Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 35,869 miles, with a variance of 12,194,0601. What is the probability that the sample mean would differ from the population mean by less than 375 miles in a sample of 269 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

Answer:0.2888

Step-by-step explanation:

Given

mean $\bar{x}=35,869$

$variance=12,194,0601$

standard deviation$=\sqrt{variance}$

$\sigma =\sqrt{12,194,0601}=11,042.67$

$n=269$

population mean by less than 375 miles in a sample

$P(z<\frac{-375}{\frac{\sigma }{\sqrt{n}}})$

$P(z<\frac{-375}{\frac{11,042.67}{\sqrt{269}}})$

$P(z<-0.5569)$

From z table Probability is 0.2888