4.15 Central limit theoremIB Maths: Applications and Interpretation HL: Revision notes

If $X_1,\dots,X_n$ are independent normal random variables, any linear combination $a_1X_1+\dots+a_nX_n$ is also normal. For independent $X$ and $Y$: $\mathrm{E}(aX\pm bY)=a\mathrm{E}(X)\pm b\mathrm{E}(Y),\qquad\mathrm{Var}(aX\pm bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y).$ Variances add even when you subtract the variables. The total of $n$ independent copies of $X\sim N(\mu,\sigma^2)$ is $N(n\mu,\ n\sigma^2)$. Example: six adults of mass $N(78,11^2)$ have total mass $N(468,\ 726)$.

The sample mean is $\bar{X}=\frac{X_1+\dots+X_n}{n}$. If $X\sim N(\mu,\sigma^2)$ and the sample is random, then $\bar{X}\sim N\left(\mu,\ \frac{\sigma^2}{n}\right).$ The mean stays at $\mu$ but the variance shrinks by a factor $n$, so the standard deviation of $\bar{X}$ is $\frac{\sigma}{\sqrt{n}}$ (the standard error). Larger samples give sample means that cluster more tightly around $\mu$. For $\sigma=0.04$ and $n=9$ the standard deviation of $\bar{X}$ is $\frac{0.04}{3}=0.0133$. This result is exact when the population is normal, for any $n$.

The central limit theorem (CLT): whatever the shape of the population (with finite mean $\mu$ and variance $\sigma^2$), the distribution of $\bar{X}$ for a random sample of size $n$ approaches $N\left(\mu,\frac{\sigma^2}{n}\right)$ as $n$ becomes large. How large $n$ must be depends on the population: a nearly symmetric population needs a small $n$, a very skewed one needs a larger $n$. In examinations, $n>30$ is considered sufficient. The result is an approximation unless the population itself is normal. Online simulations are useful: take many samples from a skewed distribution, plot the histogram of the sample means, and watch it become bell-shaped as $n$ increases.

A call centre's waiting time $T$ has mean $6$ and standard deviation $4.5$ (skewed). For $n=50>30$, the CLT gives $\bar{T}\approx N(6,\ 0.405)$. With the GDC (normal CDF, mean $6$, standard deviation $\sqrt{0.405}=0.636$): $\mathrm{P}(\bar{T}>7)=0.0581.$ Always say why the normal model is valid: either the population is normal, or $n>30$ and the CLT applies. Finding a sample size: to make $\mathrm{P}(\bar{X}<\mu-3)<0.01$ with $\sigma=15$, use the inverse normal ($z=-2.326$) and solve $\frac{3\sqrt{n}}{15}>2.326$. This gives $n>135.3$, so $n=136$. Always round $n$ up.

• Population normal: $\bar{X}$ is exactly normal for any $n$, including small samples.
• Population not normal and $n>30$: $\bar{X}$ is approximately normal (CLT).
• Population not normal and $n\le30$: no normal model is justified in IB.
• A single value $X$ uses $\sigma$; a sample mean uses $\frac{\sigma}{\sqrt{n}}$.
• Totals use $N(n\mu,\ n\sigma^2)$; means use $N\left(\mu,\ \frac{\sigma^2}{n}\right)$.