EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
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Question
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Chapter 5.3, Problem 82E

(a)

To determine

Represent this chance process with Tree diagram.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Study about media influence in young people lives aged 8 − 18:

17% were classified as light media users

62% were classified as moderate media users

21% were classified as heavy media users

Respondents with good grades:

74% light users described their grades as good

68% moderate users described their grades as good

52% heavy users described their grades as good

First level:

First level consists of three types of media users:

Light, moderate and heavy

Thus,

The first level requires three children:

Light, moderate and heavy

Second level:

In this level, there are two types of grades:

Good grades or No Good grades

Thus,

The second level consists of two children per child at the first level i.e.; Good grades or No Good grades.

The required tree diagram can be drawn as:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.3, Problem 82E

(b)

To determine

Probability for the person describes his or her grades as good.

(b)

Expert Solution
Check Mark

Answer to Problem 82E

Probability for the person with good grades is 0.6566.

Explanation of Solution

Given information:

Study about media influence in young people lives aged 8 − 18:

17% were classified as light media users

62% were classified as moderate media users

21% were classified as heavy media users

Respondents with good grades:

74% light users described their grades as good

68% moderate users described their grades as good

52% heavy users described their grades as good

Calculations:

According to the general multiplication rule,

  P(AandB)=P(AB)=P(A)×P(B|A)=P(B)×P(A|B)

According to the addition rule for mutually exclusive events:

  P(AB)=P(AorB)=P(A)+P(B)

Let

L: Light user

M: Moderate user

H: Heavy user

G: Good grades

Gc: No Good grades

Now,

The corresponding probabilities:

Probability for light user,

  P(L)=0.17

Probability for moderate user,

  P(M)=0.62

Probability for heavy user,

  P(H)=0.21

Probability for light user with good grades,

  P(G|L)=0.74

Probability for moderate user with good grades,

  P(G|M)=0.68

Probability for heavy user with good grades,

  P(G|H)=0.52

Apply general multiplication rule:

Probability for user with good grades and light user,

  P(GandL)=P(L)×P(G|L)=0.17×0.74=0.1258

Probability for user with good grades and moderate user,

  P(GandM)=P(M)×P(G|M)=0.62×0.68=0.4216

Probability for user with good grades and heavy user,

  P(GandH)=P(H)×P(G|H)=0.21×0.52=0.1092

Since each person cannot become more than one type of user.

Apply the addition rule for mutually exclusive events:

  P(G)=P(GandL)+P(GandM)+P(GandH)=0.1258+0.4216+0.1092=0.6566

Thus,

Probability for the person describes his or her grades as good is 0.6566.

(c)

To determine

Conditional probability for the randomly chosen person describes his or her grade as good is a heavy user of media.

(c)

Expert Solution
Check Mark

Answer to Problem 82E

Probability that the randomly chosen person describes his or her grade as good is a heavy user of media is approx. 0.1663.

Explanation of Solution

Given information:

Study about media influence in young people lives aged 8 − 18:

17% were classified as light media users

62% were classified as moderate media users

21% were classified as heavy media users

Respondents with good grades:

74% light users described their grades as good

68% moderate users described their grades as good

52% heavy users described their grades as good

Calculations:

According to the conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

From Part (b),

We have

Probability for the person describes his or her grades as good,

  P(G)=0.6566

Probability for user with good grades and heavy user,

  P(GandH)=0.1092

Apply the conditional probability:

  P(H|G)=P(GandH)P(G)=0.10920.6566=784690.1663

Thus,

The probability for the person describes his or her grades as good is a heavy user of media is approx. 0.1663.

Chapter 5 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.2 - Prob. 43ECh. 5.2 - Prob. 44ECh. 5.2 - Prob. 45ECh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5.3 - Prob. 84ECh. 5.3 - Prob. 85ECh. 5.3 - Prob. 86ECh. 5.3 - Prob. 87ECh. 5.3 - Prob. 88ECh. 5.3 - Prob. 89ECh. 5.3 - Prob. 90ECh. 5.3 - Prob. 91ECh. 5.3 - Prob. 92ECh. 5.3 - Prob. 93ECh. 5.3 - Prob. 94ECh. 5.3 - Prob. 95ECh. 5.3 - Prob. 96ECh. 5.3 - Prob. 97ECh. 5.3 - Prob. 98ECh. 5.3 - Prob. 99ECh. 5.3 - Prob. 100ECh. 5.3 - Prob. 101ECh. 5.3 - Prob. 102ECh. 5.3 - Prob. 103ECh. 5.3 - Prob. 104ECh. 5.3 - Prob. 105ECh. 5.3 - Prob. 106ECh. 5.3 - Prob. 107ECh. 5.3 - Prob. 108ECh. 5 - Prob. R5.1RECh. 5 - Prob. R5.2RECh. 5 - Prob. R5.3RECh. 5 - Prob. R5.4RECh. 5 - Prob. R5.5RECh. 5 - Prob. R5.6RECh. 5 - Prob. R5.7RECh. 5 - Prob. R5.8RECh. 5 - Prob. T5.1SPTCh. 5 - Prob. T5.2SPTCh. 5 - Prob. T5.3SPTCh. 5 - Prob. T5.4SPTCh. 5 - Prob. T5.5SPTCh. 5 - Prob. T5.6SPTCh. 5 - Prob. T5.7SPTCh. 5 - Prob. T5.8SPTCh. 5 - Prob. T5.9SPTCh. 5 - Prob. T5.10SPTCh. 5 - Prob. T5.11SPTCh. 5 - Prob. T5.12SPTCh. 5 - Prob. T5.13SPTCh. 5 - Prob. T5.14SPT
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