H2 Math Sampling and the Central Limit Theorem

JC2 · H2 Mathematics · Study guide

Work with sampling distributions, CLT probabilities, unbiased estimates, coded and frequency data, combined samples and required sample sizes.

Concepts · worked methods · graphs10 min read · practice linked below

Is the question about one observation, a total or a mean?

A sample mean $\overline X$ is random before the sample is drawn. The observed mean $\bar x$ is a number calculated from the data. For independent observations with common mean $\mu$ and variance $\sigma^2$ , a total has variance $n\sigma^2$ ; a mean has variance $\sigma^2/n$ . The population variance itself does not decrease when more data are collected.

The normal population case is exact

Four independent packet masses

A packet mass X has distribution $N(100,10^2)$ g. Four packets are chosen independently. Find the probability their mean mass exceeds 105 g, and express the same event in terms of their total T.

\begin{aligned}T&=X_1+X_2+X_3+X_4,\\T&\sim N(400,400),\\\overline X&=T/4,\\\overline X&\sim N(100,25).\end{aligned}

\begin{aligned}z&=\frac{105-100}{5}=1,\\P(\overline X>105)&=P(Z>1)\\&\approx0.159.\end{aligned}

Equivalently $\overline X\gt 105$ means $T\gt 420$ . Standardising T gives $(420-400)/20=1$ , the same result. Normality is exact for any sample size here because the individual masses are normal.

Use the CLT when individual observations are not normal

Average heads over fifty games

In one game, the number of fair coins tossed is 1, 2 or 3 with probabilities $1/2,1/4,1/4$ . Let X count heads. Its distribution is $P(X=0,1,2,3)=(11,15,5,1)/32$ . Independent games are played. Estimate the probability that the average over 50 games exceeds one head.

Compute the population moments from the game model, not from a sample estimate:

\begin{aligned}E(X)&=\frac{15+2(5)+3(1)}{32}\\&=\frac78,\\E(X^2)&=\frac{15+4(5)+9(1)}{32}\\&=\frac{11}{8},\\\operatorname{Var}(X)&=\frac{11}{8}-\left(\frac78\right)^2\\&=\frac{39}{64}.\end{aligned}

The sample is large enough for a CLT approximation in this bounded, independent game model:

\begin{aligned}\overline X&\sim N\left(\frac78,\frac{39}{3200}\right).\end{aligned}

This distribution is approximate. Standardise using its standard deviation:

\begin{aligned}P(\overline X>1)&\approx P\left(Z>\frac{1-7/8}{\sqrt{39/3200}}\right)\\&\approx0.129.\end{aligned}

The theorem approximates the distribution of the mean; it does not change X into a normal variable. Independence, a common underlying distribution and finite variance matter. A fixed rule such as “n is at least 20” is not a universal guarantee.

The same method applies to money

A mean net gain in repeated games

A game gives net gain Y of $10, $0 or −$2 with probabilities $1/8,1/40,17/20$ . Estimate the probability that the total net gain from 50 independent games is at most $50.

\begin{aligned}E(Y)&=10\left(\frac18\right)-2\left(\frac{17}{20}\right)\\&=-0.45,\\E(Y^2)&=100\left(\frac18\right)+4\left(\frac{17}{20}\right)\\&=15.9,\\\operatorname{Var}(Y)&=15.9-(-0.45)^2\\&=15.6975.\end{aligned}

A total of at most $50 is a mean of at most $1. By the CLT, approximately,

\begin{aligned}\overline Y&\sim N\left(-0.45,\frac{15.6975}{50}\right),\\P(\overline Y\leq1)&\approx\Phi\left(\frac{1+0.45}{\sqrt{15.6975/50}}\right)\\&\approx0.995.\end{aligned}

The high probability of staying below $50 does not make the game attractive: its expected net gain is negative. State the money conclusion using the mean net gain, rather than just a chance of reaching a selected threshold.

Unbiased estimates from coded data

Data are given as deviations from fifty

A random sample of 40 lengths x cm has $\sum(x-50)=80$ and $\sum(x-50)^2=1200$ . Find unbiased estimates of the population mean and variance.

Set $y=x-50$ . Add 50 back to the mean; a translation leaves variance unchanged:

\begin{aligned}\bar y&=\frac{80}{40}=2,\\\bar x&=2+50=52\text{ cm}.\end{aligned}

\begin{aligned}s_x^2&=s_y^2\\&=\frac1{39}\left[1200-\frac{80^2}{40}\right]\\&=\frac{1040}{39}\\&=26\frac23\text{ cm}^2.\end{aligned}

Do not treat $\sum(x-50)^2$ as $\sum(x-\bar x)^2$ : the chosen centre 50 differs from the sample mean 52. The correction $(\sum y)^2/n$ removes this difference.

An estimator is unbiased when its expectation over repeated random samples equals the population parameter. It need not equal that parameter in a particular sample.

A frequency is a count of observations

Use all ten observations, not five row labels

A sample of lengths has values 8, 9, 10, 11 and 12 cm, with frequencies 1, 2, 4, 2 and 1 respectively. Find the unbiased estimates of the mean and variance.

Length (cm)	8	9	10	11	12
Frequency	1	2	4	2	1

\begin{aligned}n&=1+2+4+2+1=10,\\\bar x&=\frac{8+18+40+22+12}{10}\\&=10\text{ cm}.\end{aligned}

The data are centred conveniently at their mean:

\begin{aligned}\sum f(x-10)^2&=1(4)+2(1)+4(0)\\&\quad+2(1)+1(4)\\&=12,\\s^2&=\frac{12}{10-1}\\&=\frac43\text{ cm}^2.\end{aligned}

A calculator standard deviation with denominator n gives $\sqrt{12/10}$ . For the unbiased variance estimate, square the standard deviation calculated with denominator n − 1, or multiply the denominator-n variance by $n/(n-1)$ . Check the calculator’s convention rather than relying on the label alone.

Combine raw totals, not sample variances

Two groups with different means

One sample has $n_1=5,\bar x_1=10,s_1^2=4$ ; another has $n_2=7,\bar x_2=14,s_2^2=9$ . Here both $s_i^2$ are unbiased variance estimates. Find the mean and unbiased variance estimate of the combined sample.

Recover each sum of squares by rearranging the variance formula:

\begin{aligned}\sum x_1&=5(10)=50,\\\sum x_1^2&=(5-1)4+5(10)^2\\&=516,\\\sum x_2&=7(14)=98,\\\sum x_2^2&=(7-1)9+7(14)^2\\&=1426.\end{aligned}

Now combine the sufficient raw totals:

\begin{aligned}n&=5+7=12,\\\sum x&=50+98=148,\\\sum x^2&=516+1426=1942,\\\bar x&=\frac{148}{12}=\frac{37}{3},\\s^2&=\frac1{11}\left[1942-\frac{148^2}{12}\right]\\&=\frac{350}{33}\\&\approx10.606.\end{aligned}

A weighted average of 4 and 9 misses variation between the group means. The combined sample is more spread out because its two groups have different centres.

Choose n for a required precision

Within two seconds with probability at least 0.95

Independent filling times have distribution $N(50,12^2)$ seconds. Find the least n such that the sample mean is within two seconds of 50 with probability at least 0.95.

\begin{aligned}\overline X&\sim N\left(50,\frac{144}{n}\right),\\P(|\overline X-50|<2)&\geq0.95.\end{aligned}

The central 95% of a standard normal lies between $\pm1.95996398$ . Therefore,

\begin{aligned}\frac{2}{12/\sqrt n}&\geq1.95996398,\\\sqrt n&\geq6(1.95996398),\\n&\geq138.2925\ldots.\end{aligned}

Thus the least n is 139. At n = 138 the probability is about 0.9498; at n = 139 it is about 0.9506. This is a central-area requirement: using the one-sided 0.95 quantile 1.645 would choose too small a sample.

How n changes the same two-second requirement

Sample size n:144

n = 144: standard error 1.000 s; |Z| < 2.000; probability 0.9545.

The graph uses a fixed Z scale. As n grows, the same ±2-second band reaches farther in standard-error units.

Variation: a one-sided requirement

For the same filling times, require $P(\overline X\lt 53)\geq0.90$ . Use the one-sided cutoff $\Phi^{-1}(0.9)=1.2815516$ :

\begin{aligned}\frac{3\sqrt n}{12}&\geq1.2815516,\\n&\geq26.278,\\n_{\min}&=27.\end{aligned}

The chosen side and area decide the quantile. Both examples are exact because the individual filling times are normal.

Before you finish

Keep $\sigma^2$ , $s^2$ , and the variance of the mean separate. For coded data undo the shift in the mean; for frequency data count observations; for combined samples combine sums and sums of squares. For an approximation, say why it is appropriate and name the sampling distribution being approximated.