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:
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
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