Question

A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with a mean of 42,100 miles and a standard deviation of 2,510 miles. Find the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles. Be certain that you round your z-values to two decimal places. Round your answer to 4 decimal places.(A) 0.1685 (B) 0.8315 (C) 0.1591 (D) 0.3315 (E) None of these are correct.

Answer 1

Answer:  (C) 0.1591

Step-by-step explanation:

Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

$\mu=42,100\text{ miles}$

$\sigma=2,510\text{ miles}$

Let x be the random variable that represents the lifetime of the tires .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 44,500 miles

$z=\dfrac{44500-42100}{2510}\approx0.96$

For x= 48,000 miles

$z=\dfrac{48000-42100}{2510}\approx2.35$

Using the standard normal distribution table , we have

The p-value : $P(44500<x<48000)=P(0.96<z<2.35)$

$P(z<2.35)-P(z<0.96)= 0.9906132-0.8314724=0.1591408\approx0.1591$

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles =  0.1591