The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 5.3, Problem 67E

a.

To determine

To find:the probability that the result for “democrats” in the given table.

a.

Expert Solution
Check Mark

Answer to Problem 67E

  P(D|F)=0.7647 .

Explanation of Solution

Given:

The two-way table describes the members of the U.S. Senate in a recent year.

Consider event D is democrat, even R is republic, F is female.

    Male Female
    Democrats 47 13
    Republics 36 4

Concept used:

Probability is the number of favourable outcomes divided by the number of possible outcomes.

Thus, Definition condition probability:

  P(A|B)=P(A and B)P(B)

Calculation:

According to the question, the person is female. Now to find the probability that the result for “democrats”.

Therefore,

  P(D and F)=favourable outcomespossible outcomes=1347+13+36+4=13100

  P(F)=favourable outcomespossible outcomes=13+4100=17100

Thus, the conditional probability becomes:

  P(D|F)=P(D and F)P(F)=13/10017/100=13170.7647

Thus,the probability that the result for “democrats” is P(D|F)=0.7647 .

b.

To determine

To find: the probability that the result for “female” in the given table.

b.

Expert Solution
Check Mark

Answer to Problem 67E

  P(F|D)=0.2167 .

Explanation of Solution

Given:

The two-way table describes the members of the U.S. Senate in a recent year.

Consider event D is democrat, even R is republic, F is female.

    Male Female
    Democrats 47 13
    Republics 36 4

Concept used:

Probability is the number of favourable outcomes divided by the number of possible outcomes.

Thus, Definition condition probability:

  P(A|B)=P(A and B)P(B)

Calculation:

According to the question, the person is democrats. Now to find the probability that the result for “female”.

Therefore,

  P(D and F)=favourable outcomespossible outcomes=1347+13+36+4=13100

  P(D)=favourable outcomespossible outcomes=47+13100=60100

Thus, the conditional probability becomes:

  P(F|D)=P(D and F)P(D)=13/10060/100=13600.2167

Hence, the probability that the result for “female” P(F|D)=0.2167 .

Chapter 5 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10ECh. 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. 5.1 - Prob. 31ECh. 5.1 - Prob. 32ECh. 5.1 - Prob. 33ECh. 5.1 - Prob. 34ECh. 5.1 - Prob. 35ECh. 5.1 - Prob. 36ECh. 5.1 - Prob. 37ECh. 5.1 - Prob. 38ECh. 5.2 - Prob. 1.1CYUCh. 5.2 - Prob. 1.2CYUCh. 5.2 - Prob. 1.3CYUCh. 5.2 - Prob. 2.1CYUCh. 5.2 - Prob. 2.2CYUCh. 5.2 - Prob. 2.3CYUCh. 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. 5.2 - Prob. 61ECh. 5.2 - Prob. 62ECh. 5.3 - Prob. 1.1CYUCh. 5.3 - Prob. 1.2CYUCh. 5.3 - Prob. 2.1CYUCh. 5.3 - Prob. 2.2CYUCh. 5.3 - Prob. 2.3CYUCh. 5.3 - Prob. 3.1CYUCh. 5.3 - Prob. 3.2CYUCh. 5.3 - Prob. 4.1CYUCh. 5.3 - Prob. 4.2CYUCh. 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. 5.3 - Prob. 109ECh. 5 - Prob. 1CRECh. 5 - Prob. 2CRECh. 5 - Prob. 3CRECh. 5 - Prob. 4CRECh. 5 - Prob. 5CRECh. 5 - Prob. 6CRECh. 5 - Prob. 7CRECh. 5 - Prob. 8CRECh. 5 - Prob. 9CRECh. 5 - Prob. 10CRECh. 5 - Prob. 11CRECh. 5 - Prob. 1PTCh. 5 - Prob. 2PTCh. 5 - Prob. 3PTCh. 5 - Prob. 4PTCh. 5 - Prob. 5PTCh. 5 - Prob. 6PTCh. 5 - Prob. 7PTCh. 5 - Prob. 8PTCh. 5 - Prob. 9PTCh. 5 - Prob. 10PTCh. 5 - Prob. 11PTCh. 5 - Prob. 12PTCh. 5 - Prob. 13PTCh. 5 - Prob. 14PT
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