Sampling Distributions

Distribution of \bar{X}:

• The sampling distribution we’ll use for this class.

Why \bar{X}:

• Sampling is inherently random.
• \bar{X} is a R.V. with useful properties.

Key Facts

If X_1 , ... , X_n is a random sample:

1: Estimation

\boxed{ \begin{aligned} \mu_{\bar{X}} &= \mu \\~\\ \sigma_{\bar{X}} &= \frac{\sigma}{\sqrt{n}} \end{aligned} } \\~\\ \small\textit{if $X_i$s have mean $\mu$ and SD $\sigma$}

2: Normal Distribution

\boxed{ \begin{aligned} \bar{X} \sim N (\mu,\frac{\sigma}{\sqrt{n}}) \end{aligned} } \\~\\ \small\textit{if $X_i$s are from $N(\mu,\sigma)$}

3: Central Limit Theorem

\boxed{ \begin{aligned} \bar{X} \sim N (\mu,\frac{\sigma}{\sqrt{n}}) \end{aligned} } \\~\\ \small\textit{if $X_i$s have sufficient $n$}

Example

Q: It’s known that the rainfall has a distribution with a mean of 0.53 and SD of 0.32 on winter days (inches). We take 50 winter days at random and compute the sample mean of 50 observations. What is the probability that the sample mean is less than 0.5 in?

A:

We want P(\bar{X} < 0.5).

\bar{X} Mean: 0.53

\bar{X} SD: \frac{0.32}{\sqrt{50}} = 0.0453

We were able to derive \bar{X}’s mean and SD from key fact #1

n is big-enough to apply CLT (central limit theorem) (key fact #3).

• Thus, \bar{X} is approximately normal, which means we can transform it to standard normal (Ch. 5)

\begin{aligned} P(\bar{X} < 0.5) &= P(z < \frac{0.5 - 0.53}{0.453} \\ &= P(z < -0.66) \\ &= 0.2546 \end{aligned}