Question

The annual frequency of claims against a single policy in a certain portfolio follows the distribution: P(N-0) 0.6, P(N-1) 0.25, P(N-2)-0.10, P(N-3) 0.0:5 There are 900 policies in this portfolio. Current staffing levels can handle as many as 580 claims against this portfolio in a year. If the annual claim count exceeds 580, then additional help wll be hired. The number of claims against different policies are independent. Calculate the probability that current staffing levels are adequate for next year.

Answer 1

From the discrete probability distribution we calculate the mean and variance of the number of claims against a single policy.

So, E(N) = 1*0.25+....+3*0.05 = 0.6

Variance (N) = E(N^2) - E(N)^2

E(N^2) = 1^2*0.25 +.....+3^2*0.05 = 0.55

So variance (N) = 0.55- 0.6^2 = 0.19

Now mean no of claims of portfolio = 900*0.6 = 540

Variance = Variance (900N) = 900^2*0.19 = 153900

So S.D of the no .of claims of the portfolio = 392.301

Now we will use normal approximation,

Now we want P (X<580) = P(Z<580-540/392.301) = P(Z<0.10) = 0.5398

is the required probability.