Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
8th Edition
ISBN: 9781464158933
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 4.3, Problem 55E

(a)

To determine

To find: The probability for selecting a single person.

(a)

Expert Solution
Check Mark

Answer to Problem 55E

Solution: Theprobability for selecting a single twitter user or personis 0.19.

Explanation of Solution

Given: According to Question 4.54, a random sample in which 19% are Twitter users who use Twitter to update their post or see another person's post.

Calculation: Consider the sample space (S) for choosing one person is,

S=[Y,N]

Here, Y represents the Twitter user in the sample and N represents those users who do not use Twitter.

The probabilities for S=[Y,N] are,

P(Y)=P(selecting a single person)=0.19

(b)

To determine

To find: The probability for selecting three people.

(b)

Expert Solution
Check Mark

Answer to Problem 55E

Solution: The probabilitiesare:P(TTT)=0.0069,P(TTTc)=P(TTcT)=P(TcTT)=0.0292P(TcTTc)=P(TcTcT)=P(TTcTc)=0.1247,P(TcTcTc)=0.5314

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

P(T)=0.19P(Tc)=10.19=0.81

The probabilities for sample space are,

P(TTT)=0.19×0.19×0.19=0.0069

P(TTTc)=P(TTcT)=P(TcTT)=0.19×0.19×(10.19)=0.0292

P(TcTTc)=P(TcTcT)=P(TTcTc)=0.19×(10.19)×(10.19)=0.1247

P(TcTcTc)=(10.19)×(10.19)×(10.19)=0.5314

(c)

To determine

To find: The probability for selecting a single person.

(c)

Expert Solution
Check Mark

Answer to Problem 55E

Solution: The possible values are 0, 1, 2, 3 and their probabilities are 0.0069, 0.0876, 0.3741 and 0.5314 respectively.

Explanation of Solution

Calculation: Consider the sample space (S) for choosing three person is,

S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc]

Here, T represents the Twitter user in the sample and Tc represents those users who do not use Twitter. Now, the sample space for the random variable X, which expresses the number of Twitter users in a sample size 3, is

X={0,1,2,3}

The probabilities for S=[TTT,TTTc,TTcT,TcTT,TTcTc,TcTTc,TcTcT,TcTcTc] are,

P(T)=0.19P(Tc)=10.19=0.81

Now, the probabilities are,

P(X=3)=P(TTT)=0.0069

P(X=2)=P(TTTc)+P(TTcT)+P(TcTT)=0.0292+0.0292+0.0292=0.0876

P(X=1)=P(TcTTc)+P(TcTcT)+P(TTcTc)=0.1247+0.1247+0.1247=0.3741

P(X=0)=P(TcTcTc)=0.5314

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Chapter 4 Solutions

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card

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. 17UYKCh. 4.2 - Prob. 18UYKCh. 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. 4.2 - Prob. 42ECh. 4.2 - Prob. 43ECh. 4.2 - Prob. 44ECh. 4.2 - Prob. 45ECh. 4.3 - Prob. 46UYKCh. 4.3 - Prob. 47UYKCh. 4.3 - Prob. 48UYKCh. 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. 4.3 - Prob. 63ECh. 4.3 - Prob. 64ECh. 4.3 - Prob. 65ECh. 4.3 - Prob. 66ECh. 4.4 - Prob. 66UYKCh. 4.4 - Prob. 67UYKCh. 4.4 - Prob. 68UYKCh. 4.4 - Prob. 69UYKCh. 4.4 - Prob. 70UYKCh. 4.4 - Prob. 71UYKCh. 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. 4.4 - Prob. 89ECh. 4.4 - Prob. 90ECh. 4.4 - Prob. 91ECh. 4.4 - Prob. 92ECh. 4.4 - Prob. 93ECh. 4.4 - Prob. 94ECh. 4.5 - Prob. 95UYKCh. 4.5 - Prob. 96UYKCh. 4.5 - Prob. 97UYKCh. 4.5 - Prob. 98UYKCh. 4.5 - Prob. 99UYKCh. 4.5 - Prob. 100UYKCh. 4.5 - Prob. 101UYKCh. 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. 4.5 - Prob. 124ECh. 4.5 - Prob. 125ECh. 4.5 - Prob. 126ECh. 4.5 - Prob. 127ECh. 4.5 - Prob. 128ECh. 4.5 - Prob. 129ECh. 4.5 - Prob. 130ECh. 4.5 - 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. 139ECh. 4 - Prob. 140ECh. 4 - Prob. 141ECh. 4 - Prob. 142ECh. 4 - Prob. 143ECh. 4 - Prob. 144ECh. 4 - Prob. 145ECh. 4 - Prob. 146ECh. 4 - Prob. 147ECh. 4 - Prob. 148ECh. 4 - Prob. 149ECh. 4 - Prob. 150ECh. 4 - Prob. 151ECh. 4 - Prob. 152E
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