Question

An automobile insurer has found that repair claims are Normally distributed with a mean of $510 and a standard deviation of $460.

(a) Find the probability that a single claim, chosen at random, will be less than $490. ANSWER:

(b) Now suppose that the next 50 claims can be regarded as a random sample from the long-run claims process. Find the probability that the average x¯ of the 50 claims is smaller than $490. ANSWER:  

Answer 1

Solution :

(a)

P(x < $490) = P[(x - ) / < (490 - 510) / 460]

= P(z < -0.0435)

= 0.4827

Answer = 0.4827

(b)

= / n = 460 / 50 = 65.0538

P( < $490) = P(( - ) / < (490 - 510) / 65.0538)   

P(z < -0.3074)

= 0.3793

Answer = 0.3793