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CNNBC recently reported that the mean annual cost of auto insurance is 1048 dollars. Assume the standard deviation is 282 dollars. You take a simple random sample of 55 auto insurance policies. Find the probability that a sample of size n =55 is randomly selected with a mean less than 997 dollars.

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Answer: 0.0899.

Explanation:

Given: CNNBC recently reported that the mean annual cost of auto insurance is 1048 dollars, the standard deviation is 282 dollars.

Sample size : n= 55

Let
\overline{X} be the sample mean.

The probability that a sample of size n =55 is randomly selected with a mean less than 997 dollars:


P(\overline{X}<997)=P(\frac{\overline{X}-\mu}{(\sigma)/(√(n))}<(997-1048)/((282)/(√(55))))


=P(Z<-1.3412)\ \ \ [Z=\frac{\overline{X}-\mu}{(\sigma)/(√(n))}]\\\\=1-P(Z<1.3412)\\\\=1-0.9101\ \ \ \ [\text{By z-table}]\\\\ =0.0899

Hence, the required probability = 0.0899 .

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