For a population with mean $\mu$ and standard deviation $\sigma$, when drawing samples of size $n$:

1. Mean of the sampling distribution: $\mu_{\bar{x}} = \mu$ The sample mean is an unbiased estimator of the population mean.

2. Standard error (standard deviation of $\bar{x}$): $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ Larger samples produce less variability in $\bar{x}$.

3. Shape of the sampling distribution:

   • If the population is Normal, then $\bar{x}$ is exactly Normal for any $n$.
   • If the population is not Normal, then $\bar{x}$ is approximately Normal for large (Central Limit Theorem, typically ).

Central Limit Theorem (CLT)

The Central Limit Theorem states: No matter the shape of the population distribution, the sampling distribution of $\bar{x}$ approaches a Normal distribution as $n$ increases.

$\bar{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Practical implications:

• For $n \geq 30$, even if the population is skewed or multimodal, $\bar{x}$ is approximately Normal.
• Smaller samples may suffice if the population is already approximately Normal.
• The approximation improves as $n$ increases.

The Standard Error and Sample Size

The standard error decreases with the square root of $n$:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \propto \frac{1}{\sqrt{n}}$

This means:

• To cut the standard error in half, you must increase $n$ by a factor of 4.
• Larger samples yield more precise estimates of $\mu$.

Worked Example 1: Normal Population

A population has $\mu = 100$ and $\sigma = 15$ (Normal). Draw samples of size $n = 25$. What is the sampling distribution of $\bar{x}$?

Solution:

• $\mu_{\bar{x}} = \mu = 100$
• $\sigma_{\bar{x}} = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3$
• Shape: Normal (population is Normal)

So $\bar{x} \sim N(100, 3)$.

Find $P(\bar{x} > 103)$: $Z = \frac{103 - 100}{3} = 1$ $P(Z > 1) \approx 0.1587$

About 15.87% of sample means exceed 103.

Worked Example 2: Non-Normal Population and CLT

A population is heavily skewed with $\mu = 50$ and $\sigma = 20$. Draw samples of size $n = 100$. Describe the sampling distribution of $\bar{x}$.

Solution:

• Population is skewed, but $n = 100 \geq 30$ is large.
• By CLT, $\bar{x}$ is approximately Normal.
• $\mu_{\bar{x}} = 50$
• $\sigma_{\bar{x}} = \frac{20}{\sqrt{100}} = \frac{20}{10} = 2$

So $\bar{x} \sim N(50, 2)$ approximately.

Find $P(49 < \bar{x} < 51)$: $Z_{low} = \frac{49 - 50}{2} = -0.5; \quad Z_{high} = \frac{51 - 50}{2} = 0.5$ $P(-0.5 < Z < 0.5) \approx 0.3830$

About 38.3% of sample means fall between 49 and 51.

Conditions and Assumptions

Common Pitfalls

⚠️ Confusing $\sigma$ and $\sigma_{\bar{x}}$: The population SD is $\sigma$; the standard error is $\sigma_{\bar{x}} = \sigma/\sqrt{n}$. They are very different! Standard error decreases as $n$ increases; population SD does not.

⚠️ Forgetting the Square Root of $n$: When computing standard error, always divide $\sigma$ by $\sqrt{n}$, not by $n$.

⚠️ CLT Threshold: While $n \geq 30$ is a rule of thumb, it's not a hard cutoff. If the population is already Normal, CLT applies for any $n$. If the population is very skewed, you may need $n > 30$.

Calculator Tip

💡 TI-84 / TI-Nspire: Use normalcdf() to find probabilities for $\bar{x}$. For $P(\bar{x} < 103)$ with $\mu_{\bar{x}} = 100$ and $\sigma_{\bar{x}} = 3$, enter: normalcdf(-999999, 103, 100, 3).

Standardize: $Z = \frac{190 - 200}{6} = \frac{-10}{6} \approx -1.67$

Find probability: $P(Z < -1.67) \approx 0.0475$

About 4.75% of sample means are below 190.

Find $P(56 < \bar{x} < 64)$: $Z_{low} = \frac{56 - 60}{2} = -2; \quad Z_{high} = \frac{64 - 60}{2} = 2$

Probability: $P(-2 < Z < 2) \approx 0.9545$

About 95.45% of sample means fall within $\pm 4$ units (or within $\pm 2$ standard errors).