Question

A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. Five such tires are manufactured and tested. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal

Answer 1

For simplicity we use units of thousands of miles. Then the sample mean $\bar{X}$  has mean $\mu_{\bar{X}}=\mu=38.5$ thousands of miles and standard deviation $\sigma_{\bar{X}}=\sigma/\sqrt{n}=2.5/\sqrt{5}$

thousands of miles.

Since the distribution of lifetimes of such tires is normal, then $\bar{X}\sim N(\mu, \sigma^2/n).$

$$P(\bar{X}<36)=P(Z<\dfrac{36-\mu}{\sigma/\sqrt{n}})$$

$$=P(Z<\dfrac{36-38.5}{2.5/\sqrt{5}})= P(Z<-\sqrt{5})$$

$$\approx P(Z<-2.236068)\approx 0.01267366$$