Answer to Question #209986 in Statistics and Probability for nena

Question

A sample of size 60 is taken from a population of chocolate balls. The weights of chocolate

balls are normally distributed with a mean of 40 g and a standard deviation of 8 g.

a. Find the parameters for this distribution of the sample mean.

b. Find the probability that the sample mean is between 38 g and 42

Answer 1

Let $\bar{X}=$ the sample mean: $\bar{X}\sim N(\mu, \sigma^2/n).$

a.

$$\mu_{\bar{X}}=\mu=40\ g, \sigma_{\bar{X}}=\sigma/\sqrt{n}=8/\sqrt{60}\ g$$

b.

$$P(38<\bar{X}<42)=P(\bar{X}<42)-P(\bar{X}\leq38)$$

$$=P(Z<\dfrac{42-40}{8/\sqrt{60}})-P(Z\leq\dfrac{38-40}{8/\sqrt{60}})$$

$$\approx P(Z<1.93649)-P(Z\leq-1.93649)$$ $$\approx 0.973596-0.026404\approx0.9472$$

The probability that the sample mean is between 38 g and 42 g is 0.9472.