Question

a)     An insurance company attends 50 clients a day. On average 3 in 100 require special services. On a certain day it is found that there are 3 special service providers available. Assuming that 50 clients will be attended to, find the probability that there are more than three clients require special services?                                                 

Answer 1

The appropriate distribution for this question is Poisson distribution.

$P(X=k) = \frac{λ^k \times e^{-λ}}{k!} \\ P(X>3) = 1 -P(X≤3) \\ = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)]$

Mean $λ= \frac{3}{2} = 1.5$ (for 50 clients)

By Excel function

P(X≤3) = POISSON(3,1.5,1)

= 0.93435

P(X>3) = 1 -0.93435 = 0.06565