Probability and Statistics 2 sample

Cambridge International AS & A Level Mathematics:

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Probability & Statistics 2

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Worked Solutions Manual

Elevate Original material Š Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

Dean Chalmers

Cambridge International AS & A Level Mathematics:

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Probability & Statistics 2

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A

Worked Solutions Manual

Original material Š Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

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FT

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This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published 2019

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D

All worked solutions within this resource have been written by the author.

In examinations, the way marks are awarded may be different.

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Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

Contents

Contents How to use this book

v

1 Hypothesis testing

1 1

Exercise 1B

4

Exercise 1C

6

End-of-chapter review exercise 1

9

2 The Poisson distribution

14

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Exercise 1A

Exercise 2A

14

Exercise 2B

15

Exercise 2C

17 18

Exercise 2D Exercise 2E

21

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End-of-chapter review exercise 2

3 Linear combinations of random variables Exercise 3A Exercise 3B

29 29 30 32

Exercise 3D

34

End-of-chapter review exercise 3

36

Cross-topic review exercise 1

40

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Exercise 3C

23

4 Continuous random variables

46

5 Sampling

70

Exercise 4A

46

Exercise 4B

50

Exercise 4C

54

End-of-chapter review exercise 4

60

Exercise 5A

70

Exercise 5B

71

End-of-chapter review exercise 5

75

iii

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

78

Exercise 6A

78

Exercise 6B

79

Exercise 6C

82

Exercise 6D

83

End-of-chapter review exercise 6

85

Cross-topic review exercise 2

90

Practice exam-style paper

98

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6 Estimation

iv

Original material Š Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

How to use this book

How to use this book This book contains worked solutions to the questions in the Cambridge International AS & A Level Mathematics: Probability & Statistics 2 Coursebook. This includes all questions in the chapter exercises, end-of-chapter review exercises, cross-topic review exercises and practice exam-style paper.

FT

Each solution shows you step-by-step how to solve the question. You will be aware that often questions can be solved by multiple different methods. In this book, we provide a single method for each solution. Do not be disheartened if the working in a solution does not match your own working; you may not be wrong but simply using a different method. It is good practice to challenge yourself to think about the methods you are using and whether there may be alternative methods.

Additional guidance is included in Commentary boxes throughout the book. These boxes often clarify common misconceptions or areas of difficulty.

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All worked solutions in this resource have been written by the author. In examinations, the way marks are awarded may be different.

v

Original material Š Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

FT A R

D Original material Š Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

Chapter 1 Hypothesis testing EXERCISE 1A 1 a Let X be the number of nature programmes, then X ∼ B(20, 0.25).

Null and alternative hypotheses are statements about population parameters, not sample values. It is acceptable to use the letter p instead of the words ‘population proportion’.

FT

H0: population proportion = 0.25 and H1: population proportion < 0.25.

b Test statistic is P(X < 2) = 0.7520 + 20 × 0.251 × 0.7519 + 20C 2 × 0.252 × 0.7518 = 0.0913 0.0913 < 10% or 0.10, so H0 is rejected.

Alternatively, H0: p = 0.25 and H1: p < 0.25.

A statement like ‘H0 is rejected’ is part of the test result, it is not a conclusion.

A

The test result leads Yolande to conclude that there is insufficient evidence to accept the television channel’s claim. c 0.0913 > 5% or 0.05, so H0 is accepted.

R

Yolande would now conclude that the proportion of nature programmes is likely to be 25%, as the television channel claims.

2

Let X be the number of professional footballers that cannot explain the offside rule, then X ∼ B(15, 0.4).

H0: population proportion = 0.4 and H1: population proportion < 0.4.

D

Test statistics is P(X < 2) = 0.615 + 15 × 0.41 × 0.614 + 15C 2 × 0.42 × 0.613 = 0.0271 0.0271 < 0.05, so H0 is rejected.

There is insufficient evidence to accept the claim that 40% of professional footballers cannot explain the offside rule.

3 a Let X be the number of sheep deficient in a particular mineral, then X ∼ B(80, 0.3).

H0: population proportion = 0.3 and H1: population proportion < 0.3.

Approximating distribution is N(80 × 0.3, 80 × 0.3 × 0.7) = N(24, 16.8) with a continuity correction at 19.5.

1

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

(

19.5 − 24 √16.8 = 1 − Φ(1.098) = 0.136

P(X < 19) ≈ P z <

) Provided np and nq are both greater than 5, the distribution B(n, p) can be approximated

by N(np, npq) . Remember to make continuity corrections when approximating a discrete distribution by a continuous distribution.

0.136 > 0.10, so H0 is accepted.

There is no evidence to suggest that the percentage of sheep deficient in a particular mineral has decreased.

(c + 0.5) − 24 < −1.282 √16.8 c < 18.245...

The critical value is 18.

FT

b With a continuity correction at c + 0.5, we need to find the largest integer c for which P(X < c + 0.5) < 0.10.

4 a For X ∼ B(9, 0.5), we need to find the largest integer n for which P(X > n) > 0.05.

A

P(X > 8) = 9C 8 × 0.51 × 0.58 + 0.59 = 0.0195 < 0.05

P(X > 7) = 9C 7 × 0.52 × 0.57 + 9C 8 × 0.51 × 0.58 + 0.59 = 0.0898 > 0.05

R

Critical region is X > 7.

H0: population proportion of tails = 0.5 and H1: population proportion of tails > 0.5.

0.0898 > 0.05, so H0 is accepted.

There is insufficient evidence to support the claim that the coin has bias towards tails.

D

b Let the number of heads obtained be Y , then Y ∼ B(9, 0.5).

H0: population proportion of heads = 0.5 and H1: population proportion of heads < 0.5.

Critical region is Y < 2.

P(Y < 2) = 0.59 + 9C 1 × 0.51 × 0.58 + 9C 2 × 0.52 × 0.57 = 0.0898 0.0898 > 0.05, so H0 is accepted.

There is insufficient evidence to support the claim that the coin has bias towards tails.

2

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Chapter 1: Hypothesis testing

c Let the number of tails obtained be T , then T ∼ B(180, 0.5).

H0: m = 90 and H1: m > 90.

Approximating distribution is N(180 × 0.5, 180 × 0.5 × 0.5) = N(90, 45) with a continuity correction at 101.5.

(

P(T > 102) ≈ P z > 101.5 − 90 √45 = 1 − Φ(1.714) = 0.0432

)

FT

0.0432 < 0.05, so H0 is rejected.

There is evidence to suggest that the number of tails obtained with 180 flips is significantly more than 90, which supports the claim that the coin has bias towards tails.

5

X ∼ B(120, 0.25).

The null and alternative hypotheses here use m as an acceptable alternative to the words ‘population mean’.

H0: population mean = 30 and H1: population mean > 30.

A

Approximating distribution is N(120 × 0.25, 120 × 0.25 × 0.75) = N(30, 22.5) with a continuity correction at 39.5.

(

)

R

P(X > 40) ≈ P z > 39.5 − 30 √22.5 = 1 − Φ(2.003) = 0.0227

0.0227 > 0.02, so H0 is accepted.

D

The claim is not justified, as there is insufficient evidence to suggest that the spinner has bias towards blue.

6

Let X be the number of times the darts player hits the bullseye, then X ∼ B(17, 0.6).

H0: p = 0.6 and H1: p > 0.6.

P(X > 15) = 17C 15 × 0.615 × 0.42 + 17C 16 × 0.616 × 0.41 + 0.617 = 0.0123 X = 15 is the critical value at the smallest integer significance level above 1.23%, which is 2%.

3

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

EXERCISE 1B 1 a Let X be the number of nature programmes, then X ∼ B(20, 0.25).

H0: population proportion = 0.25 and H1: population proportion ≠ 0.25. b The test statistic is P(X < 2) = 0.7520 + 20C 1 × 0.251 × 0.7519 + 20C 2 × 0.252 × 0.7518 = 0.0913

In a two-tailed test at S% significance, a critical value of 1 S% is used for each of the 2 two rejection regions.

0.0913 > 0.05, so H0 is accepted.

FT

The critical value is 0.05, as a two-tail test is required.

Xavier concludes that the proportion of nature programmes is likely to be 25%, as claimed.

2

Note that Xavier’s conclusion concerns the proportion of nature programmes, not the number of nature programmes.

Let X be the number of professional footballers that cannot explain the offside rule, then X ∼ B(15, 0.4).

A

H0: population proportion = 0.4 and H1: population proportion ≠ 0.4.

P(X < 2) = 0.615 + 15C 1 × 0.41 × 0.614 + 15C 2 × 0.42 × 0.613 = 0.0271 0.0271 > 0.025, so H0 is accepted.

R

There is sufficient evidence to accept the claim that 40% of professional footballers cannot explain the offside rule.

It would be a mistake to make a conclusion about professional footballers not understanding the offside rule. Always refer back to the wording used in a question before writing a conclusion.

3 a Let X be the number of Americans that believe in horoscopes, then X ∼ B(16, 0.8).

D

H0: population proportion = 0.8 and H1: population proportion ≠ 0.8.

P(X > 15) = 16C 15 × 0.815 × 0.21 + 0.816 = 0.141 0.141 > 0.05, so H0 is accepted.

There is sufficient evidence to accept the claim that 80% of Americans believe in horoscopes.

P(X > 16) = 16C 16 × 0.816 × 0.20 = 0.0281 < 0.05, so X = 16 is the critical value.

b Let X be the number of Americans that believe in horoscopes, then X ∼ B(160, 0.8).

H0: population proportion = 0.8 and H1: population proportion ≠ 0.8.

Approximating distribution is N(128, 25.6) with a continuity correction at 116.5.

4

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

Chapter 1: Hypothesis testing

H0: m = 128 and H1: m ≠ 128.

(

116.5 − 128 √25.6 = 1 − Φ(2.273) = 0.0115

P(X < 116) ≈ P z <

) An alternative way of writing this conclusion is ‘There is insufficient evidence to accept the claim that 80% of Americans believe in horoscopes’.

0.0115 < 0.025, so H0 is rejected.

There is sufficient evidence to reject the claim that 80% of Americans believe in horoscopes.

H0: p = 0.12 and H1: p ≠ 0.12

FT

4 a Let X be the number of people who can correctly spell ‘onomatopoeia’, then X ∼ B(180, 0.12).

Approximating distribution is N(21.6, 19.008) with a continuity correction at 13.5.

H0: population mean = 21.6 and H1: population mean ≠ 21.6.

The significance level for the two-tail test, S%, must be such that the rejection region contains X = 13 and the acceptance region contains X = 14.

(

13.5 − 21.6 √19.008 = 1 − Φ(1.858) = 0.0316

)

A

P(X < 13) ≈ P z <

(

14.5 − 21.6 √19.008 = 1 − Φ(1.629) = 0.0517

)

R

P(X < 14) ≈ P z <

D

3.16 < 1 S < 5.17, so 6.32 < S < 10.34.

2

The smallest integer significance level is 7%, but it could be any from 7 to 10%, inclusive.

Magnus concludes that there is insufficient evidence to support the claim that 12% of people can correctly spell onomatopoeia.

b The test statistic of 0.0316 is less than each of 7 , 8 , 9 and 10 %, so H0 will be rejected in all cases.

2 2 2 2 Magnus would conclude that there is evidence to suggest that the proportion of people who can spell onomatopoeia is less than 12%.

5

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

5 a Let X be the number of red marbles, then X ∼ B(20, 0.3).

H0: population proportion = 0.3 and H1: population proportion ≠ 0.3.

P(X < 3) = 0.720 + 20C 1 × 0.31 × 0.719 + 20C 2 × 0.32 × 0.718 + 20C 3 × 0.33 × 0.717 = 0.107 0.107 > 0.05, so H0 is accepted.

There is sufficient evidence to accept the manufacturer’s claim that 30% of the marbles are red.

FT

P(X < 2) = 0.720 + 20C 1 × 0.31 × 0.719 + 20C 2 × 0.32 × 0.718 = 0.0355 < 0.05.

H0 is rejected at X = 2, but accepted at X = 3. The critical value is X = 2. b H0: population proportion = 0.3 and H1: population proportion < 0.3.

0.107 > 0.10, so H0 is accepted.

Ginny’s conclusion will not change.

6

Let X be the number of mixed-handed people, then X ∼ B(600, 0.01).

A

H0: p = 0.01 and H1: p ≠ 0.01.

Approximating distribution is N(6, 5.94) with a continuity correction at 11.5.

H0: population mean = 6 and H1: population mean ≠ 6.

(

)

R

P(X > 11) ≈ P z > 11.5 − 6 √5.94 = 1 − Φ(2.257) = 0.0120

0.0120 < 0.02, so H0 is rejected.

D

There is sufficient evidence to suggest that the proportion of mixed-handed people worldwide is not 1% as claimed originally (and that Amie’s belief is correct).

EXERCISE 1C

1 a L et the random variable X be the number of arrows that hit the bullseye, then X has a binomial distribution with n = 12 and p = 0.4, i.e., X ∼ B(12, 0.4). b H0: population proportion = 0.4 and H1: population proportion < 0.4.

6

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

Chapter 1: Hypothesis testing

c Using p = 0.4, we need to find the smallest value of n for which P(X < n) < 0.05.

P(X < 2) = 0.612 + 12C 1 × 0.41 × 0.611 + 12C 2 × 0.42 × 0.610 = 0.0834 > 0.05 P(X < 1) = 0.612 + 12C 1 × 0.41 × 0.611 = 0.0196 < 0.05 The rejection region is X < 1. d If H0: p = 0.4 is true, then X = 1 lies in the rejection region, which means that the archer’s claim that p = 0.4 will be rejected, thus a Type I error occurs.

FT

P(Type I error) = P(reject H0 | H0 is true)

2 a Let X be the number of eggs that produce ducklings, then X ∼ B(16, 0.8). b H0: population proportion = 0.8 and H1: population proportion > 0.8.

c Using p = 0.8, we need to find the smallest n for which P(X > n) < 0.10.

A

P(X > 15) = 0.816 + 16C 15 × 0.815 × 0.21 = 0.1407 > 0.10 P(X > 16) = 0.816 = 0.0281 < 0.10 The rejection region is X > 16.

d P(Type I error) = P(X > 16) | p = 0.8) = 0.0281

Let X be the number of teenagers who can tell the difference between this chocolate and the leading brand, then X ∼ B(300, 0.2).

R

3

H0: p = 0.2 and H1: p < 0.2.

Approximating distribution is N(60, 48).

D

H0: population mean = 60 and H1: population mean < 60.

Let the critical value be a then, with a continuity correction at a − 0.5, we have

(

P z<

)

(a − 0.5) − 60 < 0.05 √48 a − 60.5 < −1.645 √48 a < 49.103...

A Type I error occurs if 49 or fewer of the 300 teenagers can tell the difference.

7

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

4 a H0: population mean height = 1.8 m and H1: population mean height > 1.8 m. b P(Type I error) = P(X > 2.25 | m = 1.8)

(

=P z>

2.25 − 1.8 0.32

= 1 − Φ(1.406) = 0.0800

)

c P(Type II error) = P(X < 2.25 | m = 2.4) 2.25 − 2.4 0.32

= 1 − Φ(0.469) = 0.320

)

P(Type II error) = P(accept H0 | H0 is false) or P(accept H0 | H1 is true), where H1 offers an alternative population parameter value to the one given in H0.

FT

(

=P z<

5 a A Type I error is made when Janina concludes that the die has bias towards the number 1 when in fact it does not.

Let X be the number of 1s rolled, then X ∼ B(20, 0.125).

H0: p = 0.125 and H1: p > 0.125.

A

P(Type I error) = P(X > 4 | p = 0.125) = 1 − P(X < 3 | p = 0.125) = 1 − [ 0.87520 + 20C 1 × 0.1251 × 0.87519 + 20C 2 × 0.1252 × 0.87518 + 20C 3 × 0.1253 × 0.87517] = 0.235

R

b A Type II error is made when Janina concludes that the die does not have bias towards the number 1, when in fact it does.

D

We would need to be told the true (or an alternative) probability for rolling a 1 with this die.

8

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

End-of-chapter review exercise 1

END-OF-CHAPTER REVIEW EXERCISE 1 1 a Let X be the number of bowls with a defect, then X ∼ B(25, 0.16).

H0: p = 0.16 and H1: p < 0.16.

P(X < 1) = 0.8425 + 25C 1 × 0.161 × 0.8424 = 0.0737 0.0737 > 0.05, so H0 is accepted.

FT

There is no evidence to suggest that the new clay is more reliable than the old clay. b Let the number of bowls with a defect be Y , then Y ∼ B(150, 0.16).

H0: p = 0.16 and H1: p < 0.16.

Approximating distribution is N(24, 20.16) with a continuity correction at 18.5.

H0: m = 24 and H1: m < 24.

(

18.5 − 24 √20.16 = 1 − Φ(1.225) = 0.1103

)

A

P(X < 18) ≈ P z <

0.1103 > 0.05, so H0 is accepted.

R

There is still no evidence to suggest that the new clay is more reliable than the old clay.

The discrete value X = 18 is represented by the range of continuous values 17.5 < X < 18.5, so a continuity correction must be used to find the approximate value of P(X < 18).

2

Let X be the number of days on which the land is flooded, then X ∼ B(1000, 0.2).

Approximating distribution is N(200, 160) with a continuity correction at 217.5.

D

H0: m = 200 days and H1: m > 200 days.

(

217.5 − 200 √160 = 1 − Φ(1.383) = 0.0833

P(X > 218) ≈ P z >

)

We have accepted H0 because P(X > 218) > significance level, i.e., 0.0833 > 0.05.

We can, therefore, accept H0 because P(X < 218) < 1 − significance level, i.e., 0.9167 < 0.95.

0.0833 > 0.05 (or 0.9167 < 0.95), so H0 is accepted.

There is evidence to suggest that the meteorologist’s suspicion that the land will be flooded more often is not justified.

9

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

3 a Let X be the life time of a light bulb, then X ∼ N(800, 2500).

H0: m = 800 hours and H1: m ≠ 800 hours.

(

720 − 800 50 = 1 − Φ(1.6) = 0.0548

P(X < 720) ≈ P z <

)

FT

0.0548 > 0.025, so H0 is accepted.

Although this light bulb lasts a much shorter time than expected, it actually provides evidence in support of the manufacturer’s claim, but does not prove that the claim is true.

(

920 − 800 50 = 1 − Φ(2.4) = 0.00820

P(X > 920) ≈ P z <

)

A

0.00802 < 0.025, so H0 is rejected.

This light bulb lasts much longer than expected, and provides evidence to reject the manufacturer’s claim, but does not prove that the claim is untrue. b Let the necessary number of hours be t.

(

)

R

P z < t − 800 < 0.025 50

t − 800 < −1.960 50 t < 702

D

John's light bulb would last for up to 702 hours.

4

Let X be the number of wrongly recorded digits, then X ∼ B(40, 0.08).

H0: p = 0.08 and H1: p < 0.08.

We need to find the largest n for which P(X < n) < 0.10.

P(X < 0) = 0.9240 = 0.0356 < 0.10 P(X < 1) = 0.9240 + 40 × 0.081 × 0.9239 = 0.1594 > 0.10 The critical region is X < 0, and the critical value of X is 0.

10

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

End-of-chapter review exercise 1

5 a Let X be the number of people with red hair, then X ∼ B(850, 0.012).

Approximating distribution is N(10.2, 10.0776) with a continuity correction at 14.5.

H0: m = 10.2 and H1: m > 10.2.

(

)

FT

14.5 − 10.2 √10.0776 = 1 − Φ(1.355) = 0.0877

P(X > 15) ≈ P z >

0.0877 > 0.02, so H0 is accepted.

There is insufficient evidence to support the claim that the proportion of red-haired people in Scotland is greater than 1.2%. b Let the critical value be x with a continuity correction at x − 0.5.

(

6

i

)

(x − 0.5) − 10.2 < 0.02 √10.0776 x − 10.7 > 2.054 or 2.055 √10.0776 x > 17.22..., so the critical value is 18.

A

P z>

H0: P(free gift) = 0.3 and H1: P(free gift) < 0.3.

R

ii P(X < 2) = 0.720 + 20C 1 × 0.719 × 0.31 + 20C 2 × 0.718 × 0.32 = 0.03548... < 5% P(X < 3) = 0.03548... + 20C 3 × 0.717 × 0.33 = 0.1070... > 5%

D

X = 2 is in the rejection region, but X = 3 is in the acceptance region.

P(Type I error) = P[ X < 2 | P(free gift) = 0.3] = 0.0355.

iii P(X < 3) = 0.107 and 0.107 > 0.05, so H0 is accepted.

There is insufficient evidence to suggest that the proportion of packets containing free gifts is less than 30%.

7

i

H0: P(correct) = 1 and H1: P(correct) > 1 .

8 8

ii A Type I error occurs if Marie is correct in 3 or more races using p = 1 .

8

11

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Cambridge International AS & A Level Mathematics: Worked Solutions Manual

( (( )

) ( )( ))

P(Type I error) = 1 − P correct in 0, 1 or 2 races | p = 1 8 =1−

7 8

10

( )( )

+ 10C 1 1 8

1

7 8

9

+ 10C 2 1 8

= 0.120

2

7 8

8

iii 0.120 = 12% i

H0: P(five) = 0.2 and H1: P(five) < 0.2.

FT

8

Let X be the number of fives, then X ∼ B(25, 0.2).

P(X < 1) = 0.825 + 25C 1 × 0.824 × 0.21 = 0.0274 0.0274 > 0.025, so H0 is accepted.

There is insufficient evidence to support Max’s claim.

A

ii The approximating distribution for B(1000, 0.2) is a normal distribution with m = np = 200 and s 2 = np(1 − p) = 160, i.e., N(200, 160).

iii Concluding that one-fifth of the digits produced are fives when in fact the proportion produced is significantly smaller than that. i

P(X < 1) = 0.7520 + 20C 1 × 0.7519 × 0.251 = 0.02431... < 0.05

R

9

P(X < 2) = 0.02431... + 20C 2 × 0.7518 × 0.252 = 0.09126... > 0.05

D

The critical region is where fewer than 2 sample packets contain a free gift.

ii P(Type I error) = P(X < 1 | p = 0.25) = 0.0243. iii X = 2 is in the acceptance region, so she concludes that there is evidence that the manufacturer’s claim is true.

10 The number of correct answers follows B(100, 0.2).

H0: P(correct) = 0.2 and H1: P(correct) > 0.2.

Approximating distribution is N(20, 16) with a continuity correction at 26.5.

H0: population mean score = 20 and H1: population mean score > 20.

12

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.

End-of-chapter review exercise 1

(

26.5 − 20 √16 = 1 − Φ(1.625) = 0.0521

P(X > 27) ≈ P z >

)

The decision to accept or reject H0 can be made by comparing the test statistic (1.625) with the critical value for an upper one-tail test at 5% significance, which is Φ−1 (0.95) = 1.645.

That is, we know that P(X > 27) > 0.05 because 1.625 < 1.645, so H0 is accepted.

0.0521 > 0.05, so H0 is accepted.

11 i

FT

Ashok’s claim is not justified; his general knowledge does not enable him to get significantly more correct answers than by just guessing.

H0: P(six) = 1 and H1: P(six) > 1 .

6 6

( ) ) ( ( )( ) ( )( )

ii Let X be the number of sixes, then X ∼ B 10, 1 .

6

(

P(Type I error) = P X > 4 | P(six) = 1 = 1 − P X < 4 | P(six) = 1 6 6

()

+ 10C1 1 6

1

5 6

9

+ 10C2 1 6

2

5 6

8

)

( )( )

+ 10C3 1 6

A

⎡ =1−⎢ 5 ⎣ 6 = 0.0697

10

3

7 5 ⎤ 6 ⎦⎢

iii The die is in fact biased in favour of six, but fewer than 4 sixes are thrown, so there is no evidence that the die is biased in favour of six.

(

iv P(Type II error) = P X < 4 | P(six) = 1 2 10

( )( ) 1

R

()

= 1 2

+ 10C 1 1 2

1 2

9

)

( )( )

+ 10C 2 1 2

2

1 2

8

( )( )

+ 10C 3 1 2

3

1 2

7

= 11 or 0.172 64

D

12 H0: m = 18.5 mm and H1: m < 18.5 mm

(

)

Let the mean diameter of 20 women’s rings be X , then X ∼ N 18.5, 1.1 . √20 ⎛ ⎞ P(X < 18.1) = P z < 18.1 − 18.5 1.1 ⎝ ⎠ √ 20 = 1 − Φ(1.626) = 0.0520

⎜

⎜

The data concern ring diameters, but the claim concerns finger sizes, so certain assumptions about ring diameter and finger size are being made in reaching this conclusion.

0.0520 > 0.025, so H0 is accepted.

The researchers’ claim that women in Jakarta have smaller fingers is not justified because there is evidence to suggest that the two groups of women wear rings of the same diameter.

13

Original material © Cambridge University Press 2019. This material is not final and is subject to further changes prior to publication.