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Suppose the average waiting time for a customer's call to be answered by a company representative is four minutes. (a) Find the probability that a call is answered during the first two minutes. (b) Find the probability that a customer waits more than four minutes to be answered.

Suppose the average waiting time for a customer's call to be answered by a company representative is four minutes. (a) Find the probability that a call is answered during the first two minutes. (b) Find the probability that a customer waits more than four minutes to be answered.

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.4: Expected Value
Problem 20E
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EXAMPLE Suppose the average waiting time for a customer's call to be answered by a company representative is four minutes. (a) Find the probability that a call is answered during the first two minutes. (b) Find the probability that a customer waits more than four minutes to be answered. SOLUTION (a) We are given that the mean of the exponential distribution is μ = 4 min and so, from the result of this example, we know that the probability density function is f(t) = Thus the probability that a call is answered during the first two minutes is [² P(0 ≤ T ≤ 2) = (0.25e-t/4 So about 39 P(T> 4) = = = = = = [" About 36.78 1² f(t) dt lim X→O 0.25 0.25e-t/4 dt 0.393 (b) The probability that a customer waits more than four minutes is Sof f(t) dt lim (00 0.25e-t/4 -0.25.4 e 30/+ 0.368 if t < 0 if t ≥ 0. e :(1/ -(4) 1 (₁-(²)_e-0) ~ - I (rounded to three decimal places.) % of the customers' calls are answered during the first two minutes. dt 0.25e-t/4 dt X (round to three decimal places). % of customers wait more than four minutes before their calls are answered.
Transcribed Image Text:EXAMPLE Suppose the average waiting time for a customer's call to be answered by a company representative is four minutes. (a) Find the probability that a call is answered during the first two minutes. (b) Find the probability that a customer waits more than four minutes to be answered. SOLUTION (a) We are given that the mean of the exponential distribution is μ = 4 min and so, from the result of this example, we know that the probability density function is f(t) = Thus the probability that a call is answered during the first two minutes is [² P(0 ≤ T ≤ 2) = (0.25e-t/4 So about 39 P(T> 4) = = = = = = [" About 36.78 1² f(t) dt lim X→O 0.25 0.25e-t/4 dt 0.393 (b) The probability that a customer waits more than four minutes is Sof f(t) dt lim (00 0.25e-t/4 -0.25.4 e 30/+ 0.368 if t < 0 if t ≥ 0. e :(1/ -(4) 1 (₁-(²)_e-0) ~ - I (rounded to three decimal places.) % of the customers' calls are answered during the first two minutes. dt 0.25e-t/4 dt X (round to three decimal places). % of customers wait more than four minutes before their calls are answered.
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