Question

1) During busy times, 60 potential customers per hour arrive at the booth (assume a Poisson distribution). A booth worker takes 3.75 minutes, on average, to meet the information needs of a potential customer (assume an exponential distribution). How many booth workers are required for the average potential customer to spend no more than 5 minutes waiting and being served

Answer 1

Booth workers required = 4

Explanation:

As given ,

Arrival rate = λ = 60 customers per hour

Time taken = 3.75 minutes

∴ we get

Average service rate = μ =
`(60)/(3.75)` = 16 customers per hour

Now,

Minimum numbers of booth workers required =
`(60)/(16)` = 3.75

As workers can not be in fraction

∴ Minimum number of booth workers required = 4