Question

A call center receives an average of 4.5 calls every 5 minutes. Each agent can handle one of these calls over the 5 minute period. If a call is received, but no agent is available to take it, then that caller will be placed on hold. Assuming that the calls follow a Poisson distribution, what is the minimum number of agents needed on duty so that calls are placed on hold at most 10% of the time

Answer 1

Answer:

7 agents

Step-by-step explanation:

According to poisson distribution,  

P (x=k) = λ^k e^-λ / k!    

λ = 4.5 calls

Hence, for 10% calls to be maximum on hold, meaning 90% of the calls need to be taken, hence, P(x=k) >0.9

P(x=0) = 4.5⁰ e^-4.5/0! = 0.01111

P(x=1) = 4.5¹ e^-4.5 /1! = 0.0495

P(x=2) = 0.173

P(x=3) = 0.342

P(x=4) = 0.532

P(x=5) = 0.703

P(x=6) = 0.831

P(x=7) = 0.913

Hence, P(x=7) > 0.9 ,

7 agents should be used for the job