = 1 1 15k = 1 ⇒ k = 15 1 59 P(C 2  1) = 1 − P(C 2 = 0) = 1 − = 60 60



Exercise 3.2A

4 a i 0.2 + 0.15 + 0.35 + a + 0.1 = 1

1 a E(X) = −2 × 0.13 − 1 × 0.27 + 0 × 0.1 + 1 × 0.18 + 2 × 0.22 + 3 × 0.1

a = 1 − (0.2 + 0.15 + 0.35 + 0.1)



ii E(X) = 0  × 0.2 + 1 × 0.15 + 2 × 0.35 + 3 × 0.2 + 4 × 0.1

= 0.39 b E(Y) = 2 × 1 + 4 × 5 + 6 × 1 + 8 × 1 3 6 12 12 31 = 5.17 (3 s.f.) = 6



 = 1.85 iii

x2

2 a E(Z) = 5 × 0.4 + 6 × 0.3 + 7 × 0.2 + 8 × 0.1



P(X = x)

=6

b



z2

25

36

49

64

P(Z = z)

0.4

0.3

0.2

0.1

c (E(Z))2 ≠ E(Z 2) since (E(Z))2 = 62 = 36 ≠ 37 = E(Z 2).

1

2

3

4

5

6

1

0

1

2

3

4

5

2

1

0

1

2

3

4

3

2

1

0

1

2

3

4

3

2

1

0

1

2

5

4

3

2

1

0

1

6

5

4

3

2

1

0

d

0

1

2

3

4

5

P(D = d)

1 6

5 18

2 9

1 6

1 9

1 18



=

9

16

0.2

0.1

b i k + 2k + 3k + k = 1 7k = 1 k = 1 = 1.14 (3 s.f.) 7 1 3 1 2 ii E(Y ) = −2 × 7 + 0 × + 2 × 7 + 4 × 7 7 8 = 7 1 2 3 1 2 iii E(Y  ) = 4 × 7 + 0 × 7 + 4 × 7 + 16 × 7 32 = 4.57 (3 s.f.) = 7 5 a

x

4 5

2 50 = 0.04

2 3

1 5 2 1 +1× +2× +3× 1 +4× 6 9 9 18 6 1 +5× 18 35 = 1.94 (3 s.f.) 18

P(X = x) 7 50 22 50 18 50 1 50

1

b E(D) = 0 ×

4 0.35

E(X2) = 0 × 0.2 + 1 × 0.15 + 4 × 0.35 + 9 × 0.2 + 16 × 0.1

a



1 0.15



= 37

3

0 0.2

= 4.95

E(Z 2) = 25 × 0.4 + 36 × 0.3 + 49 × 0.2 + 64 × 0.1



= 0.2

= 0.14 = 0.44 = 0.36 = 0.02

b E(X) = 1 × 0.14 + 2 × 0.44 + 3 × 0.36 + 4 × 0.02 + 5 × 0.04

= 2.38

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3 

WORKED SOLUTIONS

1 1 1 4 1 b E(X) = 1 × + 2 × + 3 × + 4 × + 5 × 10 8 8 10 8 1 37 or 3.7 +6× = 8 10

c P(X > 2.38) = P(X = 3) + P(X = 4) + P(X = 5) = 0.36 + 0.02 + 0.04 = 0.42 6

x

P(X = x)

0

0.73 = 0.343

1

3 × (0.72)(0.3) = 0.441

2

3 × (0.7)(0.32) = 0.189

3

0.33 = 0.027

c

= 0.9 7 a + 4b + 2b + a = 1 2a + 6b = 1 −1 × a + 0 × 4b + 1 × 2b + 2 × a = 5 12 −a + 2b + 2a = a + 2b =

10 a

5 12

5 12

12a + 24b = 5

(2)

8a + 24b = 4

(1) × 4

4a = 1

(2) – 4(1)

a=



(1)

b

()

6b = 1 − =

1 2

6

2x

2

4

6

8

10

12

P(X = x)

1 10

1 8

1 8

4 10

1 8

1 8

y

2

4

6

8

P(Y = y)

2 20

4 20

6 20

8 20

2

4

6

8

y–1

2–1

4–1

6–1

8–1

P(Y = y)

2 20

4 20

6 20

8 20

E (Y–1) = 2





P(Z = z)

x

1−x

4x + 12(1 − x) = 7 4x + 12 − 12x = 7 5 = 8x 5 x = = 0.625 8

2

4

12

0.625

0.375

z



×

2 4 6 + 4 −1 × + 6 −1 × 20 20 20 8 1 = 20 5

= 2.60

b Var(Y) = 22 ×

z

P(Z = z)

−1

1 a Var(X) = (−2)2 × 0.13 + (−1)2 × 0.27 + 02 × 0.1 + 12 × 0.18 + 22 × 0.22 + 32 × 0.1 – 0.092

12



5

Exercise 3.3A

4



4

+ 8 −1 ×

1 ÷6 2 1 = 12

9 a

3

1 2

b=

8

2

y

1 4

Substitute a into (1): 2 1 + 6b = 1 4

1

E(2X) = 2 × 1 + 4 × 1 + 6 × 1 + 8 × 4 + 10 × 1 10 8 8 10 8 1 + 12 × = 7.4 8 Alternative method: 37 E(2X) = 2 E(X) = 10 × 2 = 7.4



E(X) = 0 × 0.343 + 1 × 0.441 + 2 × 0.189 + 3 × 0.027

x

 =

2

107 = 2.97 36

d

1

2

3

4

5

6

P(D = d )

1 6

1 6

1 6

1 6

1 6

1 6

E(D) = 1 ×

1 1 1 5 31 + 42 × + 62 × 3 + 82 × 6 −   12 12  6

1 1 1 1 1 1 +2× +3× +4× +5× +6× 6 6 6 6 6 6

= 3.5

1 1 1 1 1 + 22 × +32 × + 42 × + 52 × 6 6 6 6 6 1 + 62 × – 3.52 6

∑p = 1

Var(D) = 1  2×

k + 4k + 4 = 1 8 1 k= 10

=

35 = 2.92 12

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3 Discrete random variables

3 a

1

2

3

4

1

2

3

4

5

2

3

4

5

6

3

4

5

6

7

4

5

6

7

8

s P(S = s)

6 a E(A) = 10 × 0.6 + 25 × 0.25 + −5 × 0.15

3

4

5

6

7

8

1 16

2 16

3 16

4 16

3 16

2 16

1 16

3 2 1 + 6 × 16 + 7 × 16 + 8 × 16 =5 1 2 3 4 Var(S) = 22 × 16 + 32 × 16 + 42 × 16 + 52 × 16 3 2 1 + 62 × 16 + 72 × 16 + 82 × 16 − 52

= 2.5 4 a 0.1 + p + 0.2 + 0.35 = 1 p = 1 − (0.1 + 0.2 + 0.35)



  = 0.35 4.75 + 0.2a = 5.95



0.2a = 5.95 − 4.75



   



= 716

c The expectation of stock B is higher, so on average, he would get better returns on his investment. The variance of B is much higher so there is more risk. The stockbroker may make a higher profit in stock B but could also lose a lot more than in stock A. 7 a P(R = 3) means having two rolls that are not 1 and then a third roll that is a 1, or 3 rolls that are not 1. =  5  × 1 +  5   6 6  6



=

1.2 0.2

=6

= 3.35 5 a a + 3b + 2a + b = 1 3a + 4b = 1



1 × a + 3 × 3b + 5 × 2a + 7 × b = 3.8



11a + 16b = 3.8

(2)



12a + 16b = 4

(1) × 4



Subtracting:



a = 0.2



Substitute into (1):



3(0.2) + 4b = 1



4b = 0.4



b = 0.1

b Var(Y) = 1  2 × 0.2 + 32 × 0.3 + 52 × 0.4 + 72 × 0.1 – 3.82

r

1

2

3

P(R = r)

1 6

5× 1= 5 6 6 36

25 36



=

1 5 25 +2× +3× 6 36 36

91 = 2.53 36

( )

1 5 25 91 + 32 × – + 22 × 6 36 36 36 = 0.583

Var(R) = 12 ×

(1)

a + 9b + 10a + 7b = 3.8

3

25 36

c E(R) = 1 ×





 = 3.36

= 87.75

Var(B) = 1  02 × 0.55 + 502 × 0.25 + (−30)2 × 0.2 − 122



b

c Var(X) = 2  2 × 0.1 + 52 × 0.35 + 62 × 0.2 + 82 × 0.35 − 5.952



= 12%





= 1.2 a=

E(B) = 10 × 0.55 + 50 × 0.25 + −30 × 0.2

2

b 2 × 0.1 + 5 × 0.35 + a × 0.2 + 8 × 0.35 = 5.95



= 11.5%

2





b Var(A) = 1  02 × 0.6 + 252 × 0.25 + (−5)2 × 0.15 – 11.52

1 2 3 4 b E(S) = 2  × 16 + 3 × 16 + 4 × 16 + 5 × 16







2

8 a Let G be the number of girls who are appointed. P(G = 0) = 8 × 7 = 14 20 19 95 P(G = 1) = 8 × 12 + 12 × 8 = 48 20 19 20 19 95 P(G = 2) = 12 × 11 = 33 20 19 95



g P(G = g )

0

1

2

14 95

48 95

33 95

b E(G) = 0 × 14 + 1 × 48 + 2 × 33 = 6 = 1.2 95 95 5 95

Rounding to the nearest integer, one girl is expected to be appointed.

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3 

WORKED SOLUTIONS

9 Let A be the profit from new car Model A:

a P(A = a)

3000

2500

–500

0.5

0.3

0.2

E(A) = 3000 × 0.5 + 2500 × 0.3 + (−500) × 0.2 = 2150 Let B be the profit from new car Model B:

b P(B = b)

3500

3000

–1000

0.4

0.4

0.2



0 5  5 P(H = 0) =    1   1  = 1 32  0  2   2 



P(H = 3) = 1 × 1 + 5 × 1 = 7 8 3 16 3 48

b

P(H = h)

h

( 21 + 81 + 321 ) × 31 ( 21 + 83 + 325 ) × 31 ( 83 + 165 ) × 31 ( 81 + 165 ) × 31

0

E(B) = 3500 × 0.4 + 3000 × 0.4 + (−1000) × 0.2 = 2400

1

E(B) > E (A), so the finance director would support new car Model B.

2

10 a Let X be the number showing on the spinner: 3

P(X = x)

1 3

1 3

1 3

4

5 ×1 32 3

5 96

5

1 ×1 32 3

1 96

7 + 1 × 11 + 2 × 11 + 3 × 7 + 4 × 5 48 48 96 32 32 1 3 +5× = or 1.5 96 2

c E(H) = 0 ×

If the spinner shows '3': 1 2  3 P(H = 1) =    1   1  = 3 8  1  2   2 

P(H = 2) =

 3  1  2  1  1 3  2  2   2  = 8

3 0  3 P(H = 3) =    1   1  = 1 8  3  2   2 

P(H = 0) = 1 − 3 − 3 − 1 = 1 8 8 8 8 If the spinner shows '5':



1 4  5 P(H = 1) =    1   1  = 5 32  1  2   2 



P(H = 2) =

 5  1  2  1  3 5  2  2   2  = 16



3 2  5 P(H = 3) =    1   1  = 5 16  3  2   2 



4 1  5 P(H = 4) =    1   1  = 5     2 2 32  4



7 48

5





11 48

3

P(H = 1) = 1 2 P(H = 0) = 1 − 1 = 1 2 2



11 32

1

If the spinner shows '1':



7 32

x





P(H = h)

5 0  5 P(H = 5) =    1   1  = 1 32  5  2   2 



Var (H) =  0 2 × 7 + 12 × 11 + 22 × 11 + 32 × 7 + 4 2 × 5 + 52 × 1  −  32 32 48 48 96 96 

 0 2 × 7 + 12 × 11 + 22 × 11 + 32 × 7 + 4 2 × 5 + 52 × 1  −  3   32 32 48 48 96 96   2 

2

= 15 or 1.42 12

Exercise 3.4A 1 a  i P(X = 2) = (1 − 0.2) × 0.2 = 0.16  ii P(X = 5) = (1 − 0.2)4 × 0.2 = 0.081 92 iii P(X = 10) = (1 − 0.2)9 × 0.2 = 0.027 (3 d.p.) b  i P(X = 2) = (1 − 0.8) × 0.8 = 0.16  ii P(X = 5) = (1 − 0.8)4 × 0.8 = 0.001 28 iii P(X < 3) = P(X = 1) + P(X = 2)

 

 = 0.8 + (1 − 0.8) × 0.8



 

 = 0.8 + 0.16 = 0.96

c  i P(X = 4) =

( 78 ) ( 18 ) = 0.084 (3 d.p.) 3

 ii P(X  4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) =

( )( ) ( ) ( 18 ) + ( 78 ) ( 18 )

7 1 7 1 + 8 8 + 8 8

2

3

= 0.414 (3 d.p.) iii P(3  X < 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

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3 Discrete random variables

( ) ( 18 ) + ( 78 ) ( 18 ) + ( 78 ) ( 18 ) 7 1 + (8) (8)

7 = 8

2

3

4

5

= 0.317 (3 d.p.) d  i P(X < 4) = P(X = 1) + P(X = 2) + P(X = 3) = 0.4 + 0.6(0.4) + 0.62(0.4)



= 0.784  ii P(X = 5) = 0.64 × 0.4 = 0.051 84 iii P(X > 5) = (1 − 0.4)5 = 0.077 76 e  i P(X  3) = P(X = 1) + P(X = 2) + P(X =3)

= 0.16 + (1 − 0.16)(0.16) + (1 − 0.16)2(0.16) = 0.407 (3 d.p.)  ii P(X > 6) = (1 − 0.16)6 = 0.351 (3 d.p.) iii P(X  4) = P(X > 3) = (1 − 0.16)3 = 0.593 (3 d.p.) 2 X ~ Geo

(101 )

P(X = 6) =  1 − 1   1  = 0.0590 (3 s.f.)  10   10 

3 X ~ Geo(0.5) P(X = 4) = 0.53 × 0.5 = 0.0625 4 a X ~ Geo(0.25) P(X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

=0  .25 + 0.75 × 0.25 + 0.752 × 0.25 + 0.753 × 0.25



= 0.684

b P(X  7) = P(X > 6) = (1 − 0.25)6 = 0.178 5 a X ~ Geo(0.35) P(X = 6) = 0.655 × 0.35

 = 0.0406 (3 s.f.)

b P(X  10) = P(X > 9)

= 0.659 = 0.0207 (3 s.f.)

6 X ~ Geo(0.4) There is time for three attempts in 1 h 30 min (90 ÷ 25 = 3.6) P(X  3) = P(X = 1) + P(X = 2) + P(X = 3) = 0.4 + 0.6 × 0.4 + 0.62 × 0.4 = 0.784

() 4 1 P(X = 4) = ( ) ( ) = 0.1024 5 5 4 b P(X > 5) = ( ) = 0.328 (3 s.f.) 5

1 7 a X ~ Geo 5

c P(X = 2) = 0.21

3

5

(1 − s)s = 0.21

s − s2 = 0.21 s2 − s + 0.21 = 0 s = 0.7 or 0.3

Since the probability of a person answering the phone on the first call is greater than 0.5, s = 0.7.

9 a X ~ Geo  1   12 

P ( X  20 ) = P(X > 19) =

= 0.191

x

x ln

11 = ln0.1 12



x = 26.46



She must collect at least 27 beetles.

10 a Y ~ Geo  1   20 

( 1920 ) ( 201 ) = 0.031 5 b P (Y > 25) = ( 19 ) = 0.277 20 c P (Y  15) = 1 − ( 19 ) = 0.537 20 P (Y = 10 ) =

9

25

15

Exercise 3.4B  10 1 a  i P(X = 2) =   × 0.22 × 0.88 = 0.302  2  10  ii P(X = 5) =   × 0.25 × 0.85 = 0.0264  5  10 iii P(X = 10) =   × 0.210 × 0.80 = 1.02 × 10−7  10 b  i P(X = 2) =  ii P(X = 5) =

 7 × 0.82 × 0.25 = 0.004 30  2  7 × 0.85 × 0.22 = 0.275  5

iii Since the binomial distribution is discrete, P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) =

The probability of the phone calls getting answered is constant.

32

19

 11  = 0.1  12 

8 a Each phone call is independent.

(1211 )

b P(X > x) = 0.1

5



b P(X = n) = (1 − s)n − 1s

 7  7 × 0.80 × 0.27 +   × 0.81 × 0.26  0  1

 7 +   × 0.82 × 0.25 = 0.004 67  2

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WORKED SOLUTIONS

 20 c  i P(X = 18) =   × 0.12518 × 0.8752  18   ii From cumulative binomial probabilities,

 17 +   × 0.161 × 0.8416  1

P(X  3) = P  (X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

 17 +   × 0.162 × 0.8415  2



( ) × ( 78 )

 20 1 = × 8 0 ×

( 78 )

19

 17 e  i P(X  2) =   × 0.160 × 0.8417  0

 = 8.08 × 10−15



( ) × ( 78 )

 20 1 +  × 8  2

2

18

0

20

() ()  20 1 7 +  ×( ) × ( ) 8 8  3 1  20 1 × 77 +  × × 8 8 8  1 3

19 19

3 

 = 0.473

( ) ×× ( 7878)

20 11  20 ++   ×× 8  22  8

17

22

18 18

 17  ii P(X = 5, 6, 7) =   × 0.165 × 0.8412 1 1  5 20 20   11 ++   ×× 8 17  17 8    11  + 6 11 7 10     6  × 0.16 × 0.84 +  7  × 0.16 × 0.84        = 0.120

()

iii P(2 < X  15) = P(X  15) − P(X  2)

= 0.765

  17 iii P(3 < X  18) =  P(X  18) − P(X  3) 16 1  17  17 0 = 1 −   16 × 0.16 × 0.84 +  17 × 0.16 × 0.84  = (1 − P(X = 19 or 20)) − 0.765 33 19 1 20   20  20 1 7 1    17  1 −    × 8 × 8 +   × 8  17   20 × 0.1617 × 0.840  − 0.473… × 0.1616 × 0.841 +    19      17    16   − 0.765 33 = 0.527  = 0.235 2 Number of trials is eight, therefore n = 8. d  i Since the binomial distribution is discrete, 1 P(X < 4) = 1 − P(X = 4 or 5) Probability of rolling a six is 1 , therefore p = . 6 6   5   5 Since there are a set number of independent 5 0 4 1 =1−   × 0.4 × 0.6 +  4 × 0.4 × 0.6  trials with a consistent probability, the problem   5  can be modelled using a binomial distribution. = 0.913

() ()

()



Alternatively, P(X = 0, 1, 2 or 3)



5  5 × 0.41 × 0.64 =   × 0.40 × 0.65 +  1  0



+

5  5 × 0.42 × 0.63 +   × 0.43 × 0.62  2  3

= 0.913  5  ii P(X = 5) =   × 0.45 × 0.60 = 0.010 2  5 iii P(0  X < 3) = P(X < 3)

= P(X  2) = P(X = 0) + P(X = 1) + P(X = 2)



 5 =   × 0.40 × 0.65  0



 5 × 0.41 × 0.64  1



+



 5 +   × 0.42 × 0.63  2 = 0.683

Let X be the number of sixes rolled.  1 Therefore, X ~ B 8,   6

( ) × ( 65 )

 8 P(X = 3) =   × 1 6  3 3 P(X = 6) =

3

5

= 0.104

 10 × (s)6 × (1 − s)4  6 

4 a P(X < 3) = P(X  2)

 12  12 0 12 1 11  12 =   × 0.25 × 0.75 +   × 0.25 × 0.75 +   × 0.  0  1  2  12 +   × 0.251 × 0.7511  1  12 +   × 0.252 × 0.7510  2 = 0.391  12 b P(X  9) =   × 0.259 × 0.753  9  12 +   × 0.2510 × 0.752  10

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3 Discrete random variables

+

 12 × 0.2511 × 0.751  11

 12 +   × 0.2512  12 = 0.000 392 (3 s.f.) 5 a  Number of students selected is 30, therefore n = 30.

Probability of a student asking for paper is 3 out of 5, therefore p = 0.6.



Since there is a set number of independent trials and a fixed probability of success, the problem can be modelled as a binomial distribution.



Let X be the number of students who ask for paper.



Therefore, X ~ B(30, 0.6)

 30 P(X = 10) =   × (0.6)10 × (0.4)20  10  = 0.002 00 (3 s.f.) b P(X < 25) = 1 − P(X  25)



Therefore, X ~ B(10, 0.25)

 10 P(X = 3) =   × (0.25)3 × (0.75)7  3 = 0.250 (3 s.f.) b P(X  3) = 1 − P(X  2) = 1 − 0.5256 = 0.474 (3 s.f.) 7 a Number of students is 20, therefore n = 20.



Probability of having a birthday in January is 1 , therefore p = 1 . 12 12 Since there is a set number of independent students and a fixed probability of having a birthday in January, the problem can be modelled as a binomial distribution. Let X be the number of students with birthdays in January. 1 Therefore, X ~ B 20,   12 

( ) × (1211 )

20 P(X = 1) =   × 1  1  12

1

19

= 0.319 (3 s.f.)

b P(X  4) = 0.978 (3 s.f.)   30   30   30   30   30  30 25 5 26 4 = 1 –    × 0.6 × 0.4 +   × 0.6 × 0.48 a  +  Number × 0.6 27of × 0customers .43 +   chosen × 0.6 28 ×is05, .4 4 2therefore +   × 0.6 29 × 0.4 +   × 0.630    28  26  27  29  30   25  n = 5. Probability of being in the 18–25 age group is 0.4. Since there is a set   30 of  30   30number  25 5  30  26 4  30  27 3  30  28 29 30 4 2 +  customers 0 6 0 4 0 6 × . × . + . × 0.4 ×   25 × 0.6 × 0.4 +  26 × 0.6 × 0.4 +  27 × 0.6 × 0.4 +  28 × 0.6independent  and a fixed probability,  30    29        the problem can be modelled as a binomial distribution   30   30  30   30B(5, 0.4).  30  0.6 25 × 0.45 +   × 0.6 26 × 0.4 4 +   × 0.6 27 × 0.43 +   × 0.6 28 × 0.4 4 2 +   × 0.6 29 × 0.4 +   × 0.630  30number of customers in the age  28  29 Let X be the  27  26 group 18–25. Therefore:   30 26 4  30  27 3  30  28 2  30  29 30 0.6 × 0.4 +   × 0.6 × 0.4 +   × 0.6 × 0.4 4 +   × 0.6 × 0.4 +   × 0. 6 P(X  < 2) = P(X = 0) + P(X = 1)  29  28  30  27   5 0 5  5 1 4 =   ( 0.4 ) ( 0.6 ) +   ( 0.4 ) ( 0.6 )  30 30    30     1  0 4 2 +   × 0.6 29 × 0.4 +   × 0.630  0.6 27 × 0.43 +   × 0.6 28 × 0.4  30  29  28  = 0.337   30 b The binomial distribution will not be +   × 0.630   30  suitable. Customers chosen from the same = 1 − 0.005 659 = 0.994 (3 s.f.) 6 a  Number of flights in the winter is 10, therefore n = 10.

Probability of poor visibility is 25%, therefore p = 0.25.



Since there are a set number of independent trials with a fixed probability of success, the problem can be modelled as a binomial distribution.



Let X be the number of poor visibility encounters.

table are not independent of each other; they are more likely to be in the same age group. 9 A packet contains 10 gingerbread men biscuits, therefore n = 10. Probability of being broken is 0.03. Since there are a set of independent trials with a fixed probability, this can be modelled as a binomial distribution, X~B (10, 0.03). 10 P(X = 2) =   ( 0.03)2 ( 0.97 )8 = 0.031 7  2 

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WORKED SOLUTIONS

10 A box contains 10 duck eggs, therefore n = 10. Probability of any egg being broken is 0.1. Since there is a set number of independent duck eggs and a fixed probability, this problem can be modelled as a binomial distribution, X~B (10, 0.1). P(X  2) = 1 – P(X  1)   10 0 10    10  1 9  = 1 –   0  ( 0.1) ( 0.9)  +   1  ( 0.1) ( 0.9)       



  10   10 0 10  1 9     ( 0.1) ( 0.9)  +    ( 0.1) ( 0.9)     1     1  = 1 – 0.736 1

= 0.264

= 5 × 0.1 × (1 − 0.1) = 0.45 1 3 a X ~ Geo 6

()

E(X ) = 1 = 1 = 6 p  1  6 1  b Y ~ B 42,   6 E(Y) = np 1 = 42 × 6 =7 = 30 × 0.1

1= 1 p 0.3

= 10 3 = 3.33 (3 s.f.) 1 1 b E(X ) = = p 0.45 =

ii Var(X) = np(1 − p)

4 a E(X) = np

Exercise 3.5A 1 a E(X) =

3 

=3 Var(X) = np(1 − p) = 30 × 0.1 × (1 − 0.1) = 2.7 b P(μ  X < μ + σ) = P(3  X < 3 + 1.643) = P(X = 3) + P(X = 4)

20 9

=

= 2.22 (3 s.f.) 1 1 c E(X) = = p 3 7 7 = 3

()

= 2.33 (3 s.f.) 2 a i E(X) = np = 10 × 0.2 = 2 ii Var(X) = np(1 − p) = 10 × 0.2 × (1 − 0.2) = 1.6 b i E(X) = np = 150 × 0.3 = 45 ii Var(X) = np(1 − p) = 150 × 0.3 × (1 − 0.3)

+

= 20 ×

= 0.413

 25 =   × 0.613 × 0.412  13   25 +   × 0.614 × 0.411  14 

3 = 7.5 8

 25 +   × 0.615 × 0.410  15 

ii Var(X) = np(1 − p) 3 = 20 × × (1 − 3 ) 8 8

 25 +   × 0.616 × 0.49  16 

= 4.687 5

 25 +   × 0.617 × 0.48  17 

d i E(X) = np = 5 × 0.1 = 0.5

 30 × 0.14 × 0.926  4 

5 E(X) = np = 25 × 0.6 = 15 Var(X) = np(1 − p) = 25 × 0.6 × (1 − 0.6) =6 So σ = 6 = 2.449 4… = 2.45 (3 s.f.) P(μ − σ < X < μ + σ) = P(15 − 2.449 5 < X < 15 + 2.449 5) = P(12.550 5 < X 3) = (1 − 0.4)3 = 0.216

c E(X) = 15 × 0.1 = 1.5

4 a If S = 4 then you can have 1 and 3 or 3 and 1.

7 a P(3 fails, then 1 success) = 0.73 × 0.3 = 0.103 b P(1st success after 4th throw) = 0.74 = 0.240

so P(S = 4) = 1 × 2 + 2 × 1 4 3 4 3 1 = 3 1 1 b P(S = 1) = 2 × 4 3

c P(2nd success on 4th throw) = P(1 success in 1st 3 throws, then 1 success on 4th throw)   3  =   (0.3)1(0.7)2  × 0.3 1    

( ) 2 1 P(S = 3) = 2 ( × ) 4 3



= 0.132 d P(Lucas first scores on his 2nd shot) = 0.7 × 0.8 × 0.3 = 0.168

P(S = 6) = 2 1 × 1   4 3



s

1

3

4

6

P(S = s)

1 6

1 3

1 3

1 6

e P(Stefan first scores on his 7th attempt) = (0.7 × 0.8)6 × 0.7 × 0.2 = 0.00432 8 a E(R) = −2a + 3(1−a) = 3 – 5a Var(R) = (−2)2a + 32(1 – a) – E2(R)

1 1 1 1 c E(S) = 1 × 6 + 3 × 3 + 4 × 3 + 6 × 6 = 3.5 1 1 1 1 Var(S) = 12 × 6 + 32 × 3 + 42 × 3 + 36 × 6 – 3.52 = 2.25

= 4a + 9(1 – a) – (3 – 5a)2 = 25a – 25a2 = 25a(1 – a) b E(R) = 2

5 X ~ B(10, 0.21)  10 a P(X = 2) =   (0.21)2 (0.79)8 = 0.301 (3 s.f.)  2 b P(X  4) = 1 – P(X  3)

3 

9

1 5



Therefore: 3 – 5a = 2 ⇒

a=



Var(R) = 25 × 4 = 4 ⇒ 5 5

σ = 4=2

Y ~ B(30, 0.05)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

 30 a P(X = 2) =   (0.05)2(0.95)28 = 0.259  10   2 10 10 10       = 1 −   (0.21)0(0.79)10 +   (0.21)1(0.79)9 +   (0.21)2(0.79)8 +   (0.21)3(0.79)7  3 3) = 1 − P(X  3)  2 >  1 b P(X  0 

  10  10  10 0.21)0(0.79)10 +   (0.21)1(0.79)9 +   (0.21)2(0.79)8 +   (0.21)3(0.79)7   2  1  3 

= 1 − 0.860 8



= 0.139 (3 s.f.)

= 1 − [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

 30  30  30 0 30 1 29 = 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0 1      2 

c P(4 < X < 9) = P  (X = 5)+ P(X = 6) + P(X = 7) + P(X = 8)

 30 0 30  30  1 29  30  2 28  = 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95)  1 2 0         10   10   10  10 =   (0.21)5(0.79)5 +   (0.21)6(0.79)4 +   (0.21)7(0.79)3 +   (0.21)8(0.79)2   7   8   30 6  0  30  30 1 29  30 2 28  30   5  = 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   (0.05)3 (0.95)27  2 0 1 3        10    10  10   10  =   (0.21)5(0.79)5 +   (0.21)6(0.79)4 +   (0.21)7(0.79)3 +   (0.21)8(0.79)2  5 6 7 8           30   30 0 30  30  1 29  30  2 28 = 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   (0.05)3 (0.95)27  1  2  3   0=0.039 9 6 X ~ B(15, 0.1) = 1 − 0.939 2 …  15 = 0.060 8 a P(X = 3) =   (0.1)3(0.9)12 = 0.129  3

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3 Discrete random variables

c Y ~ B(20, 0.06077)

 20 P(Y = 5) =   (0.06077)5(0.93923)15  5 = 0.005 02 (3 s. f.)

d E(X) = 20 × 0.06077 = 1.22   10 X ∼ B5, 1   7

3

5  Blue beads: X ~ B 10,   14 



 10 P(X = 6) =   5  6  14

( ) (149 ) = 0.074 4 …

2



P(exactly six beads of same colour) = 0.000 96… + 0.007 7… + 0.029 7… + 0.074 4…



P(exactly six beads of same colour) = 0.113(3 s.f.)

 6 0 10  6  1 9  6 2 8 = 1 −   ( 0.7 ) ( 0.3) +   ( 0.7 ) ( 0.3) −   ( 0.7 ) ( 0.3)   2  1  0 

P(3 or 2 children develop the condition) = 0.021 4 … + 0.128 5 … = 0.150

 6 0 10  6  1 9  6 2 8 +   ( 0.7 ) ( 0.3) +   ( 0.7 ) ( 0.3)  = 1 −   ( 0.7 ) ( 0.3) −  2  1  0 

  11 X ∼ B6, 4   9

= 1 – 0.070 47

( ) ( 59 )

 6 a P(X = 4) =   4  4 9

4

2

= 0.930 15 a The spins are independent of each other and the order of the outcomes does not matter.

= 0.181

b P(X  4) = P(X = 4) + P(X = 5) + P(X = 6)

( )( )

 6 =  4  4 9

4

5 9

2

( )( )

 6 +  4  5 9

5

5 9

1

()

 6 +  4  6 9

6

1 1 b P( R ) = 1 , P( B ) = , P(Y ) = 6 3 2

x

= 0.181 + 0.058 + 0.008 = 0.246 P(X  1) = 1 − P(X = 0) =1−

 10 (0.01)0(0.99)10  0 

= 0.095 6 b Y ~ B(5, 0.09562)  5 P(Y = 3) = (0.09562)3(0.90438)2 = 0.00715  3

( )  10 1 6 P(R = 6) = = 0.000 96 …  6  ( 7 ) ( 7 ) 3 Yellow beads: Y ~ B(10, ) 14  10 P(Y = 6) =   ( 3 ) ( 11 ) = 0.007 7 …  6  14 14 2 Green beads: G ~ B(10, 7 )

13 Red beads:



1 R ~ B 10, 7 6

4

6

4

( ) ( 75 ) = 0.029 7 …

 10 P(G = 6) =   2  6 7

6

4

P(X = x)

2R

1× 1 = 1 2 2 4

1R and 1B

1× 1×2 1 = 3 2 3

1R and 1Y

1× 1×2 1 =6 2 6

2B

1× 1 1 3 3=9

1B and 1Y

1× 1×2 1 =9 6 3

2Y

1× 1 1 6 6 = 36

12 a X ~ B(10, 0.01)



4

P(X  3) = 1–P(X = 0) – P(X = 1) – P(X = 2)

2 3  5 P(2 children) =    1   6  = 0.128 5 …  2  7   7 



6

14 Let X be the number of bullets that hit the target. X ~ B (6, 0.7)

( ) ( 67 ) = 0.021 4 …

 5 P(3 children) =   1  3 7



c Let YB be the random spin that landed on yellow from the biased spinner and YF be the random spin that landed on yellow from the fair spinner. P(Y B ) = 1 and P(Y F ) = 1 3 6

Therefore, the probability of landing on yellow is: 1 × 1 + 1 × 1 = 1 2 6 2 3 4

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WORKED SOLUTIONS

3 

()

1 19 a The geometric distribution Geo 5 .

 1 d Y ~ Geo  4

Bob has at least three attempts before he wins, that is:



P(Y  3) = 3 4

( ) ( 15 ) = 0.081 9

b P(X = 5) = 4 5

( ) = 169

4

1

4

2

 4 c P( X  5) =   = 0.410  5

16 The number of weeks from July to October is 16, therefore n = 16. The probability of rainfall is 0.35. Let X be the number of weeks that have rainfall. This problem can be modelled as binomial distribution X ~ B(16, 0.35).

d Assume that the outcome of each kick is independent of the results of any other kicks.

Assume that the probability of scoring a goal remains constant.

20 a X ~ Geo(0.04) P(X  3) = 1 − P(X > 3) = 1 − 0.963 = 0.115  16  16 0 16  16  1 15  16  2 3 13   )114 + = 1 −   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +  b 0.E(X) 35) (= 0.65 0.35) ( 0.65)  ( ( = 25      1  2  0.04   3   0   P(X < 25) = 1 − P(X > 24)  16  0 16  16  1 15  16  2 14  16  = 1 −   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0 .35     )3(0.65)13 = 1 − (0.96)24 1 2 3 0           0.625      = 6 c Y ~ B(10, 0.115) 0 16  16  1 15  16  2 14  16  3 13   ( 0.35) ( 0.65) +  1  ( 0.35) ( 0.65) +  2  ( 0.35) ( 0.65) +  3  ( 0.35) ( 0.65)  E(Y) = 10 × 0.115 = 1.15  P(X > 3) = 1 – P(X  3)

 16 3 13  16  34 13 16   16  4 16  +   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65)   3  34   4 



Var(Y) = 10 × 0.115 × (1 − 0.115) = 1.02



Assume that whether the machine is stopped on each run is independent of the outcomes of the other runs.



or



Assume that the probability that the machine is stopped remains constant.

= 1 – 0.133 86 = 0.866 (3 s.f.)



Assume that the probability of rainfall each week is constant.



Assume that each time it rains is independent of whether it rains in the other weeks.

17 a There are nine single positive digits, which are 1, 2, 3, 4, 5, 6, 7, 8, 9 Therefore, probability of getting a six is P(6) = 1 . 9 b Let X be the number of digits that the computer program generated up to and  1 including the first 6. X ~ Geo   9 12 8   P(X 13) =   = 0.243  9 1 c P(N=1) = 0.2433 × 9 = 0.0270

Mathematics in life and work X ~ B(100, 0.999)  100 1 P(X = 99) =  (0.999)99(0.001)1 = 0.0906  99  2 P(X  98) = P(X = 98) + P(X = 99) + P(X = 100)  100 = (0.999)98(0.001)2  98   100 + (0.999)99(0.001)1  99   100 + (0.999)100(0.001)0  100

8

 8 1 d P(N = 9) = 0.2433 ×   × = 0.010 5  9 9

= 0.999 8

18 a X ~ Geo (0.65)

3 P(more than 3 inaccurate bags)

b P(X = 1) = 0.65 c P(X  5) = 1 – P(X > 5) = 1 – d 200 × 0.65 = 130 people

0.355

= 0.995



= P(X  96) = 1 − P(X  97)



= 1 − [P(X  98) + P (X = 97)]

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3 Discrete random variables



 100 (0.999)97(0.001)3 = 1 − 0.999 849 62 −   97  = 1 − 0.999 849 62 − 0.000 146 75

= 3.63 × 10−6 4 P(more than 4 inaccurate bags)

= P(X  95) = 1 − P(X  96)



= 1 − [P(X  98) + P (X = 97) + P (X = 96)]



= 1 − 0.999 849 623 6 − 0.000 146 744 7



 100 − (0.999)96(0.001)4  96  = 1 − 0.999 849 623 6 − 0.000 146 744 7 − 0.000 003 562 1

= 6.96 × 10−8

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4 

WORKED SOLUTIONS

4 Normal distribution Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. Where values from the Cambridge Assessment International Education statistical tables are used, the same level of accuracy has been used in workings unless stated otherwise. Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication of the answer.

Prerequisite knowledge

6 a Random variable A: The red curve Random variable B: The blue curve Random variable C: The black curve

1 a Mean = 3.75 b Standard deviation = 2.12 c New mean = 3.32, new standard deviation = 2.24

b The mean of random variable B is 4.8.

2 a E(Y ) = 11 3 σ  = 1.31 (3 s.f.) b P(Y > m + σ ) = P(Y > 4.98) = 1 6

Exercise 4.1A



2.4

3.6

4.8

6

7.2

8.4

9.6

10.8

7 a Always true.

1 a The distribution is symmetrical about the mean. b The mode, median and mean are all the same. c The total area below the curve is 1. d The distribution is defined by two parameters: the mean and the standard deviation. 2 Standard deviation is equal. Mean is different for the two distributions. 17 3 a 50% b 50 c 47.5% d 47.5% + (99.75 ÷ 2) = 779 800 4 Curve A has a higher mean compared to curve B. Curve A has a smaller standard deviation compared to curve B.

b Never true. The point of inflexion is located one standard deviation on either side of the mean. c Never true. (It is possible but the probability is 0.) d Always true. 8 a Mean = 2.5 mm Standard deviation = 0.1 mm b 6  8% of the screws have a diameter measurement within one standard deviation. 9 a 68% of the data is located within one standard deviation. 167 ± 1.8 = (165.2, 168.8)



5 X

16%

16%

165.2 cm

Y 42.5

Standard deviation of blue curve, X, is 1.5 Standard deviation of red curve, Y, is 2.5

167.0 cm

168.8 cm

b To find how many standard deviations away from the mean, 169.7 − 168 = 1.5 standard 1.8 deviations, this is about 6.75%  16% − 2.5%  on   2 the right of 1σ.

Probability = 0.025 + 0.0675 = 0.0925, so about 9.25%. 41

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4 NORMAL DISTRIBUTION

6

Adam is wrong because both random variable X and random variable Y have the same variance, 1.52 = 2.25

16%

Billy is wrong because it is possible to have two different random variables which have the same mean and variance.

2.5% m

s

Neither of them were right.

2s

7

10

i Graph b and graph c are normal distribution curves because they show:

› ›

symmetry about the centre a bell-shaped curve.

ii Graph b – N(21, 1.52) 16%

Graph c – N(16.7, 2.32) 8 m

16% of the students failed the exam, which means that these students did not pass the bench mark. From the graph, it shows that this is 1 standard deviation to the left of the mean. Therefore, the pass mark for the test = 68 – 15 = 53

b Sometimes true. Many continuous random variables can be modelled by a normal distribution, e.g. heights of people, blood pressure, errors in measurements, etc. However, not every continuous random variable is normally distributed.

Exercise 4.2A 1 2

3

4

A ~ N(3, 4), B ~ N(2, 3²) Let A represent Year 1 scores and B represent Year 2 scores. A ~ N(0, 5), B ~ N(−1, 2²) The ‘N’ and ‘X’ are in incorrect places; the mean should be 7 and the standard deviation should be 11. X ~ N(7, 11²) a Student A: mean = 5, variance = 49, standard deviation = 7

a Sometimes true. In the case where t = 1, the variance 12 is equal to the standard deviation 1 and so the statement is true. Otherwise, the statement is untrue.

9

a Experiment B b Experiment B has the shortest average reaction time; it also has the smallest variance. This indicates that the reaction times are less varied.

10 a Let X be the mass of mixed nuts, X ~ N(50, 2.32) b

Student B: mean = 6, variance = 81, standard deviation = 9

16%

16%

b A B

47.7 g

50 g

a

0

52.3 g

Exercise 4.2B 1

a

5

X

Z

Y m

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4

WORKED SOLUTIONS

b

5

a

Z

0

0

c

1.064

P(Z > 1.064) = 1 − Φ(1.064) = 1 − 0.8563 = 0.1437 a

0

b

Z

6

d

–2.872

a

b

P(Z < −2.872) = Φ(−2.872)

Z

0

0

= 1 − Φ(2.872)

e

= 1 − 0.9979 = 0.0021 7 0

2

a

Z

‘N’ and ‘Z’ are written in the incorrect positions. The mean should equal 0. The standard deviation should equal 1. Z ~ N(0, 1²)

–1.326

0

P(Z > −1.326) = 1 − Φ(−1.326) = 1 − [1 − Φ(1.326)]

3

= Φ(1.326) = 0.9076 8 0

1.423

P(Z < 1.423) = Φ(1.423) = 0.9226 4 0

1.1

2.1

Z

P(1.1 < Z < 2.1) = Φ(2.1) − Φ(1.1) = 0.9821 − 0.8643 = 0.1178 0

0.87

Z

P(Z < 0.87) = Φ(0.87) = 0.8078

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4 NORMAL DISTRIBUTION

P(0.24 < Z < 1.102) = Φ(1.102) – Φ(0.24) = 0.8647 – 0.5948 = 0.2699

9

Exercise 4.3A 1

a P(0 < Z < 1.2) = Φ(1.2) − Φ(0) = 0.8849 − 0.5

–1.325

0

= 0.3849

1.218

b P(−2.25 < Z < 0) = Φ(0) − Φ(−2.25)

P(−1.325 < Z < 1.218) = Φ(1.218) − Φ(−1.325)

= Φ(0) − [1 − Φ(2.25)]

= Φ(1.218) − [1 − Φ(1.325)] = Φ(1.218) + Φ(1.325) − 1

= Φ(0) + Φ(2.25) − 1

= 0.8884 + 0.9074 − 1

= 0.5 + 0.9878 − 1

= 0.7958

= 0.4878 c P(−0.5 < Z < 1.5) = Φ(1.5) − Φ(−0.5)

10

= Φ(1.5) − [1 − Φ(0.5)] = Φ(1.5) + Φ(0.5) − 1 = 0.9332 + 0.6915 − 1 = 0.6247 d P(Z > 2) = 1 − Φ(2) 0

–2.651 –1.43

= 1 − 0.9772 = 0.0228

P(−2.651 < Z < −1.43)

(

= [1− Φ(1.43)] − [1 − Φ(2.651)]

35 − 30 10 = P(Z < 0.5)

= Φ(2.651) − Φ(1.43)

= Φ(0.5)

= 0.9960 − 0.9236

= 0.6915

= Φ(−1.43) − Φ(−2.651)

= 0.0724

2

a P(X < 35) = P Z
38.6) = 1 − P(X < 38.6)

(

11

=1−P Z
0.541) = 1 − P(Z < 0.541) = 1 − 0.7057 = 0.2943

c P(X > 20) = 1 − P(X < 20) 20 − 30 =1−P Z < 10

(

)

= 1 − P(Z < −1) = 1 − Φ(−1)

12

= 1 − [1 − Φ(1)] = Φ(1) = 0.8413

0

1.102

d P(35 < X < 40) = P(X < 40) − P(X < 35) 40 − 30 35 − 30 =P Z < −P Z < 10 10

(

) (

)

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WORKED SOLUTIONS







= Φ(1) − Φ(0.5)





= 0.8413 − 0.6915





= 0.1498

= P(Z < 1) − P(Z < 0.5)



= P( Z > 2.938 )  



= 1 – Φ (2.938) = 1 – 0.9883 = 0.0017

4 

10 − 10.8  = P Z  −0.5 = 1 −  0.5  e P(15 < X < 32) = P(X < 32) − P(X < 15) b P( X  10 ) = P  Z  ( ) ( 5) = 1 − 0.691  1.6  32 − 30 15 − 30 =P Z < − P Z <   10 − 10.8  = P Z  −0.5 = 1 −  0.5 10P( X  10 ) = P  Z10 ( ) ( 5) = 1 − 0.6915 = 0.3085 1.6  = P(Z < 0.2) − P(Z < −1.5) c P(9.2 < X < 11.2) = P( X < 11.2) – P( X < 9.2) = Φ(0.2) − Φ(−1.5) 11.2 − 10.8 9.2 − 10.8 =P Z < −P Z < 1.6 1.6 = Φ(0.2) − [1− Φ(1.5)] = Φ(0.2) + Φ(1.5) − 1 = P(Z < 0.25) – P( Z < – 1)

) (

(

)

(

) (





= 0.5793 + 0.9332 − 1

= Φ (0.25) – (1 – Φ (1))





= 0.5125



f P(17 < X < 19) = P(X < 19) − P(X < 17) = P  Z < 19 − 30  − P  Z < 17 − 30    10  10 





= Φ(−1.1) − Φ(−1.3)

= P(Z < −1.1) − P(Z < −1.3) = [1 − Φ(1.1)] − [1 − Φ(1.3)]

= 0.5987 – (1 – 0.8413)

5 a X ~ N(6, 1²)

= 0.44

(

4.5 − 6 1 = P(Z < −1.5)



P(X < 4.5) = P Z
8) = 1 − P(X < 8)

3 a X ~ N(177, 7²)

(



172 − 177 P(X < 172) = P Z < 7 = P(Z < −0.714)



= Φ(−0.714)





= 1 − Φ(0.714)





= 1 − 0.7623





= 0.2377



)

b P(X > 180) = 1 − P(X < 180)



(

180 − 177 7 = 1 − P(Z < 0.429) =1−P Z
15.5) = P  Z > ( )   1.6

)

) (

65 − 60 55 − 60 −P Z < 4 4







= Φ(1.25) − Φ(−1.25)

)

= P(Z < 1.25) − P(Z < −1.25)

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4 Normal distribution



= Φ(1.25) − [1 − Φ(1.25)]





= 1 − Φ(1.7)



= Φ(1.25) + Φ(1.25) − 1





= 1 − 0.9554





= 0.8944 + 0.8944 − 1





= 0.0446





= 0.7888

c P(X > 65) = 1 − P(X < 65) 65 − 60 =1−P Z < 4 = 1 − P(Z < 1.25)

(





= 1 − Φ(1.25)





= 1 − 0.8944





= 0.1056

c P(X > 85) = 1 − P(X < 85)

)

= 1 − P  Z < 85 − 68.5    5





= 1 − P(Z < 3.3)





= 1 − Φ(3.3)





= 1 − 0.9995





= 0.0005 = 0.05%

9 X ~ N(2.025, 0.0152)

d P(65 < X < 68) = P(X < 68) − P(X < 65)

(

) (

68 − 60 65 − 60 =P Z < −P Z < 4 4





= Φ(2) − Φ(1.25)





= 0.9772 − 0.8944





= 0.0828

)

= P(Z < 2) − P(Z < 1.25)

2.05 − 2.025  = P Z < 1.667  a P( X < 2.05) = P  Z < ( )  0.015 

= Φ(1.667 ) = 0.9515 + 0.0007 = 0.9522

b P( X > 2) = P  Z > 2 − 2.025   0.015 

= P( Z > −1.667 ) = Φ(1.667 ) = 0.9522

10 a X ~ N (250, σ 2)

7 X ~ N (168, 4.62)

158 − 168  = P Z < −2.174  a P( X < 158 ) = P  Z < ( )  4.6 

P (X < 235) = 0.005

=1–  Φ (2.174) = 1 –  (0.9850 + 0.0002) = 0.0148

z = – 2.576      235 − 250 = −2.576 σ σ = 5.82

b P(165 < X < 170 ) = P( X < 170 ) − P( X < 165) 68  = P  Z < 170 − 168  − P  Z < 165 − 16   4.6  4.6 

= P  (Z  70) = 1 − P(X < 70) 70 − 68.5 =1−P Z < 5 = 1 − P(Z < 0.3)





= 1 − Φ(0.3)





= 1 − 0.6179





= 0.3281

(

)

z = Φ−1(0.805) = 0.859

d P(Z < z) = 1− 0.375

z = Φ−1(0.625) = 0.318

e P(Z > z) = P(Z < −z) = 0.8531

−z = Φ−1(0.8531) = 1.05



z = −1.05

f P(Z > z) = P(Z < −z) = 0.6

−z = Φ−1(0.6) = 0.253

b 10% increase of 70 km/h gives 77 km/h





g P(Z < z) = 0.372 so P(Z < −z) = 1 − 0.372



P(X > 77) = 1 − P(X < 77)

(

=1−P Z
x) = 0.1949, P(X < x) = 0.8051

0

0.8051 = Φ(z) z = 0.86 = x − 30 10 x = 38.6

z

P(Z < z) = 0.95

d 0.75 = Φ(z)

z = Φ−1(0.95) = 1.645

z = 0.674 = x − 30 10 x = 36.74

i P(Z < −z) + P(−z < Z < z) = 0.4 + 0.5 z = Φ−1(0.9) = 1.282 (using the critical value for the normal distribution table) or 1.281 0.4

e

P(X < x) = 0.35 P(Z < −z) = 0.65

0.4

Φ−1(0.65) = −z

z = −0.385 = x − 30 10 x = 26.15

0.1

f If P(X > x) = 0.05, P(X < x) = 0.95 −z 2

0

z = 1.645 =

Underweight: 0.15 = Φ(z) z = −1.036 =

3

0.95 = Φ(z)

z x − 498.7 7.3

x = 491.14 Therefore, bags of sugar are rejected when the weight is less than 491.1 g. Overweight: 0.9 = Φ(z) z = 1.282 = x − 498.7 7.3 x = 508.06 Therefore, bags of sugar are rejected when the weight is more than 508.1 g. a X ~ N(19, 2.4²) P(X > a) = 0.432, therefore, P(X < a) = 1 − 0.432 = 0.568

x = 46.45 5

X ~ N(4000, 375²), P(X < x) = 0.9 0.9 = Φ(z) z = 1.282 = x − 4000 375 x = 4480 hours

6

X ~ N(1.753, 0.1²), P(X < x) = 0.929 0.929 = Φ(z) z = 1.468 = x − 1.753 0.1 x = 1.90 m

7

X ~ N(68.5, 5²), P(a < X < b) = 0.8 0.8 = P(X < b) − P(X > a) However, since the normal distribution is symmetrical about the mean, 0.9 = P(X < b), such that a = m − (b − m) Therefore, 0.9 = Φ(z)

0.568 = Φ(z)

z = 0.171= a − 19 2.4 a = 19.41

b

z = 1.282 = b − 68.5 5 b = 74.91, a = 62.09 The range of speeds is between 62.1 mph and 74.9 mph.

P(X < b) = 0.205 P(Z > −z) = 0.205 1 − 0.205 = Φ(−z) Φ−1(0.795) = −z z = −0.824 = b − 19 2.4 b = 17.02

4

a 0.99 = Φ(z) z = 2.326 = x = 53.26

x − 30 10

x − 30 10

8

a Marks scored by students are independent. b X ~ N(80, 12²), P(X > x) = 0.25, P(X < x) = 0.75 0.75 = Φ(z)

z = 0.674 = x − 80 12 x = 88.088

Therefore, the lowest mark needed is 89.

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4 Normal distribution

b $45 ÷ $1.55 = 29.03 litres

c P(a < X < b) = 0.95, and since the normal distribution is symmetrical, P(X < b) = 0.975

0.975 = Φ(z)











Therefore, the middle 95% of marks will be between 56 and 104.



z = 1.96 = x − 80 12 b = 103.52, a = 56.48

d P(X > 90) = 1 − P(X < 90)

(

90 − 80 =1−P Z < 12



= 1 − P(Z < 0.833)





= 1 − Φ(0.833)





= 1 − 0.7975





= 0.2025



200 × 0.2025 = 40.5



Therefore, 40 students are likely to score 90 or more.

9 X ~ N(190, 2.52) a P (X > x) = 0.2 1 – Φ(z) = 0.2 Φ(z) = 0.8 z = 0.842 190 x −      = 0.842 2.5

x = 192.1 g

b P( –z < Z < z) = 0.4

Φ(z) – (1 – Φ(z)) = 0.4



Φ(z) = 0.7



z = 0.524 x − 190 = 0.524 2.5

x = 191.3, 188.7

x − 190 = −0.524 2.5

Therefore, the range in weights is from 188.7 g to 191.3 g. c P( X  195) = P  Z  195 − 190  = P( Z  2)  2.5  = 1 − Φ( 2) = 1 − 0.9772 = 0.0228



100 × 0.0228 = 2.28



At least two tins are likely to be 195 g or more.

10 X ~ N(35, 82)

20 − 35  = P Z < −1.875  a P( X < 20 ) = P  Z < ( )  8  = 1– Φ(1.875) = 1 – (0.9693 + 0.0004) = 0.0303

)

= P ( Z > −0.746 ) = Φ ( 0.746 ) = 0.7722

c Lorries may also fill up at the petrol station, which will affect the normal distribution parameters, as will the amount of fuel remaining in the petrol station. Some people may just visit the counter at the petrol station rather than purchase petrol. This will also affect the normal distribution parameters.

)



(

P ( X > 29.03) = P Z > 29.03 − 35 8

Exercise 4.3C 1 2

X ~ N(43, σ ²) P(X > 48) = 0.2 Therefore, P(X < 48) = 0.8 0.8 = Φ(z) z = 0.842 = 48 − 43 s σ = 5.94 X ~ N( m, 1.3²) P(X > 47) = 0.11 Therefore, P(X < 47) = 0.89 0.89 = Φ(z) 47 − µ z = 1.227 = 1.3 m = 45.4 cm

3 X ~ N(m, σ ²) P(X > 332.91) = 0.18, P(X > 325.42) = 0.72 Therefore, P(X < 332.91) = 0.82 332.91 − m = Φ(0.82) s 332.91 − m = 0.915 s 332.91 − m = 0.915σ Also P(X < 325.42) = 0.28 325.42 − m = Φ(0.28) s 325.42 − m = −0.583 s 325.42 − m = −0.583σ 1 − 2: 7.49 = 1.498σ σ = 5 ml Therefore, m = 325.42 + 0.583(5) = 328 ml

1

2

4 X ~ N(m, σ ²) P(X < 9.92) = 0.25, P(X > 12.24) = 0.25 P(X < 9.92) = 0.25

9.92 − m = Φ(0.25) s 9.92 − m = −0.674 s 9.92 − m = −0.674σ

1

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WORKED SOLUTIONS





Also P(X < 12.24) = 0.75. 12.24 − m = Φ(0.75) s 12 . 24 − m = 0.674 s  12.24 − m = 0.674σ 2 − 1: 2.32 = 1.348σ σ = 1.721 Therefore, m = 12.24 − 0.674(1.721) = 11.08

Alternative method to find the mean:

LQ = 9.92 and UQ = 12.24 Therefore, µ = 9.92 + 1 (12.24 – 9.92) 2 = 9.92 + 1.16 = 11.08

2





= 1 − Φ(1)





= 1 − 0.8413





= 0.1587

c P(87 < X < 95) = P(X < 95) − P(X < 87)



= 0.9522 − 0.1587





= 0.7935

d X ~ N(μ, σ 2)



5 a X ~ N(15.5,

P(X >18) = 0.01 Φ(z) = 0.001

z = 2.326

18 − 15.5 = 2.326 σ σ = 1.07 cm b P( X < 14 ) = P  Z < 14 − 15.5  = P( Z < −1.395)  1.075  =1 − Φ(1.395) = 1 − (0.9177 + 0.0008) = 0.0815

8.15% pebbles are smaller than 14 cm.

6 X ~ N(500, σ 2)

(

b P(X > 508) = P Z >

)

508 − 500 = P ( Z > 0.506 ) 15.60



= 1 – (0.6915 + 0.002) = 0.3065

7 a X ~ N(90, 3²) P(X > 95) = 1 − P(X < 95) 95 − 90 =1−P Z < 3

(

= 1 − Φ(1.667) = 1 − 0.9522 = 0.0478

(

87 − 90 3 = P(Z < −1)







= Φ(−1)

)

)

) ( σ µ)

P(X < 65) = 0.2



85 −

= 0.9









85 – μ = 1.281 σ



(

) ( σ µ)











1

65 − µ = 0.2 σ

P Z
85) = 0.1



a P(X < 480) = 0.1



σ 2)

4 

20 = 2.123 σ

σ = 9.42 minutes

Therefore, μ = 85 – 1.281 × 9.42 = 72.9 minutes

8 X ~ N( m, σ ²) P(X ⩾ 140) = 0.2, P(X ⩾ 130) = 0.8 Therefore, P(X < 140) = 0.8 140 − m = Φ−1(0.8) s 140 − m = 0.842 s 140 − m = 0.842 σ Also P(X > 130) = 0.8 P(Z > z) = 0.8 P(Z < −z) = 0.8 130 − m = −Φ−1(0.8) s 130 − m = −0.842 s 130 − m = −0.842σ 1 − 2: 10 = 1.684σ σ = 5.94 kg Therefore, m = 140 − 0.842(5.94) = 135 kg

1

2

  Kk

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4 Normal distribution

9 X ~ N(12, 22)

Exercise 4.4A

a P( X < 11) = P  Z < 11 − 12   2 

1 a m = np = 20 × 0.5 = 10 b Variance: σ 2 = np(1 − p) = 20 × 0.5 × 0.5 = 5

= P( Z < −0.5) = 1 − Φ( 0.5)



= 1 − 0.6915 = 0.3085

(

13.5 − 12 b P ( X > 13.5) = P Z > 2

)

Standard deviation: σ = 5 = 2.236

c P(15.5 < X < 16.5) = P

= P ( Z > 0.75) = 1 − Φ ( 0.75) = 1 − 0.7734 = 0.2266

(15.52.236− 10 < Z < 16.52.236− 10 )







= Φ(2.907) – Φ(2.460)

= P(2.460 < Z < 2.907)

= 0.9982 − 0.9931 ( X) =P13 = 0.0051  2  10−13 12− 12  12 10 −   10 13 P X < + P X > = P Z < ) ) =+( PP13  −)12   P( X < 10 ) +( P( X > 13 Z< 5) < −1) + P( Z > 0.5) 2 a µ = 108 × 0.88 = 95.04  2 2= P( Z+PP0(.Z σ2 = 108 × 0.88 × 0.12 = 11.4048  13 − 12  = P Z < −1 + P Z > 0.5 − 13 12   + P ) .5) ( ) = P Z 0 σ = 3.377 +P   2   (2  ) ( 99.5 − 95.04 P(X < 99.5) = P Z < = (1 – Φ (1)) + (1 – Φ (0.5)) = (1 – 0.8413) + 0.3085 3.377          = Φ(1.321) = 0.4672

(

d P(X > 13) = 0.05    and   P(X < 10) = 0.12



1 – Φ(z1) = 0.95   and   1 – Φ(–z2) = 0.88

b P(85.5 < X < 104.5)



13 − µ = 1.645 σ 13 – µ = 1.645σ 10 − µ = −1.171    σ –10 + µ = 1.171σ

          = 0.9067

(

= P 85.5 − 95.04 < Z < 104.5 − 95.04 3.377 3.377

z1 = 1.645          –z2 = 0.1171





)

1

2



= P(Z < 2.801) − P(Z < −2.825)



= Φ(2.801) – (1 − Φ(2.825))



= 0.9975 − (1 − 0.9976)



= 0.9951

)



1 − 2:



3 = 2.816σ

3 m = np = 200 × 0.32 = 64 Variance: σ 2  = np(1 − p) = 200 × 0.32 × 0.68 = 43.52



σ = 1.07 minutes



Therefore, µ = 13 – 1.645 × 1.07 = 11.2 minutes

Standard deviation: σ = 43.52 = 6.597 129.5 − 64 a P(X < 129.5) = P Z < 6.597

(

)

= Φ(9.93) 12 − 16 P ( X < 12) + P12 18 ) = P Z < ( X −>16 ≈1 4.5 a P ( X < 12) + P ( X > 18 ) = P Z < 4.5 12 − 16 b P(X > 50.5) = P  Z > 50.5 − 64  = P(Z > − 2.046) 18 − 16 > ) +)P=( PX 18 )12= −P16Z 0.444 ) > 18 Z 18 < ( X < 12) P+(PX( X< 12 − 16 4.5 ) + P ( Z > 0.444 ) 4.5 +P = P ( Z < −0.889 4.5 = P( Z < 2.046) 18 − 16+ P 18 − 16 = P ( Z < −0.889) + P ( Z > 0.444 ) P = P (4.5 Z < −0.889) + P ( Z > 0.444 ) 4.5 = Φ(2.046) = (1 – Φ (0.889)) + (1 – Φ (0.444)) = 0.9796 = (1 – (0.8106 + 0.0025)) + (1 – (0.67 + 0.0014)) 75.5 − 64 c P(X > 75) = P Z > 6.597 = 0.5155 = 1 − Φ(1.743) b P(X < x) = 0.9, Φ(z) = 0.9, z = 1.282 10 X ~ N(16,

(

)

)(

4.52)

((

( )() ( ) )

) (

)

(

x − 16 = 1.282 4.5 x = 21.769, so 22 minutes c X ~ N(16, σ 2)

P(X < 19) = 0.99, Φ(z) = 0.99, z = 2.326 19 − 16 = 2.326 σ σ = 1.29 minutes







)

= 1 − 0.9594 = 0.0406

d P(75 < X < 130) = P(X < 129.5) − P(X < 75.5)



= 1 − 0.9594

= 0.0406 4 a  You may use the normal distribution as an approximation for the binomial B(n, p) (where n is the number of trials each having probability p of success) when:

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4 

WORKED SOLUTIONS

b P ( 44.5  X  45.5) = P ( X  45.5) − P ( X  44.5)



n is large



np > 5 and n(1 – p) > 5



This ensures that the distribution is reasonably symmetrical and not skewed at either end.

b P(T < 5) = 0.2173

(

)

= P(Z , 0.972) – P(Z , 0.795) = Φ(0.972) – Φ(0.795)

c Mean = m = 15 × 0.4 = 6

) (

= P Z  45.5 − 40 − P Z  44.5 − 40 5.657 5.657

= (0.8340 + 0.0005) – (0.7852 + 0.0015) = 0.0478

8 a X ~ B (30, 0.12)



Variance = σ 2 = np(1 − p) = 15 × 0.4 × 0.6 = 3.6



Standard deviation =



P(T < 5) = P Z
−2.033) = Φ ( 2.033) = 0.9790

(

=1−P Z
900) = P Z >





= 1 − Φ(1)





= 1 − 0.8413





= 0.1587



= Φ(0.909)









c P(1000 < X < 1007) = 1 − (0.1587 + 0.0062)





So the probability that they both contain between 1000 g and 1007 g is 0.83512 = 0.697.

= 0.8351

(

2 a P(taller than 160 cm) = P Z > 160 − 164.5 8.75 = Φ(0.514)



= 0.6964

)

= 0.8182

P(Z < z) = 0.45



P(Z < −z) = 0.55

)

(

b P(X > 1200 hours) = P Z > 1200 − 1000 110 = P(Z > 1.818)



= 1 − Φ(1.818)





= 0.0345

)

c P(900 < X < 1000) = P(X < 1000) − P(X < 900) 1000 − 1000 = P Z < 110

(

b P(H < h) = 0.45

)

(

−P X
320) = 1 – P(X < 320)  320 − 4( 72)   =1−Φ   4 × 10 2    = 1 − 0.9452 = 0.0548





= 0.3182

4 a X ~ N(82, 62)

x − 1000 = −2.326   110 x = 744.14 hours



z = −0.125 = h − 164.5 8.75 h = 163.4 cm

(



(

P(X > 90) = P Z >

90 − 82 6

)

)

= P(Z < 0) − P(Z < −0.909)

d P(X < x) = 0.01

P(Z < z) = 0.01



P(Z < −z) = 0.99



−z = 2.326



or 31 days to the nearest day.

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WORKED SOLUTIONS

b X ~ N(m, 0.13²)

6 a X ~ B(30, 0.2)

Approximated by Y ~ N(6, 4.8) since np = 6 > 5 and nq = 24 > 5.

b P( X  4) = P(Y > 3.5) 3.5 − 6   = P Z >   4.8 



3.54 − µ  0.13 

z = −Φ−1(0.96)





m = 3.77 litres

c X ~ N(3.58, σ ²)



 





= P Z < 

= Φ(1.141) = 0.8731









−211.8 + m = 1.175σ 1 + 2: 0.8 = 1.615σ σ = 0.495 Therefore, m = 212.6 – 0.44(0.495) = 212.38 P(X > 212) = P  Z > 212 − 212.38   0.4954    = P(Z > −0.767)





= P( Z > −1.141)

7 a X ~ N(m, σ ²) P(X > 212.6) = 0.33 Therefore, 0.33 = 1 − P(X < 212.6) = 1 − P  Z < 212.6 − µ    σ 212.6 − µ Φ−1(0.67) = σ 212.6 − µ 0.440 = σ 212.6 − m = 0.44σ 1 Also 0.12 = P(X < 211.8) P(Z < z) = 0.12 is the same as P(Z < −z) = 0.88. Φ−1(0.88) = −  211.8 − µ    σ



P(X < 3.54) = 0.04

 3.54 − µ  = −1.751  0.13  3.54 − m = −1.751× 0.13

= 0.8731 × 0.8731 = 0.7623







c P(X  4 on each of two days)



2



P(X < 3.54) = 0.04

















(

3.54 − 3.58 s z = −Φ−1(0.96) =P Z
400) = 0.99 Therefore, P Z > 400 − 416 = 0.99 2.828s

(



(

P Z
1000 − 950   589 





= P(Z > 2.06)





= 1 − P(Z < 2.06)





= 1 − 0.9803



= 0.0197

11 X ~ N(75, 122) a From the tables: P(Z < 1.281) = 0.9 x − 75 Therefore, 1.281 =   ⇒   x = 90.372 12 Assuming that only whole marks are awarded, students must score at least 91 marks to be in the top 10%.

53

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4 Normal distribution

b From the tables: P(z < 0.674) = 0.75 Therefore, 0.674 = x − 75   ⇒   x = 83.1 12 By symmetry, the middle 50% of the students score between 67 marks and 83 marks (to nearest mark).

(

c P(X < 40) = P Z < 40 − 75 12

)



From the tables: P(z < 1.724) = 0.9577



Therefore, 1.724 = 15 − 8.75   ⇒   σ = 3.63 minutes s

b P(X < 5) = P z < 5 − 8.75   3.625 



= P(z < −1.034)





= 1 − P(z < 1.034)





= P(Z < −2.917)





= 1 − 0.8494





= 1 − P(Z < 2.917)





= 0.1506

c 15 000 × 0.1506 = 2259 patients





= 1 − 0.9982





= 0.0018

14 M ~ B(200, 0.44) np = 200 × 0.44 = 88

12 X ~ N(55, 172) a Equal proportions means equal probabilities 42 − 55 P(X < 42) = P Z < 17 = P(Z < −0.765)

nq = 200 × 0.56 = 112





= 1 − P(Z < 0.765)

M ~ N (88, 49.28)





= 1 − 0.7779





= 0.2221

P ( 75.5 < M < 99.5) = P ( M  99.5) − P ( M  75.5)



The 0.7779 remaining for standard eggs and large eggs needs to be divided equally. This means that P(X < x) = 1 − 0.7779 = 0.6111 2 From the tables: P(z < 0.282) = 0.6111 Therefore, 0.282 = x − 55   ⇒  x = 59.79 17 So eggs weighing more than 59.8 grams should be considered to be large eggs.

(



)

(

)

b X ~ B(1000, 0.01)

np = 1000 × 0.01 = 10 and nq = 1000 × 0.99 = 990.



Since np ≥ 5 and nq ≥ 5, a normal approximation can be used.

X ~ N(10, 9.9)

To calculate P(X = 10), we need to use a continuity correction.



P(X = 10) ≈ P(9.5 < X < 10.5)





≈ P(X < 10.5) − P(X < 9.5)





  10.5 − 10  9.5 − 10  ≈ PZ <  − P  Z <   9.9  9.9 





≈ P(Z < 0.159) − P(Z < −0.159)





≈ P(Z < 0.159) − [1 − P(Z < 0.159)]





≈ 2P(Z < 0.159) − 1





≈ 2 × 0.5632 − 1





= 0.1264

13 P(X > 15) = 0.0423 and X ~ N(8.75, σ 2) a P(X < 15) = 0.9577

Since 88 > 5 and 112 > 5, a normal approximation can be applied. np(1 – p) = 200 × 0.44 × 0.56 = 49.28

    = P  Z  99.5 − 88  − P  Z  75.5 − 88    49.28  49.28  = P(Z , 1.638) – P (Z , – 1.781) = Φ (1.638) – (1 – Φ (1.781)) = (0.9484 + 0.0008) – 1 + (0.9625 + 0.0001) = 0.912 15 a X ~ B(12 000, 0.003) b μ = np = 12 000 × 0.003 = 36

σ 2 = np (1 – p) = 12000 × 0.003 × 0.997 = 35.892

c np = 36

nq = 12 000 × 0.997 = 11 964

Since 36 > 5 and 11964 > 5, a normal approximation can be applied.

Y ~ N (36, 35.892) P(29.5  Y  30.5) = P(Y  30.5) − P(Y  29.5)



 30.5 − 36  P  Z 29.5 − 36  = P Z   −     35.892  35.892 



= P(Z  −0.918) − P(Z  −1.085) = (1 − Φ(0.918)) − (1 − Φ(1.085))



= Φ(1.085) − Φ(0.918) = (0.8599 + 0.0012) − (0.8186 + 0.002) = 0.0405

16 T ~ N(75, 152)

(

a i P (T < 60 ) = P Z
5, a normal approximation can be applied.

Since 6 > 5 and 294 > 5, a normal approximation can be applied.

np(1 – p) = 50 × 0.48 × 0.52 = 12.48

np (1 – p) = 300 × 0.04 × 0.98 = 5.88

X ~ N (24, 12.48)

X ~ N (6, 5.88)

  P( X < 4.5) = P  Z < 4.5 − 6   5.88 



= 1 – (0.7291 + 0.0029) = 0.268

  P( X > 30.5) = P  Z > 30.5 − 24   12.48 

= P( Z < −0.619) = 1 − Φ( 0.619)

b The cost to produce 300 dolls: 1.50 × 300 = $450



np = n × 0.07 = 8    n = 114.25    Therefore, 115 people b np = n × 0.07 = 2    n = 29

Therefore, the expected profit = $2940 – $450 = $2490 18 a W ~ N (500, 152)

σ 2 = np(1 – p) = 29 × 0.07 × 0.93 = 1.8879 Therefore, σ = 1.8879 = 1.374



Mathematics in life and work 1 P(X > 308 g) = 1 − P(X < 308) 308 − 300 =1−Φ 11

z = − 1.282 w − 500 = −1.282 15 w = 480.8

(

b P ( 505  W  515) = P (W  515) − P (W  505)

(

= 1 – 0.9671 = 0.0329

a E(X) = 8

The revenue from the dolls = (300 – 6) × 10 = $2940

Φ (–z) = 0.9

= P( Z > 1.840 ) = 1 − Φ(1.840 )

20 X ~ B (n, 0.07)

The expected number of damaged dolls = 300 × 0.02 = 6

P(W < w) = 0.1

np = 50 × 0.48 = 24

nq = 50 × 0.52 = 26

nq = 300 × 0.98 = 294



 12  12 +  ( 0.48 ) = 0.0137  12 

c n = 50, p = 0.48

np = 300 × 0.02 = 6



P(X > 9) = P(X = 10) + P(X = 11) + P(X = 12)  12  10 2  12  11 1 =  ( 0.48 ) ( 0.52) +  11  ( 0.48 ) ( 0.52) 10    





4 

) (





)

= 1 − 0.7663 = 0.2337

)

2 Find m such that P(X < m) = 0.02 P(Z < z) = 0.02 P(Z < −z) = 0.98 = P(Z < 1) – P(Z < 0.333) = Φ(1) – Φ(0.333) −z = Φ−1(0.98) = 0.8413 – (0.6293 + 0.0011) = 0.2109 z = −2.054  Let Y be the number of Chinese leaf cabbages m − 300 = −2.054 that Amy picked, Y ~ B (5, 0.211) 11  5 m = 277.4 = 277 g (3 s.f.) 3 2  5 4 1  5 5 P (Y  3) =   ( 0.211) ( 0.789) +   ( 0.211) ( 0.789) +   ( 0.211) = 0.0667 3  5Y ~ N(m, σ ²),  3  4 P(Y > 290) = 0.95, P(Y < 305) = 0.97  5  5  5 3 290 − m P (Y  3) =   ( 0.211) ( 0.789 )2 +   (0.211)4 (0.789)1 +   (0.211)5 = 0.0667 Therefore, Φ−1(0.95) = −1.645 = s  3  4  5 =P Z
3500 hours) = × + × 10 5 10 5 10 18 3 21 = + = 40 50 50 50

T  2000

95

T  2500

167

A Level Questions

T  3500

200

1

)

(109 × 52 ) + (101 × 35 )

X ~ N(μ, 0.7142) and the area we are interested in is:

ii Be sure to plot cumulative frequency against the upper class boundaries. 200

T

Lifetime (hours)

c P(exactly one bulb > 3500 hours) =

T  1000

Cumulative frequency

500

Therefore, the number of bulbs lasting more than 2200 hours is 200 – 120 = 80 bulbs.

μ ≈ 1999 hours

0.475 ×

×

150 100

× 3.2

50

× 0

× ×

µ ≈ 399 750 200

b i

×

50

Therefore, Sammy is more likely to win because he has the higher probability. 3

× ×

× × 500

1000 1500 2000 2500 3000 3500

Lifetime (hours)

T

μ

Using the cumulative normal distribution tables in reverse, we can say that P(Z < 1.96) = 0.975. Therefore:

57

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Summary REVIEW

     

−1.96 =

3.2 − µ 0.714

4 i For type A, there are 61 smartphones, so the median is the 31st item in the ordered list.

 −1.3994 … = 3.2 − μ μ = 4.5994 … = 4.60 m (3 s.f.)



The median is 0.52 seconds.



The lower quartile is between the 15th and 16th items in the ordered list.



The lower quartile is: 0.40 + 0.42 = 0.41 seconds 2 The upper quartile is between the 15th and 16th items from the end of the ordered list.

2 X ~ B(11, 0.76) P(X < 10) = 1 − [P(X = 10) + P(X = 11)]  11   11 = 1 −   × 0.7610 × 0.24 +   × 0.7611   11  10   11   11 = 1 −   × 0.7610 × 0.24 +   × 0.7611  11 10       = 0.781 3 i P(1st die divisible by 3 and 2nd die not divisible 2 4 2 by 3) = × = 6 6 9 But to find P(A), we need to consider that this could happen the other way around, i.e. with the 1st die not divisible by 3 and the 2nd die divisible by 3. Therefore, P( A) = 2 × 2 = 4 9 9



We know that the product of two odd numbers is an odd number. So the probability of an odd product is: 3 3 9 1 × = = 6 6 36 4 1 3 Therefore, P(B) = 1 − = 4 4 4 3 1 So P(A) × P(B) = × = 9 4 3 To find P(A ∩ B) it is useful to draw a sample space diagram. 1

2

3

4

5

6

1 2 3 4 5 6



represents event A represents event B Yellow highlighting represents A ∩ B



Therefore, P(A ∩ B) = 12 = 1 . 36 3



Because P(A ∩ B) = P(A) × P(B), we can conclude that A and B are independent.

ii Since P(A ∩ B) > 0, we can say that A and B are not mutually exclusive.



The upper quartile is: 0.78 + 0.80 = 0.79 seconds 2

ii Type A Type B 0

0.2

0.4

0.6

0.8

1

Time (s) iii The median for B is lower than for A, so smartphone B is quicker because it takes less time on average to load the apps. The interquartile range for B is narrower too, so smartphone B is more consistent. 5 a i If we treat the 4 Es as though they are one block, then we get (EEEE)***** and there are 6! arrangements of this list. However, there are 2 Ns and 2 Ss, so we need to divide by 2! twice to remove the duplicate arrangements. The number of arrangements is 6! = 180 . 2!2! ii Suppose that S is fixed in place at the beginning of the list and that T is fixed in place at the end of the list: S*******T. There are 7! arrangements, but there are 2 Ns and 4 Es, so we need to divide by 4!2!. We get 7! = 105 . But we 4!2! could repeat the process with S and T reversed, which doubles the number of arrangements to 210. b There are 3 Es available. Assume that one of them is selected and that the others are then unavailable. We then need to choose 3 letters from the remaining 6 letters, which is 6C = 20. 3

There are 20 possible selections that contain exactly one E.

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WORKED Solutions

6 i

1st Attempt

iii The lower quartile is approximately the 28th person, so in class 6–20.

3rd Attempt

2nd Attempt

The upper quartile is approximately the 83rd person, so in class 61–80.

S 0.4 U 0.6

S 0.4

S = P (successful) U = P (unsuccessful)

IQRmin = UQmin − LQmax



IQRmin = 61 − 20 = 41

8 µ =

S 0.4

U 0.6





5 − 2 + 12 + 7 − 3 + 2 − 6 + 4 + 0 + 8 = 2.7 10

Var(X) = E(X2) − E2(X)

2 2 2 2 2 2 2 2 2 2 U Var( X ) = [5 + (−2) + 12 + 7 + (−3) + 2 + (−6) + 4 + 0 + 8 ] ÷ 10 2 Var( X ) = [52 + (−2)2 + 0.6 122 + 7 2 + (−3)2 + 2 + (−6)2     + 4 2 + 0 2 + 8 2] ÷ 10 − 2.7 2

ii P(X = 0) = 0.4

Var(X) = 35.1 − 2.72 = 27.8

P(X = 2) = 0.62 × 0.4 = 0.144



P(X = 3) = 0.63 = 0.216



So the completed table is:

9 i The upper quartile is in the class ‘5.5–7.0’ cm. ii Firstly, find the class widths, then each frequency density as shown in the table below.

x

0

1

2

3

P(X = x)

0.4

0.24

0.144

0.216

Length (cm)

iii E(X) = (0 × 0.4) + (1 × 0.24) + (2 × 0.144) + (3 × 0.216)

E(X) = 1.176

7 i Firstly, find the class widths, then each frequency density as shown in the table below. 1–5 6–20 21–35 36–60 61–80

Frequency ( f )

24

9

21

15

42

Class width (cw)

5

15

15

25

20

Frequency density ( f/cw)

4.8

0.6

1.4

0.6

2.1

Frequency density

Number of typing errors

8

25

28

31

12

Class width (cw)

1.5

1

1

1.5

2

Frequency density ( f/cw)

5.33…

25

28

20.666…

6

30 25 20 15 10 5 0

5

1

2

3

4

5

6

7

8

9

Length (cm)

4 3 2 1 0



2.0–3.5 3.5–4.5 4.5–5.5 5.5–7.0 7.0–9.0

Frequency (f )

Frequency density



10 20 30 40 50 60 70 80

Number of errors

10 i There must be a fixed number of trials. Each trial must have exactly two discrete possible outcomes, one of which is defined as ‘success’. The trials must be independent with the probability of success the same in each trial.

ii µ ≈ (3 × 24) + (13 × 9) + (28 × 21) + (48 × 15) + (70.5 × 42) = 40.2 errors 111 1 9) + (28 × 21) + (48 × 15) + (70.5 × 42) = 40.2 errors   111 1

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SUMMARY REVIEW

ii X ~ B(18, 0.15)

Using the cumulative normal distribution tables in reverse, we can say that P(Z < 1.406) P(X  3) = 1 − P(X  2) = 0.92. Therefore: = 1 − [P(X = 2) + P(X = 1) + P(X = 0)] c − 14.2 −1.406 =  18 3.6 2 16  18  17  18  18  = 1 −   × 0.15 × 0.85 +   × 0.15 × 0.8 85 +   × 0.85  1 2 0       −5.0616 … = c − 14.2 

 18   18  18 = 1 −   × 0.152 × 0.8516 +   × 0.15 × 0.8 8517 +   × 0.8518  1 2 0         = 1 − 0.4796 … = 0.520 (3 s. f.) 11 i P(2 paperback and 2 hardback) = 6 5 2 1 1 × × × = 8 7 6 5 28

c = 9.14 cm (3 s.f.) ii Converting from x-values to z-values, we get: 16 − 14.2 z1 = = 0.5 and z 2 = 15 − 14.2 = 0.2 3.6 3.6 P(15 < X < 16) = P(X < 16) − P(X < 15) = P(Z < 0.5) − P(Z < 0.22…) = 0.6915 − 0.5879

4C 2

But there are = 6 ways of selecting the two hardback books within the selection of 4. Therefore, P(X = 2) = 1 × 6 = 3 28 14

= 0.1036 So X ~ B(7, 0.1036) P(X  2) = 1 − P(X  1)

= 1 − [P(X = 0) + P(X = 1)] Alternative method: There are 8C4 = 70 different 6 2 possible selections. C2 × C2 = 15 of them  7   7 = 1 −   × 0.89647 +   × 0.1036 × 0.89646  involve exactly 2 hardback books. So the 1 0      probability is 15 = 3 70  7  14  7 = 1 −   × 0.89647 +   × 0.1036 × 0.89646  ii P(X = 0) = P(X = 1) = 0 as they are not  1  0  possible. = 1 − 0.8413 … P(X = 2) = 3 from part (i) 14 = 0.159 (3 s.f.) 6 5 4 3 360 3 P(X = 4) = × × × = = 13 i If there are more women than men, then the 8 7 6 5 1680 14 number of women must be 4, 5 or 6. Since probabilities sum to 1, we can say that 8C × 5C = 700 For 4 women and 2 men: 4 2 P(X = 3) = 8 . 8C × 5C = 280 14 For 5 women and 1 man: 5 1 For 6 women and 0 men:

So the probability distribution table is: x P(X = x)

2

3

4

3 14

8 14

3 14

iii Var(X) = E(X2) − E2(X) =  22 × 3 + 32 × 8 + 4 2 × 3  − 32  14 14 14  132 −9 14 = 0.429 (3 s.f.) =

12 i

6

× 5C0 = 28

Therefore, the total number of ways is: 700 + 280 + 28 = 1008 ii The committee could include one of the two males (M1 or M2) or neither of them. For M1 selected and M2 unavailable, we must then select 2 men from the remaining 3, and 3 women from the remaining 8. So there are 3C × 8C = 168 arrangements. 2 3 There are another 168 arrangements for M2 selected and M1 unavailable. If neither of the men are selected, we must then select 3 men from the remaining 3, and 3 women from the remaining 8. So there are 3C × 8C = 56 arrangements. 3 3

8%

60

8C

So the total number of ways is: 168 × 2 + 56 = 392

c

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WORKED Solutions

15 i There must only be two possible outcomes (success and failure). Trials must be independent. The probability of success must be the same in each trial.

iii If we assume that both ends of the line are occupied by women, then there are 5P2 ways of choosing those women. There are 4! ways of arranging the middle 4 committee members regardless of their gender.

ii P  (banana on the 4th day) = 0.723 × 0.28 = 0.105 (3 s.f.)

P2

5

  16 P( 25 < X < 30 ) = P  25 − 28.3 < Z < 30 − 28.3   4.5 4.5 

4!



So the number of ways = 5P2 × 4! = 480



Alternatively, 5! × 4 = 480

ii To find P(X = 3), we need to consider (H, 1, 2), (H, 2, 1) and (T, 3). 1 1 1 1 1 1 1 1 P(X = 3) = 2 × 4 × 4 + 2 × 4 × 4 + 2 × 4

)(



= 1 + 1 + 1 32 32 8



= 3 16

)(



)(

(

)

(

)

P(X = 4) = 1 × 1 × 1 × 3 + 1 × 1 = 7 2 4 4 2 4 32 Since probabilities sum to 1, P(X = 6) = 3 32

= P(Z < 0.801) – (1 – P(Z < 1.556))



= P(z < 0.801) + P(Z < 1.556) – 1



= 0.7884 + 0.9401 – 1 = 0.729 (3 s.f.)

17 i The probability of arriving on time is 0.2 so X~B(96, 0.2). E(X) = 96 × 0.2 = 19.2 and Var(X) = 96 × 0.2 × 0.8 = 15.36 So a normal approximation is X~N(19.2, 15.36)



)

)





iii To find P(X = 2), we need to consider (H, 1, 1), and (T, 2). 1 1 1 1 1 P(X = 2) = 2 × 4 × 4 + 2 × 4 1 1 = + 32 8 5 = 32 To find P(X = 4), we need to consider (H, 1, 3), (H, 2, 2), (H, 3, 1) and (T, 4).

(

= P(– 1.556 < Z < 0.801)



14 i If the coin shows heads, then we must use the sum of the two dice, each of which has a minimum score of 1, so X = 1 is impossible with heads. With tails, X = 1 is possible if the die shows a score of 1. P(X = 1) = 1 × 1 = 1 2 4 8

(



For the normal approximation of P(X < 20), the continuity correction means that we use X = 19.5   P( X < 20 ) = P  Z < 19.5 − 19.2  =  15.36 

P(Z < 0.0765) = 0.531 (3 s.f.) ii Let E be early, OT be on time, L be late and B be eat a banana.

P(E) = 0.05



P(B | E) = 0.7



∴ P(B′ ∩ E) = P(B′|E) P(E) = 0.3 × 0.05 = 0.015



P(B|OT) = 0.4



∴ P(B′ ∩ OT) = P(B′|OT) P(OT) = 0.6 × 0.2 = 0.12



P(B|L) = 0



∴ P(B′ ∩ L) = P(B′|L) P(L) = 1 × 0.75 = 0.75



P(B′) = P(B′ ∩ E) + P(B′ ∩ OT) + P(B′ ∩ L)  = 0.015 + 0.12 + 0.75 = 0.885



P(OT ∩ B′) = 0.2 × 0.6 = 0.12



P (OT |B ′ ) =

⇒ P(B′ | E) = 0.3 ⇒ P(B′|OT) = 0.6 ⇒ P(B′|L) = 1

P(OT ∩ B ′) 0.12 = = 0.136 P(B ′) 0.885

So the completed table is: x

1

2

3

4

5

6

7

8

P(X = x)

1 8

5 32

3 16

7 32

1 8

3 32

1 16

1 32

iv P(Q ∩ R) = 0 since if the coin lands on tails, then the 1 remaining die cannot score 7. Therefore Q and R are mutually exclusive.

61

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Summary REVIEW



Alternative method 0.7

B

0.3 0.05 E 0.75 L 0.2 OT

B’ 0 1 B’ B 0.6

 n  0 n  0  × 0.2 × 0.8 < 0.01  



0.8n < 0.01



log 0.8n < log 0.01

nlog 0.8 < log 0.01 n > log0.01 (change of sign since log0.8 < 0) log0.8

B

0.4



n > 20.63 …

Therefore, the lowest possible value of n is 21. 20 X~Geo(0.35) ⇒ E ( X ) = 1 = 2.85… ⇒ 0.35

B’ P(OT ∩ B ′) 0.2 × 0.6 Daniel should expect to ask 3 people. P(OT | B ′) = = P( ) B (0.05 × 0.3) + 0.75 + (0.2 × 0.6) ′ 21 i X ~ N(μ, σ 2) P(OT ∩ B ′) 0.2 × 0.6 P(OT | B ′) = = P(B ′ ) (0.05 × 0.3) + 0.75 + (0.2 × 0.6) 63 = 0.105 ⇒ P(X < 6) = 0.105 600 = 0.12 = 0.136 (3 sf . .) ∴ From the tables, z = – 1.253 0.885 ∴ 6 − µ = −1.253 (1) 2 18 i E ( x − c ) = 1957.5 = 65.25 σ 30 155 = 0.2583 ⇒ P(X > 12) = 0.2583 600 E( x − c ) = 234 = 7.8 ∴ From the tables, z = 0.648 30 ∴ 12 − µ = 0.648 (2) Var(x) = Var(x – c) = 65.25 – 7.82 = 4.41 ⇒ σ = 2.1 σ

(

)

ii ∑ ( x − c ) = ∑x − ∑c

234 = ∑x − 30c



∑x = 234 + 30c



234 + 30c = 86 30



234 + 30c = 2580

30c = 2346 c = 78.2 Alternatively: E(X) = E(X – c) + c

86 = 7.8 + c ⇒ c = 78.2

X~B(12, 0.2)

(2) – (1)



6 = 1.901σ



Substitute in (2)



P(Z > 1) = 1 – 0.8413 = 0.1587



P(Z < –1) or P(z > 1) is 2 × 0.1587 = 0.3174



0.3174 × 1000 = 317.4

P(3  X  5) = 0.236 … + 0.132 … + 0.053 … = 0.422

iii X~B(n, 0.2)

P(X = 0) < 0.01

⇒ 317 feathers

22 a i  There are 7! ways to arrange the 7 couples. 7! = 5040





Each couple has two possible arrangements and there are 7 couples.



Therefore, the total number of arrangements is 5040 × 27 = 645 210. ii There are 7! arrangements of friends, 7! arrangements of partners and the whole group can be arranged as (partners and 5 × 0.2friends) × 0.87 or (friends and partners). So the number of different arrangements is 7! × 7! × 2 = 50 803 200.



 12  12  12 P ( 3  X  5) =   × 0.23 × 0.89 +   × 0.24 × 0.88 +    3  4  5



σ = 3.156... = 3.16 (3 s.f.)

ii P(Z < – 1) = 1 – 0.8413 = 0.1587

ii P(3  X  5) = P(X = 3) + P(X = 4) + P(X = 5)

 12  12  12 3  X  5) =   × 0.23 × 0.89 +   × 0.24 × 0.88 +   × 0.25 × 0.87  5  3          4 

⇒

μ = 12 – 0.648 × 3.156 = 9.95 (3 s.f.)

19 i T  here are 3 multiples of 5 in the given range and 15 possible numbers to choose from. P(multiple of 5) = 3 = 0.2 15



b i



7! = 7 × 6 = 21 ways 2!5! 2

ii Adults all in team ⇒ only 1 way Adults all not in team ⇒ 5! = 5 ways 4!1!

So there are 6 ways in total.

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WORKED Solutions



iii At least 2 girls means either 2 girls or 3 girls. 27 i There are x yellow balls in box B (numerator). 3! 6! 6 × 5 The total number of balls in box B is 5 white 2 girls ⇒ × = 3× = 3 × 3 × 5 = 45 2!1! 2!4! 2 balls, x yellow balls and 1 additional white ball 3 girls ⇒ 1 of the remaining 6 people = x + 6 balls (denominator). with the 3 girls ⇒ 6 ways So there are 51 ways in total. ii 6 White x+6 23 ∑ ( x − 36 ) = −60 White 4 5 x ∑x − ∑36 = −60 Yellow x+6 ∑x − 24 × 36 = −60 5 White x+6 1 x = 804 ∑ 5 Yellow

∑ ( x − 36 )2 = 227.76

x+1 x+6

∑x 2 − 72∑x + ∑362 = 227.76 ∑x

2

iii 6 = 1 ⇒ x + 6 = 18 ⇒ x = 12 x+6 3 iv P(Y2) = P(W1 ∩ Y2) + P(Y1 ∩ Y2) = 4 × 12 + 1 × 13 = 48 × 13 = 61 5 18 5 18 90 90 90 1 13 13 P(Y1 ∩ Y2) 5 × 18 90 13 P (Y1|Y2 ) = = = = 61 61 P(Y2) 61 90 90

2

− 72 × 804 + 24 × 36 = 227.76

∑x 2 = 27 011.76 24 i X~Geo(0.32)

P(X = 7) = 0.686 × 0.32 = 0.0316

ii P(X < 3) = P(X = 1) + P(X = 2) = 0.32 + 0.68 × 0.32 = 0.5376

28 i

iii P(X > 3) = 0.683 = 0.314 (3 s.f.)   2  25 i Y ~N  µ, µ  P (Y < 0 ) = P  Z < 0 − µ  = P Z < − 2µ µ  µ  4   2     µ2  0 − µ 2µ  Y ~N  µ,  P Y < 0 = P Z < ( ) µ  = P Z 2.1) = = 0.03163 8000

Yellow

Flat screen

) 6

4

Conventional 6

5

7

9

6

7

1

4

5

9

5

8

5

6

2

1

9

7

4 10

Key:

6|7|1

7

means 0.76 m for flat screen 0.71 m for conventional

ii The median is 0.74 m

∴ From the tables, z = 1.858 IQR = UQ − LQ =  0.77 + 0.85  −  0.67 + 0.69  = 0.81 − 0.68 = 0.13 m     2 2 ∴ 2.1 − 2.04 = 1.858 ⇒ σ = 0. 0323 (3 s.f.) σ IQR = UQ − LQ =  0.77 + 0.85  −  0.67 + 0.69  = 0.81 − 0.68 = 0.13 m    2            2 26 i P(X < 73) = 0.15 ∴ From the tables, z = –1.036 iii Using x to represent the diagonal lengths of flat 73 − 75 = −1.036 ∴ ⇒ σ = 1.93 (3 s.f.) screen TVs: ∑x = 8.34   ⇒  µ = 8.34 = 0.927m σ 9 ii P(X > 77) = 0.15 (by symmetry) ⇒ Y~B(8, 0.15) 7.7984 ∑x 2 = 7.7984 ⇒ σ = − 0.9272 = 0.0882m 9 P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)  8  8  8 There must only be two possible outcomes 0 8 1 7 29 i  0.152 × 0.856 P (Y < 3) =  0 × 0.15 × 0.85 +  1  × 0.15 × 0.85 +  2 × (success and failure). Trials must be independent. The probability of success must be  8  8  8 the same in each trial. P (Y < 3) =   × 0.15 × 0.858 +   × 0.151 × 0.857 +   × 0.152 × 0.856 0     1  0  2 ii P(buys on the nth call) = (1 – p)n–1p P(Y < 3) = 0.272 … + 0.384 … + 0.237 … = 0.895

63

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Summary REVIEW

⇒

ii If the number of doctors taking their final examinations is an even number, then the 0.2016 = p – statement is always true, as the median is p2 – p + 0.2016 = 0 positioned such that 50% of the data is below 1 ± 1 − 4(1)(0.2016) 1 ± 0.44 the median and 50% is above the median. p= = 2 2 If the number of doctors is odd, then the p = 0.28 or p = 0.72 number of doctors above and below the So the maximum probability is 0.72. mean is equal, but not quite 50%, since the data is discrete and the middle item of data 30 i X~B(20, 0.05) is effectively discounted. However, as the P(X > 1) = 1 – P(X  1) = 1 – P(X = 0) – P(X = 1) number of doctors increases, each ‘half’ of the  20  20 0 20 1 19 P ( X > 1) = 1 −   × 0.05 × 0.95 −   × 0.05 × 0.95 data approaches 50%. For a large number of 0 1  doctors, the data can be treated as continuous 20    20      and so we would say that the statement is true. > 1) = 1 −   × 0.050 × 0.9520 −   × 0.051 × 0.9519 1  0 iii This scenario could be true in many situations. P(X > 1) = 1 – 0.358 … – 0.377 … = 0.264 For example, suppose every doctor scored a different mark, it would require two or more ii Income = 450 × 10 = $4500 doctors just above 50% of the data to score Cost = 24 × 20 = $480 the same mark and the statement is true. If X > 1, there is a ‘profit’ of –$480 The mode is not an appropriate measure of If X  1, there is a profit £4020 central tendency for this type of statement. 3 Typically, this would be a binomial distribution So the expected profit with X ~ B(275, 0.9). However, the large numbers = (0.736 × 4020) + (0.264 × – 480) = $2832 make it impractical to use the binomial formula. We know that: Extension Questions np = 275 × 0.9 = 247.5  and  npq = 275 × 0.9 × 0.1 1 Consider the arrangement GGBBB. If the two girls = 24.75 stay in the same position, there are 3! arrangements Since np  5 and nq  5, we can approximate the of the boys. We also know that the following binomial distribution by a normal distribution, arrangements of the girls will mean they are X ~ N(247.5, 24.75). together: BGGBB, BBGGB, BBBGG, GBBBG. The last  250 − 247.5  P(X > 250) = P  Z >  arrangement works because they are standing in a  24.75  circle and the two ends meet. In each case, there will = P(Z > 0.502 518 …) be 3! arrangements for the boys. = 1 − P(Z < 0.502 518 …) Therefore, the number of arrangements for the two = 1 − 0.6924 girls standing together is (3! × 5). iii When n = 2

0.2016 = (1 – p)p

p2

The total number of arrangements for the 5 children is 5! Therefore, the probability of the girls standing 3! × 5 1 next to each other is = . 5! 4 2 i The spread of the data is the key factor if the average referred to is the mean. This statement is true if the distribution is symmetrical. It is also true if the average distance from the mean for top 50% of the data is equal in magnitude to the average distance from the mean for the bottom 50% of the data.





= 0.3076

4 i The vertical scale begins at 51 000, which gives the impression that more than twice as many people voted for yes than no. To improve, raw data could be used or the vertical axis could start at zero. ii The diagram suggests that there is a relationship between donations to charity and mobile phone sales. The scales in each case are different, so there may not be a correlation. Also, correlation does not imply causation. To improve, the data is probably best shown on two separate diagrams with a scale.

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2

WORKED Solutions

3C × 17C = 136 possibilities for choosing 3 of  3 2 the top 3 (2040 + 136) = 0.140 (3 s.f.) 15504

iii The average referred to here is the median, which is not appropriate since it does not take into consideration sugar values that are significantly greater than the median. To improve, base the calculation on the mean for all sweets, also considering the probability of each sweet being chosen. 1 5 P(1st person is selected) = n P(2nd person is selected) = n − 1 × 1 = 1 n n −1 n P(3rd person is selected)     = n − 1 × n − 2 × 1 = 1 n n −1 n − 2 n

10 i p + 2pq + 3pq + 4q = 2.5 ⇒ p + 5pq + 4q = 2.5 (1) p + pq + pq + q = 1 p + 2pq + q = 1 p + 2pq = 1 – q p(1 + 2q) = 1– q 1−q 1 + 2q (2) Substitute (2) in (1): 1−q 1−q + 5q + 4q = 2.5 1 + 2q 1 + 2q



p=



 his pattern continues for all n people regardless T of height. Since every person has an equal chance of being selected, we can say that the probability of selection is independent of height.



6 For the first digit, there are 8 possible options. The question tells us that 5 is not allowed and we also know that to be a 5-digit number the first digit cannot be 0.

1 – q + 5q(1 – q) + 4q(1 + 2q) = 2.5(1 + 2q) 1 – q + 5q – 5q2 + 4q + 8q2 = 2.5 + 5q 3q2 + 3q – 1.5 = 0 2q2 + 2q – 1 = 0 q=



−2 ± 4 − 4(2)(−1) 4

q = −2 ± 12 = −2 ± 2 3 = −1 ± 3 4 4 2 0q1 ⇒ q = 3 −1 2 Therefore, the number of combinations is   8 × 9 × 9 × 9 × 9 = 52 488. 1 −  32− 1  2− 3 −1   3 3 = 3 3 − 3 = 3 −1 ii p = = = − 6 2 7 a + b = 2 ⇒ a + b = 4 ⇒ b = 4 – a   2 3 3 1 − 2 2 3 1 + − 2 1 + 2 2   ( a − 2)2 + (b − 2)2 = ( a − 2)2 +1(−2− a3)2− =1 a 2 − 4a + 4 + 4 − 4a + a 2 = 2a 2 − 8a + 8 σ2 = 2 − 32− 1 2 2  2  2 = p= = 3− 3 = 3 3 − 3 = 3 −1 2 2 2 2 2 6 2 a +  2 + 2 3 −      2 3 3 − 1 1 a − 2) + ( 2 − a ) ( a − 4 a + 4 + 4 − 4 a + a 2 a − 8 8 = = = 1 + 2 2  2 2 2          pq =  3 − 1   3 − 1  = 3 − 2 3 + 1 = 4 − 2 3 = 2 − 3 iii 2 2 2 2 2 4 2 4     σ = a – 4a + 4 = (a – 2)

For the remaining 4-digits, the only disallowed number is 5, so there are 9 choices in each case.

( (

( b − 2 )2

( (

) )

) )

         pq =  3 − 1   3 − 1  = 3 − 2 3 + 1 = 4 − 2 3 = 2 − 3 2 2 4 4 2    2− 3 2− 3 E X 2 = 3 −1 + 4 +9 + 16 2 2 2

σ = |a – 2|

8 i 1 × 6 × 5 × 4 × 3 = 360 7 7 7 7 2401 3 −1 ii 1 − 360 = 2041 2 2401 2401 9 2 or more of the top 3 students ⇒ 2 students or 3 2 − 3       2− 3 3 −1 E X 2 = 3 −1 + 4 +9 + 16 students in the group of 5 2 2 2 2 For 2 students: E X 2 = 3 − 1 + 8 − 4 3 + 18 − 9 3 + 16 3 − 16 = 9 + 4 3       244800 2 2 3 × 2 × 17 × 16 × 15 × 5C 2 = 20 × 19 × 18 × 17 × 16 1860480 − + − + − 3 1 8 4 3 18 9 3 + 16 3 − 16 = 9 + 4 3 2 E X = 2 2 For 3 students:  9+4 3 16320 3 × 2 × 1 × 17 × 16 5 – 2.52 = 1.71 (3 s.f.) Var ( X ) =  × C3 = 2  20 × 19 × 18 × 17 × 16 1860 480 

( )

( )

So the required probability is: 24480 26112 1632 + = = 0.140 (3 s.f.) 1860 480 1860 480 1860 480

Alternatively: 20C5 possibilities = 15 504 3C × 17C = 2040 possibilities for choosing 2 of  2 3 the top 3

(

( ) ) ( )

(

(

) (

)

(

)

)

( )

11 i Let n be the number of red balls. Then (n + 2) is the number of green balls and (2n + 2) is the total number of balls. P( different ) =  n × n + 2  +  n + 2 × n   2n + 2 2n + 2   2n + 2 2n n + 2 

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Summary REVIEW



∴ 2n(n + 2) = 195 (2n + 2)2 392 784n(n + 2) = 195(2n + 2)2



784n2 + 1568n = 195(4n2 + 8n + 4)



784n2 + 1568n = 780n2 + 1560n + 780



4n2 + 8n – 780 = 0

n2 + 2n – 195 = 0

(n + 15)(n – 13) = 0

n = –15

or

n = 13

n > 0 ⇒ n = 13 Therefore, there are 13 red balls and 15 green balls.

494 325 ii P(RGRGR) = 13 × 15 × 13 × 15 × 13 = 28 28 28 28 28 17 21 10 368 570 375 15 13 15 13 15 × × × × = and P(GRGRG ) = 28 28 28 28 28 17 21 10 368 570 375 15 13 15 13 15 P(GRGRG ) = × × × × = 10 368 28 28 28 28 28 17 21 494 325 570 375 1 064 700 38025 P ( alternating ) = + = = 17 210 368 17 210 368 17 210 368 614656 494 325 570 375 1 064 700 38025 rnating ) = + = = 17 210 368 614656 17 210 368 17 210 368                12 P( X = x ) =

n− x n! p x (1 − p ) n x x ! ! − ( )

P ( X = x − 1) = P( X = x ) = P ( X = x − 1)

(

n −(x −1) n! p x −1 (1 − p ) n − ( x − 1) !(x − 1)!

)

n− x n! px 1 − p (n − x )! x ! ( ) n −(x −1) n! p x −1 (1 − p ) (n − ( x − 1))!(x − 1)!

n ! ( n − x + 1)! ( x − 1)! p x (1 − p ) P( X = x ) = n − x +1 P ( X = x − 1) n ! ( n − x )! x ! p x −1 (1 − p )

n− x

P( X = x ) p(n − x + 1) = x(1 − p) P ( X = x − 1)

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