question archive Suppose that the mean cost of automobile insurance in NC is $959 with standard deviation $215
Suppose that the mean cost of automobile insurance in NC is $959 with standard deviation $215
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Suppose that the mean cost of automobile insurance in NC is $959 with standard deviation $215. What is the probability that the average cost for a sample of 50 NC automobile insurance policies is higher than $1000? (4 decimal places)
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For a normal distribution with given sample, we will follow the formula:
(x-mean)/(sd/sqrt(n)) = z-value then use z-table to find probability
Given:
mean=959
sd=215
n =50
P(X>1000) =(1000-959)/(215/sqrt(50)) = 1.34843618738 = P(Z>1.34843618738) = 0.0888
