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, Problem 6CRE

a.

To determine

To make:a Venn diagram to model this chances process.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

The probability that Princeton will admit him is 0.4,

The probability that Stanford will admit him is 0.5,

And The probability that Pboth of them will admit him is 0.2.

Calculation:

Draw two ellipses to overlap partly.

Name the one of the left “Princeton” and on the right “Stanford”.

Write the probability being admitted to both in the intersection of the ellipses.

The probability is being admitted to Princeton is 0.4, while the probability being admitted is 0.2, thus the probability of being admitted to Princeton alone is 0.2. This has to be written to the left of the intersection.

The probability is being admitted to Stanford is 0.5, while the probability being admitted is 0.2, thus the probability of being admitted to Stanford alone is 0.2. This has to be written to the right of the intersection.

  The Practice of Statistics for AP - 4th Edition, Chapter 5, Problem 6CRE

b.

To determine

To find: the probability that neither university admits Ramon.

b.

Expert Solution
Check Mark

Answer to Problem 6CRE

0.3

Explanation of Solution

Given:

The probability that Princeton will admit him is 0.4,

The probability that Stanford will admit him is 0.5,

And The probability that Pboth of them will admit him is 0.2.

Calculation:

The probability of not being admitted to either university is written outside of the two ellipse and thus is 0.3.

c.

To determine

To find: the probability that Ramon gets at least one of the twoschool.

c.

Expert Solution
Check Mark

Answer to Problem 6CRE

0.3

Explanation of Solution

Given:

The probability that Princeton will admit him is 0.4,

The probability that Stanford will admit him is 0.5,

And The probability that Pboth of them will admit him is 0.2.

Calculation:

General addition rule:

  P(AorB)=P(A)+P(A)P(A and B)

Thus, it obtained:

  

  P(Princton or Standford)=P(Princton)+P(Standford)P(White and Hispanic)P(Princton and Standford)=0.4+0.50.2=0.7

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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