(a) Find the probability that a single claim, chosen at random, will be less than $490. ANSWER: (b) Now suppose that the next 70 claims can be regarded as a random sample from the long-run claims process. Find the probability that the average of the 70 claims is smaller than $490. ANSWER: (c) If a sample larger than 70 claims is considered, there would be chance of getting a sample with an average smaller then $490. (NOTE: Enter "LESS", "MORE" or "AN EQUAL" without the quotes.)

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An automobile insurer has found that repair claims are Normally distributed with a mean of $530 and a standard deviation of $470.

(a) Find the probability that a single claim, chosen at random, will be less than $490.
ANSWER:
(b) Now suppose that the next 70 claims can be regarded as a random sample from the long-run claims process.
Find the probability that the average of the 70 claims is smaller than $490.
ANSWER:
(c) If a sample larger than 70 claims is considered, there would be
chance of getting a
sample with an average smaller then $490. (NOTE: Enter "LESS", "MORE" or "AN EQUAL" without the quotes.)
Transcribed Image Text:(a) Find the probability that a single claim, chosen at random, will be less than $490. ANSWER: (b) Now suppose that the next 70 claims can be regarded as a random sample from the long-run claims process. Find the probability that the average of the 70 claims is smaller than $490. ANSWER: (c) If a sample larger than 70 claims is considered, there would be chance of getting a sample with an average smaller then $490. (NOTE: Enter "LESS", "MORE" or "AN EQUAL" without the quotes.)
Expert Solution
Step 1: Introduce the given information

Given,

Mean, μ = $530

Standard deviation, σ = $470

Let X be a random variable represent  the repair claims ($).

Using formulas,

Z minus s c o r e equals fraction numerator X minus mu over denominator sigma end fraction

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