Answer:
 As the one more server is added the Average number of customers in service reduces from  ( 5.585 to 3.019  )    which is a reduction of 2.566 
Explanation:
interval time of customers ( a )  = 2 minutes 
processing time ( p ) = 5 minutes
coefficient of variation for arrival process ( Cva ) = 1
coefficient of variation for service process ( Cvp ) = 0.85
<u>Determine the impact on the average number of customers in service if one more server is added </u>
<em>For three (3) servers (  m ) </em> we will have 
flow rate = 1 / a = 1/2 = 0.5 
x =  ( 2 (m+1) -1 ) ^0.5 
where m = 3 ( number of servers ) 
hence X = 2.645 
now lets determine
i) utilization =  p / (m*a). where p = 5 , m = 3 , a = 2
hence utilization ( u ) = 0.833 ,  1 - u = 1 - 0.833 = 0.167
ii) Time spent in queue 
 time spent in queue = 6.17 minutes 
number of customers been served = mu = 3 * 0.833 = 2.500
number of customers waiting to be served = waiting time / arrival time 
= 6.17 / 2 = 3.085
iii) Average customer in queue = number of customers been serviced + number of customers waiting to be served =  2.5 + 3.085 = 5.585
For 4 servers  ( m = 4 ) 
x = ( 2 (m+1) -1 ) ^0.5  
    = ( 2 ( 5 ) - 1 ) ^ 0.5 = 3
 utilization =  p / (m*a). where p = 5 , m = 4 , a = 2
    hence utilization = 0.625 ,  
Time spent in queue =  1.039 minutes  ( using equation 2 )
number of customers been serviced = m* u = 2.5
number of customers waiting to be served = waiting time / arrival time
= 1.039 / 2 = 0.520
Average customer in queue = number of customers been serviced + number of customers waiting to be served = 2.5 + 0.520 = 3.019
hence as the one more server is added the Average number of customers in service reduces from  ( 5.585 to 3.019  )    which is a reduction of 2.566