Question

(5.32) A company that owns and services a fleet of cars for its sales force has found that the service lifetime of disc brake pads varies from car to car according to a normal distribution with mean 55,000 miles and the standard deviation 4500 miles. The company installs a new brand of brake pads on 8 cars.

(a) If the new brand has the same lifetime distribution as the previous type, what is the distribution of the sample mean lifetime for the 8 cars

(b) The average of the pads on these 8 cars turns out to be 51,800 miles. What is the probability that the sample mean lifetime is 51,800 miles or less if the lifetime distribution is unchanged? The company takes this probability as evidence that the average lifetime of the new brand pads is less than 55,000 miles.

Answer 1

Suppose X is the random variable denoting the service lifetime of disc brake pads of a car.

So, X ~ N(55000, 4500²)

(a)

Now, we have to find the distribution of _{$\frac{1}{8}\sum_{i=1}^{8}X_i = \bar{X}$}

We know,

if X follows Normal distribution with mean $\mu$ and variance $\sigma^2$ , then $\frac{1}{n}\sum_{i=1}^{n}X_i$ follows Normal distribution with mean $\mu$ and variance $\sigma^2$ /n.

So, here n = 8.

So, $\frac{1}{8}\sum_{i=1}^{8}X_i = \bar{X}$ follows Normal distribution with mean 55000 and variance 4500²/8 = 2531250.

(b)

Now, the sample mean is turned out 51800.

So, the required probability is P[ _{$\bar{X} \le 51800$} ] = 0.02214552. [Using R code pnorm(51800,55000,sqrt(2531250))]