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

Answer:

$P(x < -778) = 0$

Step-by-step explanation:

Given

$\bar x = 48673$

$\sigma^2 = 11282880$

$n = 143$

Required

$P(x <- 778)$

First, we calculate the z score

$z = \frac{x}{\sqrt{\sigma^2}/n}$

So, we have:

$z = \frac{-778}{\sqrt{11282880}/143}$

$z = \frac{-778}{3359.0/143}$

$z = \frac{-778}{23.49}$

$z = -33.12$

So:

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

From z score probability, we have:

$P(x < -778) = 0$