EBK INTRODUCTION TO THE PRACTICE OF STA
EBK INTRODUCTION TO THE PRACTICE OF STA
9th Edition
ISBN: 8220103674638
Author: Moore
Publisher: YUZU
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Chapter 4.5, Problem 109E

(a)

To determine

To find: The missing probabilities in the table.

(a)

Expert Solution
Check Mark

Answer to Problem 109E

Solution: The complete table is obtained as follows.

Gender

Total

Men

Women

Institution

4-year institution

0.2684

0.3416

0.61

2-year institution

0.1599

0.2301

0.39

Total

0.4283

0.5717

1

Explanation of Solution

Calculation: Let M and F denote the gender of the students as “Male” and “Female.” Let 4Y denote that the student attends a 4-year institution and 2Y denote that the student attends a 2-year institution. The conditional probability, which is defined as the occurrence of one event provided the other event has already occurred, is used in this exercise. The probability that a male student attends a 4-year institution is denoted as P(M|4Y) and the probability that a male student attends a 2-year institution is denoted as P(M|2Y).

It is provided in the question that the 4-year institutions constitute 44% males; 61% of students attend the 4-year institution and the rest 2-year institution.

The probability that a student attends a 4-year institution is as follows:

P(4-year institution)=P(4Y)=0.61

The probability that a male student attends a 4-year institution is as follows:

P(Males|4-year instituion)=P(M|4Y)=44%=0.44

The probability that a male student attends a 2-year institution is as follows:

P(Males|2year instituion)=P(M|2Y)=41%=0.41

The provided probabilities are represented in the following table:

Gender

Total

Men

Women

Institution

4-year institution

0.61

2-year institution

Total

1

The remaining or missing probabilities can be ascertained as follows.

The probability that a student attends a 4-year institution is calculated as

P(2-year institution)=P(2Y)=1P(4-year institution)=1P(4Y)=10.61

=0.39

The concept of conditional probability is the probability of an event (A) when the other event has already occurred (B), and it is calculated as

P(A|B)=P(Aand B)P(B)=P(AB)P(B)

The concept of conditional probability is used to ascertain the following probabilities.

The probability that a student is a male and also that he attends a 4-year institution is calculated as follows:

P(Males|4-year institution)=P(Males who attend 4-year institution)P(4-year institution)P(M|4Y)=P(M4Y)P(4Y)0.44=P(M4Y)0.61P(M4Y)=0.44×0.61

=0.2684

The probability that a student is a male and also that he attends a 2-year institution is calculated as follows:

P(Males|2-year institution)=P(Males who attend 2-year institution)P(2-year institution)P(M|2Y)=P(M2Y)P(2Y)0.41=P(M2Y)0.39P(M2Y)=0.41×0.39

=0.1599

The probability that the student is a female and also that she attends a 4-year institution is calculated as follows:

P(Female | 4-year institution)=(P(4-year institution)P(Males who attend 4-year institution))P(F4Y)=P(4Y)P(M4Y)=0.610.2684=0.3416

The probability that the student is a female and also that she attends a 2-year institution is calculated as follows:

P(Female|2-year institution)=(P(2-year institution)P(Males who attend 2-year institution))P(F2Y)=P(2Y)P(M2Y)=0.390.1599=0.2301

Therefore, the probability that a student is a male and belongs to the 4-year institution {P(M4Y)} is 0.2684.

The probability that a student is a male and belongs to the 2-year institution {P(M2Y)} is 0.1599.

The probability that a student is a female and belongs to the 4-year institution {P(F4Y)} is 0.3416.

The probability that a student is a female and belongs to the 2-year institution {P(F2Y)} is 0.2301.

The probabilities P(4Y) and P(2Y) sum to 1. Also, the P(M) and P(W) sum to 1.

Therefore, the complete probability table is shown as below:

Gender

Total

Men

Women

Institution

4-year institution

0.2684

0.3416

0.61

2-year institution

0.1599

0.2301

0.39

Total

0.4283

0.5717

1

(b)

To determine

To find: The probability that a randomly selected female student attends the 4-year institution.

(b)

Expert Solution
Check Mark

Answer to Problem 109E

Solution: The required probability is 0.5975.

Explanation of Solution

Calculation: The concept of conditional probability is the probability of an event (A) when the other event has already occurred (B), and it is calculated as

P(A|B)=P(Aand B)P(B)=P(AB)P(B)

The conditional probability rule is used to ascertain the required probability. It is calculated as follows.

The probability that the randomly selected student attends a 4-year institution provided the fact that the student is a female is calculated as follows:

P(4-year institution|female)=P(Female who attend 4-year institution)P(Female)=P(F4Y)P(F)=0.34160.5717=0.5975

Hence, the required probability is 0.5975.

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Chapter 4 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

Ch. 4.2 - Prob. 11UYKCh. 4.2 - Prob. 12UYKCh. 4.2 - Prob. 13UYKCh. 4.2 - Prob. 14UYKCh. 4.2 - Prob. 15UYKCh. 4.2 - Prob. 16UYKCh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.2 - Prob. 32ECh. 4.2 - Prob. 33ECh. 4.2 - Prob. 34ECh. 4.2 - Prob. 35ECh. 4.2 - Prob. 36ECh. 4.2 - Prob. 37ECh. 4.2 - Prob. 38ECh. 4.2 - Prob. 39ECh. 4.2 - Prob. 40ECh. 4.2 - Prob. 41ECh. 4.3 - Prob. 42UYKCh. 4.3 - Prob. 43UYKCh. 4.3 - Prob. 44UYKCh. 4.3 - Prob. 45ECh. 4.3 - Prob. 46ECh. 4.3 - Prob. 47ECh. 4.3 - Prob. 48ECh. 4.3 - Prob. 49ECh. 4.3 - Prob. 50ECh. 4.3 - Prob. 51ECh. 4.3 - Prob. 52ECh. 4.3 - Prob. 53ECh. 4.3 - Prob. 54ECh. 4.3 - Prob. 55ECh. 4.3 - Prob. 56ECh. 4.3 - Prob. 57ECh. 4.3 - Prob. 58ECh. 4.3 - Prob. 59ECh. 4.3 - Prob. 60ECh. 4.3 - Prob. 61ECh. 4.3 - Prob. 62ECh. 4.4 - Prob. 63UYKCh. 4.4 - Prob. 64UYKCh. 4.4 - Prob. 65UYKCh. 4.4 - Prob. 66UYKCh. 4.4 - Prob. 67UYKCh. 4.4 - Prob. 68ECh. 4.4 - Prob. 69ECh. 4.4 - Prob. 70ECh. 4.4 - Prob. 71ECh. 4.4 - Prob. 72ECh. 4.4 - Prob. 73ECh. 4.4 - Prob. 74ECh. 4.4 - Prob. 75ECh. 4.4 - Prob. 76ECh. 4.4 - Prob. 77ECh. 4.4 - Prob. 78ECh. 4.4 - Prob. 79ECh. 4.4 - Prob. 80ECh. 4.4 - Prob. 81ECh. 4.4 - Prob. 82ECh. 4.4 - Prob. 83ECh. 4.4 - Prob. 84ECh. 4.4 - Prob. 85ECh. 4.4 - Prob. 86ECh. 4.4 - Prob. 87ECh. 4.4 - Prob. 88ECh. 4.5 - Prob. 89UYKCh. 4.5 - Prob. 90UYKCh. 4.5 - Prob. 91UYKCh. 4.5 - Prob. 92UYKCh. 4.5 - Prob. 93UYKCh. 4.5 - Prob. 94UYKCh. 4.5 - Prob. 95UYKCh. 4.5 - Prob. 96ECh. 4.5 - Prob. 97ECh. 4.5 - Prob. 98ECh. 4.5 - Prob. 99ECh. 4.5 - Prob. 100ECh. 4.5 - Prob. 101ECh. 4.5 - Prob. 102ECh. 4.5 - Prob. 103ECh. 4.5 - Prob. 104ECh. 4.5 - Prob. 105ECh. 4.5 - Prob. 106ECh. 4.5 - Prob. 107ECh. 4.5 - Prob. 108ECh. 4.5 - Prob. 109ECh. 4.5 - Prob. 110ECh. 4.5 - Prob. 111ECh. 4.5 - Prob. 112ECh. 4.5 - Prob. 113ECh. 4.5 - Prob. 114ECh. 4.5 - Prob. 115ECh. 4.5 - Prob. 116ECh. 4.5 - Prob. 117ECh. 4.5 - Prob. 118ECh. 4.5 - Prob. 119ECh. 4.5 - Prob. 120ECh. 4.5 - Prob. 121ECh. 4.5 - Prob. 122ECh. 4.5 - Prob. 123ECh. 4 - Prob. 124ECh. 4 - Prob. 125ECh. 4 - Prob. 126ECh. 4 - Prob. 127ECh. 4 - Prob. 128ECh. 4 - Prob. 129ECh. 4 - Prob. 130ECh. 4 - Prob. 131ECh. 4 - Prob. 132ECh. 4 - Prob. 133ECh. 4 - Prob. 134ECh. 4 - Prob. 135ECh. 4 - Prob. 136ECh. 4 - Prob. 137ECh. 4 - Prob. 138ECh. 4 - Prob. 139E
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