Central Limit Theorem Simulator

There are two panels, both on a 0 to 10 scale. The top panel (coral) is the population you sample from, with a dashed line and a label at its mean μ. The bottom panel (teal) is the sampling distribution of x̄: every sample you draw adds one sample mean to it. A navy Normal curve with mean μ and standard deviation σ/√n is drawn over the bars, and a dashed coral line marks the average of the sample means so far.

The readout lists samples drawn, population μ, mean of x̄ (observed), SD of x̄ (observed) and σ/√n (predicted SE), all to three decimal places.

Changing the shape or moving the n slider clears every sample drawn so far, so set n first and then draw. The population's μ and σ are estimated from 40,000 random values each time you choose a shape, so the third decimal can shift slightly from one visit to the next. Draw at least 1000 samples before comparing the observed values with the predictions.

For random samples of size n from a population with mean $\mu$ and standard deviation $\sigma$:

Keep three distributions apart. The population never changes, one sample's data look roughly like the population, and only the sampling distribution of x̄ (the bottom panel) narrows and becomes bell-shaped as n grows.

Problem: Waiting times at a coffee counter are strongly skewed to the right, with mean $\mu = 2$ minutes and standard deviation $\sigma = 2$ minutes. A random sample of 30 customers is taken. Find the probability that their mean wait is more than 2.5 minutes.

Step 1: Check the conditions. The sample is random. Thirty customers is less than 10% of all customers, so the observations are close to independent. The population is skewed, but $n = 30$ is large enough for the Central Limit Theorem to say $\bar{x}$ is approximately Normal.

Step 2: Describe the sampling distribution. $\mu_{\bar{x}} = 2$ and $\sigma_{\bar{x}} = 2/\sqrt{30} = 0.365$ minutes.

Step 3: Standardize. $z = \dfrac{2.5 - 2}{0.365} = 1.37$, so $P(\bar{x} > 2.5) = P(Z > 1.37) \approx 0.085$.

Interpretation: about 8.5% of random samples of 30 customers would have a mean wait above 2.5 minutes. A Normal model would not work for one customer's wait, because individual times are skewed.

Check it in the simulator: the Right-skewed population is close to this one. It is exponential with mean about 2, with values capped at 10, so the readout shows μ ≈ 1.99 and a slightly smaller σ of about 1.93. At n = 30 the predicted SE reads about 0.353. Press Draw 1000 twice: the observed SD of x̄ should land close to that, and the bars should follow the navy curve. Then set n = 2 and draw 1000 more. The predicted SE grows to about 1.366 and the pile of means is still clearly skewed right.

This is topic 4.1, Sampling Distributions for Sample Means, the opening topic of Unit 4 (Inference for Quantitative Data: Means). The mean and standard deviation of x̄ are on the formula sheet. You will use them for probability questions about a sample mean, and the CLT later justifies the Normal/Large Sample condition for every t procedure in the unit.