Question

CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 234 dollars. You take a simple random sample of 61 auto insurance policies. Find the probability that a single randomly selected value is at least 960 dollars.

Answer 1

Answer: 0.42

Step-by-step explanation:

Given: Mean : $\mu=954\text{ dollars}$

Standard deviation : $234\text{ dollars}$

Sample size : $n=61$

The formula to calculate z score is given by :-

$z=\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For X=960

$z=\dfrac{960 -954}{\dfrac{234}{\sqrt{61}}}=0.200262812203\approx0.2$

The p-value =$P(X\geq960)=1-P(X<960)=1-P(z<0.2)=1-0.5792597=0.4207403\approx0.42$

Hence,  the probability that a single randomly selected value is at least 960 dollars = 0.42