An automobile insurer has found that repair claims are normally distributed with

Question

An automobile insurer has found that repair claims are normally distributed with a mean of $850 and a standard deviation of $790.

(a) Find the probability that a single claim, chosen at random, will be less than $810.
ANSWER:  

(b) Now suppose that the next 60 claims can be regarded as a random sample from the long-run claims process. Find the probability that the average x¯x¯ of the 60 claims is smaller than $810.
ANSWER:

(c) If a sample larger than 60 claims is considered, there would be   chance of getting a sample with an average smaller then $810. (NOTE: Enter ''LESS'', ''MORE'' or ''AN EQUAL'' without the quotes.)

Explanation / Answer

Mean is 850 and s is 790. z can be found as (x-mean)/s

a) P(x<810) =P(z<(810-850)/790)=P(z<-0.05) or 1-P(z<0.05). from normal distribution table we get 1-0.5199=0.4801

b) now we need standard error which is s/sqrt(N)=790/sqrt(60)=101.989

Thus P(xbar<810) =P(z<(810-850)/(101.989))=P(z<-0.39) or 1-P(z<0.39). from normal table we get 1-0.6517=0.3483

c) LESS since b<a

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