Question

A certain population has a mean of 500 and a standard deviation of 23. Many samples of size 41 are randomly selected and the means calculated.

(a) What value would you expect to find for the mean of all these sample means? (Give your answer correct to nearest whole number.)

(b) What value would you expect to find for the standard deviation of all these sample means? (Give your answer correct to two decimal places.)

(c) What shape would you expect the distribution of all these sample means to have? skewed right approximately normal skewed left uniform

You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

Let $\bar{X}$ denotes the mean of all sample means.

a) The mean of all these sample means

$E(\bar{X}) = E[\frac{1}{41}\sum_{i=1}^{41}X_i] = \frac{1}{41}\sum_{i=1}^{41}E(X_i) = \frac{1}{41}\sum_{i=1}^{41}500 = \frac{1}{41}*41*500 = 500$

b) The variance of all these sample means

$Var(\bar{X}) = Var[\frac{1}{41}\sum_{i=1}^{41}X_i] = \frac{1}{41^2}\sum_{i=1}^{41}Var(X_i) = \frac{1}{41^2}\sum_{i=1}^{41}23^2$

$\Rightarrow Var(\bar{X}) = \frac{1}{41^2}*41*23^2$

$\Rightarrow Var(\bar{X}) = \frac{23^2}{41}$

The standard deviation of all these sample means

$sd(\bar{X}) = \sqrt{Var(\bar{X}) }= \sqrt{\frac{23^2}{41}} = \frac{23}{\sqrt{41}} = 3.592$

c) Since sample size is large enough ( > 30), so the distribution of all these sample means is expected to have approximately normal distribution.