Answer: (C) 0.1591
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}](https://img.qammunity.org/2020/formulas/mathematics/high-school/rdh5xoe87khumjt83ydhpbnqztf6ckmovt.png)
![\sigma=2,510\text{ miles}](https://img.qammunity.org/2020/formulas/mathematics/high-school/ta1kmbj20muwme6e4fvtcdmpspou3lf6jk.png)
Let x be the random variable that represents the lifetime of the tires .
z-score :
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/high-school/10fia1p0qwvlz4zhb867kzy3u7bscognwz.png)
For x= 44,500 miles
![z=(44500-42100)/(2510)\approx0.96](https://img.qammunity.org/2020/formulas/mathematics/high-school/dw5yb63wvgoi4b8e53k16pzsqo19yweq8i.png)
For x= 48,000 miles
![z=(48000-42100)/(2510)\approx2.35](https://img.qammunity.org/2020/formulas/mathematics/high-school/z43y2aooot8eq1af9q84dfmtslmic0r7yk.png)
Using the standard normal distribution table , we have
The p-value :
![P(44500<x<48000)=P(0.96<z<2.35)](https://img.qammunity.org/2020/formulas/mathematics/high-school/m0h0yk2py156u1t9qmg9t6jckglpdtwkd0.png)
![P(z<2.35)-P(z<0.96)= 0.9906132-0.8314724=0.1591408\approx0.1591](https://img.qammunity.org/2020/formulas/mathematics/high-school/br61tb7bagxs6jqzgoch2gmyr756m6h4c8.png)
Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591