Question

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.

Answer 1

Answer: 0.0899.

Step-by-step 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:

$P(\overline{X}<997)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{997-1048}{\dfrac{282}{\sqrt{55}}})$

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

Hence, the required probability = 0.0899 .