Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,637 miles, with a variance of 11,282,880. What is the probability that the sample mean would differ from the population mean by less than 778 miles in a sample of 143 tires if the manager is correct

Answer 1

`P(x < -778) = 0`

Explanation:

Given

`\bar x = 48673`

`\sigma^2 = 11282880`

`n = 143`

Required

`P(x <- 778)`

First, we calculate the z score

`z = (x)/(√(\sigma^2)/n)`

So, we have:

`z = (-778)/(√(11282880)/143)`

`z = (-778)/(3359.0/143)`

`z = (-778)/(23.49)`

`z = -33.12`

So:

`P(x < -778) = P(z < -33.12)`

From z score probability, we have:

`P(x < -778) = 0`