The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 77 minutes. what is the probability that a person is served in less than 22 minutes on at least 55 of the next 77 days?

0.01265

Step-by-step explanation:

Since, if the time to be served has an exponential distribution with a mean of 7, then

Chance to be served in under 2 minutes:

Let A represents the number of days when a person is served in less than 2 minutes,

Hence,

the probability that a person is served in less than 2 minutes on at least 5 of the next 7 days ( using binomial distribution )

= P(A=5) + P(A=6) + P(A=7)

≈ 0.01265

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step-by-step explanation:

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