(a) Explanation Given: The number of customers waiting in a line at the express checkout of each 3 min interval between 9 A .M . and 12 noon on Saturday is shown in Table 1. Customers Frequency of Occurrence 0 1 1 4 2 2 3 7 4 14 5 8 6 10 7 6 8 3 9 4 10 1 Table 1 Calculation: The frequency of occurrence of customers in line is shown in Table 1. Consider n ( S ) as the total number of frequency of occurrence. The value of n ( S ) can be calculated as below. n ( S ) = 1 + 4 + 2 + 7 + 14 + 8 + 10 + 6 + 3 + 4 + 1 = 60 The number of customers is 0 if the frequency of occurrence is 1. Consider x as the number of customers waiting in line and P ( X = x ) as the probability of number of customers waiting in line. The probability P ( X = 0 ) can be calculated as follows: P ( X = 0 ) = 1 60 = 0 (b) To determine To graph: The histogram of the probability distribution.

10th Edition

ISBN: 9781337825597

Author: Tan

Publisher: INTER CENG

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Chapter 10.1, Problem 3TE

(a)

To determine

To find: The probability distribution of the random variable X.

(b)

To determine

To graph: The histogram of the probability distribution.

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Chapter 10.1, Problem 2TE

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The probability distribution of the random variable X .

Solution Summary: The author calculates the probability distribution of the random variable X. The number of customers waiting in a line at the express checkout is shown in Table 1.