IB Mathematics Applications & Interpretation AHL 4.15—central Limit Theorem Notes

Why The Average Of Many Is Tamer Than Any One

1. A single measurement, such as one person's daily commute or one bottle's fill volume, can land almost anywhere the underlying variable allows.
2. Take the mean of a whole sample, though, and the extremes tend to cancel, so the average clusters far more tightly around the true centre.
3. This subtopic pins down exactly how tight that cluster is and what shape it takes.
4. Two results do the work: a linear combination of independent normal variables is itself normal, and the sample mean $\bar{X}$ has a predictable normal distribution.

Linear Combinations Of Independent Normals Stay Normal

1. For a general random variable, the AHL 4.14 rules tell you the mean and variance of a combination, but they say nothing about its shape.
2. The normal distribution is special: when you add or scale independent normal variables, the result is not just some distribution with a known mean and variance, it is exactly normal again.
3. So if $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$ are independent, then for constants $a$ and $b$ the combination is normal: $$aX + bY \sim N\!\left(a\mu_X + b\mu_Y,\; a^2\sigma_X^2 + b^2\sigma_Y^2\right)$$
4. The mean uses the linear-combination rule $E(aX + bY) = aE(X) + bE(Y)$, which holds whether or not the variables are independent.
5. The variance uses $\operatorname{Var}(aX + bY) = a^2\operatorname{Var}(X) + b^2\operatorname{Var}(Y)$, which needs independence and adds the terms even when the combination subtracts.
6. A minus sign vanishes because each coefficient enters the variance squared, so $\operatorname{Var}(aX - bY) = a^2\sigma_X^2 + b^2\sigma_Y^2$ is still a sum.

Where The Booklet Helps And Where It Does Not

1. The linear-combination formulas for $E$ and $\operatorname{Var}$ are printed in the AHL section of the formula booklet (Topic 4), so you may quote them directly before substituting.
2. The fact that the combination stays normal is a stated property, not a formula, so there is no booklet line for it; you simply know it and cite it in words.
3. Once you have the mean and variance, you have fully described the normal distribution, and every probability comes from the GDC.

The Distribution Of The Sample Mean $\bar{X}$

1. Suppose you draw a random sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$.
2. Model each observation as an independent copy $X_1, X_2, \ldots, X_n$ of the population variable, each with mean $\mu$ and variance $\sigma^2$.
3. The sample mean is the linear combination in which every coefficient is $\tfrac{1}{n}$: $$\bar{X} = \frac{1}{n}\big(X_1 + X_2 + \cdots + X_n\big)$$

Deriving The Mean Of $\bar{X}$

1. Apply the linear-combination rule for the mean, with each coefficient $\tfrac{1}{n}$: $$E(\bar{X}) = \frac{1}{n}\big(E(X_1) + \cdots + E(X_n)\big) = \frac{1}{n}(n\mu)$$ $$E(\bar{X}) = \mu$$
2. So on average the sample mean sits exactly on the population mean, which is why $\bar{X}$ is an unbiased estimator of $\mu$.

Deriving The Variance Of $\bar{X}$

1. The observations are independent, so their variances add, and each coefficient $\tfrac{1}{n}$ enters squared.
2. This gives $\operatorname{Var}(\bar{X}) = \left(\tfrac{1}{n}\right)^2\big(\operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)\big) = \tfrac{1}{n^2}(n\sigma^2)$: $$\operatorname{Var}(\bar{X}) = \frac{\sigma^2}{n}$$
3. Dividing by $n$ is the whole point: averaging shrinks the variance, so larger samples give means that cluster more tightly around $\mu$.
4. The standard deviation of $\bar{X}$, called the standard error of the mean, is therefore $\operatorname{SD}(\bar{X}) = \dfrac{\sigma}{\sqrt{n}}$.

Putting The Distribution Together

1. If the population itself is normal, then because $\bar{X}$ is a linear combination of independent normals it is exactly normal for any sample size. $$\bar{X} \sim N\!\left(\mu,\; \frac{\sigma^2}{n}\right)$$
2. There is no separate booklet box for the sample mean; its mean $\mu$ and variance $\dfrac{\sigma^2}{n}$ are derived from the AHL 4.14 linear-combination formulas, which is exactly the derivation shown above.

The Central Limit Theorem: Normality Without A Normal Population

1. The result for $\bar{X}$ so far assumed the population was normal, but most real populations are skewed, discrete, or otherwise not normal at all.
2. The central limit theorem rescues the situation: for a large enough sample size, the sample mean is approximately normal whatever the shape of the population.
3. So for large $n$, drawing from a population with mean $\mu$ and variance $\sigma^2$, $$\bar{X} \approx N\!\left(\mu,\; \frac{\sigma^2}{n}\right)$$
4. The mean and standard error are identical to before; only the word approximately changes, and it disappears when the population is already normal.

Which Result Applies When

1. If the population is normal, $\bar{X}$ is exactly normal for every $n$, even $n = 2$, and no theorem is needed.
2. If the population is not normal, you need the central limit theorem, and you may only use it once $n$ is large.
3. Either way the GDC calculation is the same: enter mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$, then read off the probability.