Answer to Question #113588 in Statistics and Probability for meem

Question

Suppose that 100 items are drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a non-normal probability distribution with mean = 8 ounces and standard deviation = 3 ounces. Consider the sampling distribution of the sample mean of the weights . a. Can you describe the sampling distribution of the sample mean? Explain why. b. Find the mean and variance of the sample mean? c. Find the probability that the mean weight is greater than 8 ounces. d. Find the probability that the mean weight is less than 6 ounces. e. Find the probability that the mean weight is between 7 ounces and 9 ounces
Expert's answer

Answer 1

a. If $n>30,$ the Central Limit Theorem can be used.

Let$X_1,X_2,...,X_n$ be a random sample from a distribution with mean $\mu$ and variance $\sigma^2.$

Then if $n$ is sufficiently large, $\bar{X}$ has approximately a normal distribution with $\mu_{\bar{X}}=\mu$ and $\sigma_{\bar{X}}=\sigma/\sqrt{n}.$

b.

$$\mu_{\bar{X}}=\mu=8\ ounces$$

$$Var(\bar{X})=\sigma^2/n=3^2/100=0.09\ (ounces^2)$$

$$\sigma_{\bar{X}}=\sigma/\sqrt{n}=3/\sqrt{100}=0.3 \ ounces$$

c.

$$P(\bar{X}>8)=1-P(\bar{X}\leq8)=1-P\big(Z\leq {8-8\over 0.3}\big)=$$

$$=1-0.5=0.5$$

d.

$$P(\bar{X}<6)=P\big(Z<{6-8\over 0.3}\big)\approx P(Z<-6.6667)\approx$$

$$\approx1.31\times 10^{-11}\approx 0$$

e.

$$P(7<\bar{X}<9)=P(\bar{X}<9)-P(\bar{X}<7)=$$

$$=P\big(Z<{9-8\over 0.3}\big)-P\big(Z<{7-8\over 0.3}\big)\approx$$

$$\approx P(Z<3.3333)- P(Z<-3.3333)\approx$$

$$\approx0.999571-0.000429\approx0.999142$$