Answer:
Probability that a customer will be on hold less than 30 minutes is [tex]95[/tex]%
Step-by-step explanation:
Complete question is
The service-time distribution describes the probability P that the service time of the customer will be no more than t hours. If m is
the mean number of customers services in an hour, then [tex]P = 1 - e^{-mt}[/tex]
. a. Suppose a computer technical support representative can answer calls from 6 customers in an hour. What is the probability
that a customer will be on hold less than 30 minutes?
Solution -
Given the equation of probability -
[tex]P = 1 - e^{-mt}[/tex]
Where m is the average number of customers served in an hour
and t is the total time
Now, in this case m is equal to 6 as technical support representative ia bale to answer calls from 6 customers in an hour
and time t [tex]= 30[/tex] minutes [tex]= \frac{1}{2}[/tex] hours
Substituting the given values in above equation, we get -
[tex]P = 1 - e^{-6 *\frac{1}{2} }\\P = 1 - e^{-3}\\P = 1 - 0.04978\\P = 0.95[/tex]
Probability that a customer will be on hold less than 30 minutes is [tex]95[/tex]%
