Explanation Given: A survey of 1000 people is conducted and the following question was asked, “How much emergency living expense saving?” The answers were expressed in the number of months the person could live on savings. The result of the survey is given in the table below, Group 1 2 3 4 Time (in months) Less than 3 3 − 6 More than 6 No answer Number 480 200 290 30 Let X denote the random variable that takes on the values 1 , 2 , 3 and 4 , corresponding to the groups given in the table above. Approach: The probability of an event E is written below. P ( E ) = n ( E ) n ( S ) Here, P ( E ) is the probability of the event E , n ( E ) is the number of favorable outcomes for E and n ( S ) is the total number of outcomes. Calculation: Find the total number of outcomes from the given data as follows, n ( S ) = 480 + 200 + 290 + 30 = 1 , 000 Substitute 480 for n ( E ) , and 1 , 000 for n ( S ) to find the probability associated with group 1 X ( X = 1 ) , To determine b) The probability of a respondent that he or she had 6 or fewer months’ worth of emergency living expense savings.

Finite Mathematics for the Managerial, Life, and Social Sciences-Custom Edition

11th Edition

ISBN: 9781305283831

Author: Tan

Publisher: Cengage Learning

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Chapter 8.1, Problem 24E

To determine

a)

The probability distribution associated with the given data.

To determine

b)

The probability of a respondent that he or she had 6 or fewer months’ worth of emergency living expense savings.

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Chapter 8.1, Problem 23E

Chapter 8.1, Problem 25E

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Chapter 8 Solutions

Finite Mathematics for the Managerial, Life, and Social Sciences-Custom Edition

Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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a) The probability distribution associated with the given data.

Solution Summary: The author explains the probability distribution associated with the given data.