Question

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4959 miles, with a standard deviation of 448.If he is correct, what is the probability that the mean of a sample of 43 cars would differ from the population mean by less than111 miles? Round your answer to four decimal places.

Answer 1

Answer:

0.8958

Step-by-step explanation:

Given that:

The mean $\mu$ = 4959

The standard deviation $\sigma$ = 448

The sample size n = 43

$\mu_{\overline x} = \mu = 4959$

$\sigma __{\overline x}} = \dfrac{\sigma}{\sqrt{n}}= \dfrac{448}{\sqrt{43}}$

By applying the central limit theorem;

$\overline X \sim N \bigg( \mu_{\overline x} = 4959 , \sigma_{\overline x} = \dfrac{448}{\sqrt{43}}\bigg )$

The purpose of this question is to calculate the probability that the mean of a sample of 43 cars would differ from the population mean by less than111 miles.

This implies that;

$P( |\mu-\overline x| <111) = p( 4959-111 < \overline X < 4959 +111)$

$= P( 4848 < \overline X < 5070)$

$= P \bigg ( \dfrac{ \overline x - \mu_x }{ \dfrac{\sigma}{\sqrt{n}}} < Z < \dfrac{ \overline x - \mu_x }{ \dfrac{\sigma}{\sqrt{n}}} \bigg )$

$= P \bigg ( \dfrac{4848 -4959 }{ \dfrac{448}{\sqrt{43}}} < Z < \dfrac{5070 -4959 }{ \dfrac{448}{\sqrt{43}}} \bigg )$

$= P \bigg ( \dfrac{-111 }{ 68.319} < Z < \dfrac{111 }{68.319} \bigg )$

$= P ( -1.6247< Z < 1.6247 )$

= P ( Z < 1.627 ) - P(Z < -1.627)

From the standard normal tables

= 0.94791 -0.05208

= 0.89583

≅ 0.8958  to four decimal places.