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 5.1, Problem 19E

(a)

To determine

To state the question of interest using the language of probability.

(a)

Expert Solution
Check Mark

Explanation of Solution

By looking at the question, the question of interest made using the language of probability is as follows:

What is the probability that out of 12 first class passengers and 64 coach class passengers, a random screening of 10 passengers would include none from first class?

This is the question stated using the language of probability.

(b)

To determine

To explain how would you use random digits to imitate one repetition of the process and find out what variable you would measure.

(b)

Expert Solution
Check Mark

Answer to Problem 19E

The variable measured is the number of first class passengers.

Explanation of Solution

To use random digits to imitate one repetition of the process, we will do as;

As we know that numbers 0112 could represent the first class passengers and 1376 could represent coach class passengers. Ten two-digit numbers could then be taken from a line of the random digits table, skipping over repeated numbers and those from 77100 . Then the number of first class passengers in the selection could be measured.

Thus the variable measured is the number of first class passengers.

(c)

To determine

To use the line of random digits given to perform one repetition and explain the process.

(c)

Expert Solution
Check Mark

Explanation of Solution

As in the question given the random digits as-

  71487,09984,29077,14863,61683,47052,62224,51025 .

Select the first two digit number. If the number is between 01 and 76 , then select the correspond passenger, else ignore the number and move on to the next two digit number.

This is explained as,

  71Select passenger7148Select passenger4870Select passenger7099Ignore84Ignore29Select passenger2907Select passenger0771Repeated thus ignore48Repeated thus ignore63Select passenger6361Select passenger6168Select passenger6834Select passenger3470Repeated thus ignore52Select passenger52

Thus the numbers selected are as:

  71,48,70,29,07,63,61,68,34,52 .

(d)

To determine

To explain what conclusion would you draw.

(d)

Expert Solution
Check Mark

Explanation of Solution

As we know that numbers 0112 could represent the first class passengers and 1376 could represent coach class passengers. Ten two-digit numbers could then be taken from a line of the random digits table, skipping over repeated numbers and those from 77100 , then first class passengers could be measured.

Thus when in 100 repetitions of the simulation, there were 15 times when none of the ten passengers chosen were seated in first class. This may conclude that the ten passengers chosen could have the numbers between 1376 that represents coach class passengers or from 77100 that are skipped. This can happen as ten numbers are randomly chosen.

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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