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 7, Problem 4PT2
To determine

To explain: The way that could be used to find the number of correct answers that Jason obtains

Expert Solution & Answer
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Answer to Problem 4PT2

By simulating the two digits that show the correct answer.

Explanation of Solution

Given:

Number of multiple questions = 10

Probability of correct answer should be 15

Calculation:

Here, if one digit from the table of random digit simulates single answer using 0 or 1 = correct and all other digits = incorrect. In such case, 10 digits from the random digit table would stimulate 10 answers. So, it implies that there would be 2 digits that would depict the correct answer.

The probability of correct answer could be calculated as:

  P(Correct answer)=Number of digits that show the correct answerTotal digits=210=15

It could be said that the obtained probability is same as the required probability.

Thus, the correct option is (b).

Chapter 7 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 7.1 - Prob. 3ECh. 7.1 - Prob. 4ECh. 7.1 - Prob. 5ECh. 7.1 - Prob. 6ECh. 7.1 - Prob. 7ECh. 7.1 - Prob. 8ECh. 7.1 - Prob. 9ECh. 7.1 - Prob. 10ECh. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.2 - Prob. 1.1CYUCh. 7.2 - Prob. 1.2CYUCh. 7.2 - Prob. 1.3CYUCh. 7.2 - Prob. 1.4CYUCh. 7.2 - Prob. 27ECh. 7.2 - Prob. 28ECh. 7.2 - Prob. 29ECh. 7.2 - Prob. 30ECh. 7.2 - Prob. 31ECh. 7.2 - Prob. 32ECh. 7.2 - Prob. 33ECh. 7.2 - Prob. 34ECh. 7.2 - Prob. 35ECh. 7.2 - Prob. 36ECh. 7.2 - Prob. 37ECh. 7.2 - Prob. 38ECh. 7.2 - Prob. 39ECh. 7.2 - Prob. 40ECh. 7.2 - Prob. 41ECh. 7.2 - Prob. 42ECh. 7.2 - Prob. 43ECh. 7.2 - Prob. 44ECh. 7.2 - Prob. 45ECh. 7.2 - Prob. 46ECh. 7.2 - Prob. 47ECh. 7.2 - Prob. 48ECh. 7.3 - Prob. 1.1CYUCh. 7.3 - Prob. 1.2CYUCh. 7.3 - Prob. 1.3CYUCh. 7.3 - Prob. 1.4CYUCh. 7.3 - Prob. 49ECh. 7.3 - Prob. 50ECh. 7.3 - Prob. 51ECh. 7.3 - Prob. 52ECh. 7.3 - Prob. 53ECh. 7.3 - Prob. 54ECh. 7.3 - Prob. 55ECh. 7.3 - Prob. 56ECh. 7.3 - Prob. 57ECh. 7.3 - Prob. 58ECh. 7.3 - Prob. 59ECh. 7.3 - Prob. 60ECh. 7.3 - Prob. 61ECh. 7.3 - Prob. 62ECh. 7.3 - Prob. 63ECh. 7.3 - Prob. 64ECh. 7.3 - Prob. 65ECh. 7.3 - Prob. 66ECh. 7.3 - Prob. 67ECh. 7.3 - Prob. 68ECh. 7.3 - Prob. 69ECh. 7.3 - Prob. 70ECh. 7.3 - Prob. 71ECh. 7.3 - Prob. 72ECh. 7 - Prob. 1CRECh. 7 - Prob. 2CRECh. 7 - Prob. 3CRECh. 7 - Prob. 4CRECh. 7 - Prob. 5CRECh. 7 - Prob. 6CRECh. 7 - Prob. 7CRECh. 7 - Prob. 1PTCh. 7 - Prob. 2PTCh. 7 - Prob. 3PTCh. 7 - Prob. 4PTCh. 7 - Prob. 5PTCh. 7 - Prob. 6PTCh. 7 - Prob. 7PTCh. 7 - Prob. 8PTCh. 7 - Prob. 9PTCh. 7 - Prob. 10PTCh. 7 - Prob. 11PTCh. 7 - Prob. 12PTCh. 7 - Prob. 13PTCh. 7 - Prob. 1PT2Ch. 7 - Prob. 2PT2Ch. 7 - Prob. 3PT2Ch. 7 - Prob. 4PT2Ch. 7 - Prob. 5PT2Ch. 7 - Prob. 6PT2Ch. 7 - Prob. 7PT2Ch. 7 - Prob. 8PT2Ch. 7 - Prob. 9PT2Ch. 7 - Prob. 10PT2Ch. 7 - Prob. 11PT2Ch. 7 - Prob. 12PT2Ch. 7 - Prob. 13PT2Ch. 7 - Prob. 14PT2Ch. 7 - Prob. 15PT2Ch. 7 - Prob. 16PT2Ch. 7 - Prob. 17PT2Ch. 7 - Prob. 18PT2Ch. 7 - Prob. 19PT2Ch. 7 - Prob. 20PT2Ch. 7 - Prob. 21PT2Ch. 7 - Prob. 22PT2Ch. 7 - Prob. 23PT2Ch. 7 - Prob. 24PT2Ch. 7 - Prob. 25PT2
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