Answer
a. The expected total service time for customers = 70 minutes
b. The variance for the total service time = 700 minutes
c. It is not likely that the total service time will exceed 2.5 hours
Step-by-step explanation:
This question is incomplete. I will give the complete version below and proceed with my solution.
Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?
Reference
Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.
From the information supplied, we denote that 
 Customers that arrive within the hour
 Customers that arrive within the hour
and since  follows a Poisson distribution with mean
 follows a Poisson distribution with mean  = 7
 = 7
Therefore,
 
 

Let  the total service time for customers arriving during the 1 hour period.
 the total service time for customers arriving during the 1 hour period.
Now, since it takes approximately ten minutes to serve each customer,

For a random variable  and a constant
 and a constant  ,
,

Thus,

Therefore the expected total service time for customers = 70 minutes
and the variance for serving time = 700 minutes
Also, the probability of the distribution  is,
 is,

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

0.002 is small enough, and the function  gets even smaller when
  gets even smaller when  increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.
 increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.