Model a sample mean

Sampling distributions: exact normal and CLT cases

Start by identifying the random quantity: one observation, the sample mean, or the total. Then decide whether normality is exact or justified as an approximation.

A sampling distribution describes the values of a statistic across repeated samples of the same size. It is not the distribution of the observations within one sample. The teacher suggests a spreadsheet illustration; the comparison figure below shows the same change in shape and spread mathematically.

For iid $X_i\sim N(\mu,\sigma^2)$, the sample mean is exactly $\bar X\sim N(\mu,\sigma^2/n)$ and the total is exactly $T=\sum X_i\sim N(n\mu,n\sigma^2)$ for any positive integer $n$.

Normality is preserved under independent linear combinations; no large-sample argument is needed.

For iid observations with finite variance $0<\sigma^2<\infty$, the central limit theorem states that

$\frac{\bar X-\mu}{\sigma/\sqrt n}\xrightarrow{d}N(0,1)\quad\text{as }n\to\infty.$

For a sufficiently large sample, write $\bar X\approx N(\mu,\sigma^2/n)$ by CLT, or equivalently $T\approx N(n\mu,n\sigma^2)$.

This is a limiting distribution result used for approximate probabilities and inference, as the source’s “Limits” callout explains.

The source suggests $n\ge30$ as a working guideline; it is not a universal guarantee for highly skewed or heavy-tailed distributions. The quality depends on the population and on the event being approximated, especially a tail.

If $X$ is discrete, $\bar X$ remains discrete even when a continuous normal approximation is useful.

Increasing $n$ changes the sampling distribution, not the individual-observation distribution.

For small $n$ and an unknown population distribution, mean and variance alone do not determine an exact normal probability. State and justify any extra normality assumption rather than treating it as a fact; a near-normal population may support an approximation.

Larger $n$ gives asymptotic convergence, not a promise that every finite-sample tail error decreases monotonically.

Decide which distribution you are describing

Compare two sets of independent observations: one from a normal population and one from a strongly right-skewed population with finite variance. Both populations have mean 40 and variance 100. For each population, describe what can be said about sample means based on 4 observations and on 100 observations. Distinguish a distribution’s moments from a justified normal model.

After your attempt: compare the models

In both cases $E(\bar X)=40$. The variances are 25 for $n=4$ and 1 for $n=100$.

For the normal population, the means are exactly $N(40,25)$ and $N(40,1)$.

For the skewed population, four observations do not justify assuming a normal mean.

At 100 observations a CLT approximation may be appropriate, but adequate sample size depends on the skewness and the probability being estimated.

The individual observations remain skewed; repeated sample means are the quantities whose distribution changes.

Reference: the retained discrete-sampling comparison

For the lecture’s plants, $X\sim N(21,90)$ and $n=10$, so $\bar X\sim N(21,9)$ exactly and $P(18<\bar X<27)=\Phi(2)-\Phi(-1)\approx0.819$. In contrast, sixty iid $B(9,0.5)$ counts have $\bar X\approx N(4.5,0.0375)$ by CLT; the source’s upper tail .00491 is approximate, while the exact binomial total gives .004301767. The worked examples below retain both methods and their qualifications.

This comparison retains the lecture’s discrete sampling example and its exact-versus-approximate qualification. For normal-probability calculations in the main worked example, identify whether the quantity is an individual observation or a sample mean.