Question

> Find the mean of the population

> Find the mean of the sampling distribution of the

sample means

> find the Variance of the population

> find the Variance of the sampling distribution of the

sample mean

> find the Standard deviation of the population

> find the Standard deviation of the sampling

distribution of the sample mean

STUDENT

A

B

C

D

E

HEIGHT (IN CM)

120

130

110

125

159

Answer 1

The mean of the population:

$\mu=\sum x_i\cdot P(x_i)=\\ =120\cdot\cfrac{1}{5}+130\cdot\cfrac{1}{5}+110\cdot\cfrac{1}{5}+125\cdot\cfrac{1}{5}+\\ +159\cdot\cfrac{1}{5}=128.8.$

The mean of the sampling distribution of sample means:

$\mu_{\bar x} =\mu=128.8.$

The variance of the population:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\{120-128.8, 130-128.8, 110-128.8, \\ 125-128.8,159-128.8\}=\\ =\{-8.8,1.2,-18.8,-3.8,30.2\},$

$\sigma^2=(-8.8)^2\cdot \cfrac{1}{5}+1.2^2\cdot \cfrac{1}{5}+(-18.8)^2\cdot \cfrac{1}{5}+\\ +(-3.8)^2\cdot \cfrac{1}{5}+30.2^2\cdot \cfrac{1}{5}=271.76.$

The variance of the sampling distribution of the

sample mean (the size of samples is unknown; let, for example, n = 3):

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{271.76}{3}=90.59.$

The standard deviation of the population:

$\sigma=\sqrt{271.76}=16.49.$

The Standard deviation of the sampling distribution of the sample mean:

$\sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{16.49}{\sqrt 3}=9.52.$