Question

The operation manager at a tire manufacturer company believes that the main mileage of attire is 28,631 miles with the variance of 16,834,610. What is the probability that the sample mean would differ from the population mean by less than 219 miles and a sample of 178 tires if the manager is correct? Round your answer to four decimal places

Answer 1

The Solution:

Given:

`\begin{gathered} \mu=28631miles \\ \\ \sigma=√(16834610)=4103\text{ miles} \\ \\ n=178\text{ tires} \\ \\ \bar{x}-\mu=219 \end{gathered}`

By Z statistic formula:

`Z=\frac{\bar{x}-\mu}{(\sigma)/(√(n))}=(219)/((4103)/(√(178)))=(219√(178))/(4103)=0.71211`

From the Z-tables, the probability that the sample mean would differ from the population mean is:

[tex]P(-0.71211Answer:

0.5236