Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 44,111 miles, with a variance of 5,943,844. What is the probability that the sample mean would be less than 44,257 miles in a sample of 80 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

`z=(44257-44111)/((2438)/(√(80)))= 0.536`

And if we use the normal standard table we got this:

`P(z<0.536) =0.7040`

Explanation:

For this case we have the following info :

`\mu = 44111` represent the true mean

`\sigma= √(5943844)= 2438` represent the deviation

n= 80 represent the sample size

And we want to find the follwing probability:

`P(\bar X< 44257)`

For this case since the sample size is larger than 30 we can apply the central limit theorem and we can use the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

The distribution for the sample mean would be:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

And if we find the z score for this case we got:

`z=(44257-44111)/((2438)/(√(80)))= 0.536`

And if we use the normal standard table we got this:

`P(z<0.536) =0.7040`