help asap EXAMPLE 4 Suppose the average waiting time for a customer's call tobe answered by a company representative is five minutes.(a) Find the probability that a call is answered during the first fourminutes.(b) Find the probability that a customer waits more than fiveminutes to be answered.SOLUTION (a) We are given that the mean of the exponentialdistribution is μ = 5 min and so, from the result of this example, weknow that the probability density function isf(t) = {0.26-t/5Thus the probability that a call is answered during the first four minutes is4P(OSTS4)=Lª F= 6²°||11if t < 0if t ≥ 0.f(t) dt0.2e-t/5 dtdt-0.2So about 55during the first four minutes.419e5-e +10.551places).141.(rounded to three decimal% of the customers' calls are answered

Given: The average time taken by the company representative to pick up the call of a customer is…

EXAMPLE 4 Suppose the average waiting time for a customer's call to be answered by a company representative is five minutes. (a) Find the probability that a call is answered during the first four minutes. (b) Find the probability that a customer waits more than five minutes to be answered.

EXAMPLE 4 Suppose the average waiting time for a customer's call to be answered by a company representative is five minutes. (a) Find the probability that a call is answered during the first four minutes. (b) Find the probability that a customer waits more than five minutes to be answered.

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter13: Probability And Calculus

Section13.CR: Chapter 13 Review

Problem 61CR

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Question

help asap

EXAMPLE 4 Suppose the average waiting time for a customer's call to
be answered by a company representative is five minutes.
(a) Find the probability that a call is answered during the first four
minutes.
(b) Find the probability that a customer waits more than five
minutes to be answered.
SOLUTION (a) We are given that the mean of the exponential
distribution is μ = 5 min and so, from the result of this example, we
know that the probability density function is
f(t) = {0.26-t/5
Thus the probability that a call is answered during the first four minutes is
4
P(OSTS
4)
=
Lª F
= 6²°
||
11
if t < 0
if t ≥ 0.
f(t) dt
0.2e-t/5 dt
dt-0.2
So about 55
during the first four minutes.
419
e
5
-e +1
0.551
places).
14
1.
(rounded to three decimal
% of the customers' calls are answered

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