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 74E

(a)

To determine

Conditional probability for the 5th grade student thinksathletic ability as most important.

(a)

Expert Solution
Check Mark

Answer to Problem 74E

Conditional probability,

  P(Athletic|5thgrade)0.3333

Explanation of Solution

Given information:

Survey data summarized in two − way table:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.3, Problem 74E , additional homework tip  1

Calculations:

According to conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

Note that

The information about 335 students is provided in the table.

Thus,

The number of possible outcomes is 335.

Also note that

In the table, 108 of the 335 students are 5th graders.

Thus,

The number of favorable outcomes is 108.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  P(5thgrade)=NumberoffavourableoutcomesNumberofpossibleoutcomes=108335

Now,

Note that

In the table, 36 of the 335 students are 5th grades and think that athletics are most important. In this case, the number of favorable outcomes is 36 and number of possible outcomes is 335.

  P(Athleticand5thgrade)=NumberoffavourableoutcomesNumberofpossibleoutcomes=36335

Apply conditional probability:

  P(Athletic|5thgrade)=P(Atheticand5thgrade)P(5thgrade)=36335108335=361080.3333

Thus,

The conditional probability for the randomly selected 5th grade student thinksathletic ability as most important is approx. 0.3333.

(b)

To determine

Whether the events “5th grade” and “athletic” independent.

(b)

Expert Solution
Check Mark

Answer to Problem 74E

No, the events “5th grade” and “athletic” are not independent.

Explanation of Solution

Given information:

Survey data summarized in two − way table:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.3, Problem 74E , additional homework tip  2

Calculations:

The two events are independent, if the probability of occurrence of one event does not affect the probability of occurrence of other event.

According to conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

Note that

The information about 335 students is provided in the table.

Thus,

The number of possible outcomes is 335.

Also note that

In the table, 108 of the 335 students are 5th graders.

Thus,

The number of favorable outcomes is 108.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  P(5thgrade)=NumberoffavourableoutcomesNumberofpossibleoutcomes=108335

Also note that

In the table, 98 of the 335 students think athletics as most important.

Thus,

The number of favorable outcomes is 98 and the number of possible outcomes is 335.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  P(Athletic)=NumberoffavourableoutcomesNumberofpossibleoutcomes=983350.2925

Now,

Note that

In the table, 36 of the 335 students are 5th grades and think that athletics are most important. In this case, the number of favorable outcomes is 36 and number of possible outcomes is 335.

  P(Athleticand5thgrade)=NumberoffavourableoutcomesNumberofpossibleoutcomes=36335

Apply conditional probability:

  P(Athletic|5thgrade)=P(Atheticand5thgrade)P(5thgrade)=36335108335=361080.3333

For events “Athletic” and “5th grade” to be independent,

  P(Athletic|5thgrade)=P(Athletic)

And

  P(5thgrade|Athletic)=P(5thgrade)

But

We have

  P(Athletic|5thgrade)0.3333

And

  P(Athletic)0.2925

Both probabilities are not identical.

Thus,

Both events are not independent.

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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