Question

Management of an airline uses a normal distribution to model the value claimed for a lost piece of luggage on domestic flights. The mean of the distribution is $600 and the standard deviation is $85. Suppose a random sample of 65 pieces of luggage is to be selected.a) What is the expected value of the average claimed value of the 65 pieces of lost luggage?b) What is the standard deviation of the sampling distribution of the sample mean claimed value of the 65 pieces of lost luggage?c) What is the probability that the average claimed value of the 65 pieces is over $625?

Answer 1

Central Limit Theorem: If $\bar{X}$ is the mean of a random sample of size $n$ taken from a population with mean $\mu$ and finite variance $\sigma^2,$ then the limiting form of the distribution of

$$Z={\bar{X}-\mu \over \sigma/\sqrt{n}}$$

Given that $\mu=\$600, \sigma=\$85, n=65$

a)

$$E(\bar{X})=\mu_{\bar{X}}=\mu=\$600$$

(b)

$$\sigma_{\bar{X}}=\sigma/\sqrt{n}={\$85 \over \sqrt{65}}\approx\$10.54$$

(c)

$$P(\bar{X}>625)=P(Z>{625-600 \over 85/\sqrt{65}})=$$

$$=1-P(Z\leq{625-600 \over 85/\sqrt{65}})\approx1-0.99113578\approx0.0089$$