Suppose that the time between arrivals of customers at a bank during the noon-to

Question

Suppose that the time between arrivals of customers at a bank during the noon-to-1 p.m. hour has a uniform distribution between 0 and 180 seconds. A) What is the probability that the time between the arrivals of two customers will be less than 68 seconds? B) What is the probability that the time between the arrivals of two customers will be between 58 and 148 seconds? C) What is the probability that the time between the arrivals of two customers will be greater than 48 seconds? D) What are the mean and standard deviation of the time between the arrival of two customers?

a. The probability that the time between arrivals will be less than 68 seconds is

(Round to four decimal places as needed.)

b. The probability that the time between arrivals will be between 58 and 148 seconds is

(Round to four decimal places as needed.)

c. The probability that the time between arrivals will be greater than 48 is

(Round to four decimal places as needed.)

d. The mean of the given uniform distribution is mu equals=

(Round to four decimal places as needed.)

The standard deviation of the given uniform distribution is sigma equals=

(Round to four decimal places as needed.)

Explanation / Answer

a = 0 b = 180

b - a = 180 - 0 = 180

a. The probability that the time between arrivals will be less than 68 seconds is 68 / 180 = 0.3778.

b. The probability that the time between arrivals will be be between 58 and 148 seconds is 148/180 - 58/180 = 0.5.

c. The probability that the time between arrivals will be greater than 48 is (180-48) / 180 = 0.7333.

d. The mean of the given uniform distribution is equals = (0 + 180) / 2 = 90.0000.

The standard deviation of the given uniform distribution is equals = 180/ 12 = 51.9615.

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