The Central Limit TheoremEdexcel A-Level Further Maths: Revision notes

The Central Limit Theorem (CLT) says that if a population has mean $\mu$ and variance $\sigma^2$, then for a large sample size $n$ the sample mean $\bar{X}$ is approximately normally distributed: $\bar{X}\approx\mathrm{N}\left(\mu,\frac{\sigma^2}{n}\right).$ This holds whatever the shape of the population distribution, so the population does not have to be normal. You can use it for populations that are Poisson, binomial, geometric or negative binomial. You do not need to prove it. The larger $n$ is, the better the approximation. A sample size of about $30$ or more is usually accepted as large.

To use the CLT you need $\mu$ and $\sigma^2$ for a single observation. Use the standard results:

• Poisson $\mathrm{Po}(\lambda)$: $\mu=\lambda$, $\sigma^2=\lambda$
• Binomial $\mathrm{B}(n,p)$: $\mu=np$, $\sigma^2=np(1-p)$
• Geometric $\mathrm{Geo}(p)$: $\mu=\frac1p$, $\sigma^2=\frac{1-p}{p^2}$
• Negative binomial $\mathrm{NB}(r,p)$: $\mu=\frac rp$, $\sigma^2=\frac{r(1-p)}{p^2}$ Be careful with the symbol $n$. For a binomial population, $n$ in $\mathrm{B}(n,p)$ is the number of trials in one observation, whereas the CLT sample size is the number of observations. For example, a box of $12$ pens gives $\mathrm{B}(12,0.25)$, so $\mu=3$ and $\sigma^2=2.25$, and a sample of $40$ boxes gives $\bar{X}\approx\mathrm{N}\left(3,\frac{2.25}{40}\right)$.

Follow these steps. First, find $\mu$ and $\sigma^2$ of one observation. Second, state $\bar{X}\approx\mathrm{N}(\mu,\sigma^2/n)$ and say the CLT is being used because $n$ is large. Third, standardise with $z=\frac{\bar{x}-\mu}{\sigma/\sqrt n}$. Fourth, read the probability from the normal distribution on your calculator. Worked example: $X\sim\mathrm{Po}(4)$, $n=50$. $\bar{X}\approx\mathrm{N}(4,0.08)$. $\mathrm{P}(\bar{X}>4.5)=\mathrm{P}\left(Z>\frac{0.5}{\sqrt{0.08}}\right)=\mathrm{P}(Z>1.768)=0.0385$. A sample mean varies much less than a single observation: its standard deviation is $\sqrt{0.08}=0.283$, compared with $2$ for one observation. For a mean within a given distance of $\mu$, use $\mathrm{P}(\mu-a<\bar{X}<\mu+a)=2\Phi\left(\frac{a}{\sigma/\sqrt n}\right)-1$.

You may be asked for the smallest $n$ so that a probability condition holds. Standardise with $n$ unknown and use the inverse normal. For $\mathrm{NB}(3,0.25)$ ($\mu=12$, $\sigma=6$), $\mathrm{P}(\bar{X}>13)<0.01$ needs $\frac{13-12}{6/\sqrt n}>2.326$, so $\sqrt n>13.96$ and $n>194.8$. The smallest whole number is $n=195$. Always round $n$ up to the next whole number. To justify using the CLT, state that the sample is large, so the sample mean is approximately normal even though the population is not. Remember the approximation is poor for small $n$ when the population is very skewed, such as a geometric distribution with a small $p$, and that the observations must be independent.