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}](https://tex.z-dn.net/?f=%5Cmu%3D42%2C100%5Ctext%7B%20miles%7D)
![\sigma=2,510\text{ miles}](https://tex.z-dn.net/?f=%5Csigma%3D2%2C510%5Ctext%7B%20miles%7D)
Let x be the random variable that represents the lifetime of the tires .
z-score : ![z=\dfrac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
For x= 44,500 miles
![z=\dfrac{44500-42100}{2510}\approx0.96](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B44500-42100%7D%7B2510%7D%5Capprox0.96)
For x= 48,000 miles
![z=\dfrac{48000-42100}{2510}\approx2.35](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B48000-42100%7D%7B2510%7D%5Capprox2.35)
Using the standard normal distribution table , we have
The p-value : ![P(44500](https://tex.z-dn.net/?f=P%2844500%3Cx%3C48000%29%3DP%280.96%3Cz%3C2.35%29)
![P(z](https://tex.z-dn.net/?f=P%28z%3C2.35%29-P%28z%3C0.96%29%3D%200.9906132-0.8314724%3D0.1591408%5Capprox0.1591)
Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591
Answer:
x = 3
Step-by-step explanation:
4 + 12.5 > 11x - 3.5 - 2x
16.5 = 5.5x
16.5/5.5 = x