PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Question
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Chapter 5.3, Problem 64E

(a)

To determine

Probability for the chosen egg assigned to hot water hatched.

(a)

Expert Solution
Check Mark

Answer to Problem 64E

Probability that the randomly selected egg assigned to hot water hatched is approx. 0.7212.

Explanation of Solution

Given information:

Data on randomly assigned newly laid eggs to one of three water temperatures summarized in the two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 64E , additional homework tip  1

According to conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

Calculate the total values in the given two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 64E , additional homework tip  2

Note that

The information about 187 eggs is provided in the table.

Thus,

The number of possible outcomes is 187.

Also note that

In the table, 104 of the 187 eggs are assigned to hot water.

Thus,

The number of favorable outcomes is 104.

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

  P(Hot)=NumberoffavourableoutcomesNumberofpossibleoutcomes=104187

Now,

Note that

In the table, 75 of the 187 eggs are assigned to hot water and hatched.

In this case, the number of favorable outcomes is 75 and number of possible outcomes is 187.

  P(HatchedandHot)=NumberoffavourableoutcomesNumberofpossibleoutcomes=75187

Apply conditional probability:

  P(Hatched|Hot)=P(HatchedandHot)P(Hot)=75187104187=751040.7212=72.12%

Thus,

About 72.12% of the eggs assigned to hot water hatched and the probability is approx. 0.7212.

(b)

To determine

Probability for the chosen hatched egg was not assigned to hot water.

(b)

Expert Solution
Check Mark

Answer to Problem 64E

Probability that the randomly selected hatched egg was not assigned to hot water is approx. 0.4186.

Explanation of Solution

Given information:

Data on randomly assigned newly laid eggs to one of three water temperatures summarized in the two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 64E , additional homework tip  3

According to conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

Calculate the total values in the given two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 64E , additional homework tip  4

Note that

The information about 187 eggs is provided in the table.

Thus,

The number of possible outcomes is 187.

Also note that

In the table, 129 of the 187 eggs hatched.

Thus,

The number of favorable outcomes is 129.

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

  P(Hatched)=NumberoffavourableoutcomesNumberofpossibleoutcomes=129187

Now,

Note that

In the table, 16 hatched eggs are assigned to cold water and 38 hatched eggs are assigned to neutral water.

This implies

54 hatched eggs are not assigned to hot water.

In this case, the number of favorable outcomes is 54 and number of possible outcomes is 187.

  P(HatchedandnotHot)=NumberoffavourableoutcomesNumberofpossibleoutcomes=54187

Apply conditional probability:

  P(Hatched|notHot)=P(HatchedandnotHot)P(Hatched)=54187129187=54129=18430.4186=41.86%

Thus,

About 41.86% of thehatched eggs were not assigned to hot water and the probability is approx. 0.4186.

Chapter 5 Solutions

PRACTICE OF STATISTICS F/AP EXAM

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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