In a population of 210 women, the heights of the women are normally distributed with a mean of 64.4 inches and a standard deviation of 2.9 inches. if 36 women are selected at random, find the mean (mu sub x-bar) and standard deviation (sigma sub x-bar) of the population of sample means. assume that the sampling is done without replacement and use a finite population correction factor.

Part A:

Given that in a population of 210 women, the heights of the women are normally distributed with a mean of 64.4 inches and a standard deviation of 2.9 inches.

If 36 women are selected at random, the mean of the sample is estimated by the mean of the population.

Thefore, the mean is given by: [tex]\mu_{\bar{x}}=64.4 \ inches[/tex]



Part B:

The standard deviation using the finite population factor is given by:

[tex]\sigma_{\bar{x}}= \frac{\sigma}{\sqrt{n}} \cdot \sqrt{\frac{N-n}{N-1} } \\ \\ = \frac{2.9}{\sqrt{36}} \cdot \sqrt{\frac{210-36}{210-1} } =\frac{2.9}{6} \cdot \sqrt{\frac{174}{209} } \\ \\ =0.4833(\sqrt{0.8325}) =0.4833(0.9124) \\ \\ =0.44[/tex]