<p>Suppose ${{X}{1}},{{X}{2}},...,{{X}_{n}}$ is a random sample of size $n$ taken from a population with mean $\mu$ and variance ${{\sigma }^{2}}$. If  $\bar{X}$ is the sample mean, show that $\mathrm{E}\left( {\bar{X}} \right)=\mu $ and $\mathrm{Var}\left( {\bar{X}} \right)=\frac{{{\sigma }^{2}}}{n}$.</p>

Suppose ${{X}{1}},{{X}{2}},...,{{X}_{n}}$ is a random sample of size $n$ taken from a population with mean $\mu$ and variance ${{\sigma }^{2}}$. If  $\bar{X}$ is the sample mean, show that $\mathrm{E}\left( {\bar{X}} \right)=\mu $ and $\mathrm{Var}\left( {\bar{X}} \right)=\frac{{{\sigma }^{2}}}{n}$.

Sampling

H1 Math

Worked Example 1

**Basics of Sampling**

Note: The population is NOT necessarily normal.

$\bar{X} = \displaystyle\frac{X_1 + X_2 + X_3 + ... + X_n}{n} = \frac{1}{n}X_1 + \frac{1}{n}X_2 + ... + \frac{1}{n}X_n$

**Show $E(\bar{X}) = \mu$:**

$E(\bar{X}) = \displaystyle\frac{1}{n}E(X_1) + \frac{1}{n}E(X_2) + ... + \frac{1}{n}E(X_n)$

$= n \times \displaystyle\frac{1}{n}E(X)$

$= E(X) = \mu$

**Show $\text{Var}(\bar{X}) = \displaystyle\frac{\sigma^2}{n}$:**

$\text{Var}(\bar{X}) = \displaystyle\frac{1}{n^2}\text{Var}(X_1) + \frac{1}{n^2}\text{Var}(X_2) + ... + \frac{1}{n^2}\text{Var}(X_n)$

$= n \times \displaystyle\frac{1}{n^2}\text{Var}(X)$

$= \displaystyle\frac{1}{n}\text{Var}(X) = \frac{\sigma^2}{n}$

Worked Example 1

Worked Example 1

Video Solution: