The Central Limit Theorem

Take one measurement and it could land almost anywhere in the population. Take the mean of $40$ measurements and something remarkable happens: whatever shape the population has, the sample mean behaves like a normal variable, and it is much less spread out than a single value. This is the central limit theorem, and it’s the reason the normal distribution appears all over statistics, from quality control to opinion polls to confidence intervals.

Key ideas

Combining independent normal variables

From linear combinations of random variables, you already know how to find the mean and variance of a combination. When the variables are normal and independent, there’s a bonus: the combination is normal too.

If $X_1, X_2, \dots, X_n$ are independent and each $X_i \sim N(\mu_i, \sigma_i^2)$ , then

a_1X_1 \pm a_2X_2 \pm \dots \pm a_nX_n \sim N\big(a_1\mu_1 \pm a_2\mu_2 \pm \dots \pm a_n\mu_n,\ \ a_1^2\sigma_1^2 + a_2^2\sigma_2^2 + \dots + a_n^2\sigma_n^2\big)

So to find a probability about a sum or difference, you:

Define the new variable (for example $D = A - B$ or $T = X_1 + \dots + X_{12}$ ).
Find its mean and variance with the usual rules.
Use your GDC’s normal cdf with that mean and standard deviation (the square root of the variance).

A common trick: to find $P(A \gt B)$ , rewrite it as $P(A - B \gt 0)$ .

The distribution of the sample mean

Take a random sample of size $n$ from a population and work out its mean. Different samples give different means, so the sample mean $\bar{X}$ is itself a random variable. Its distribution is called the sampling distribution of the mean.

If the population is normal, $X \sim N(\mu, \sigma^2)$ , then

\bar{X} \sim N\left(\mu,\ \frac{\sigma^2}{n}\right)

This is exact, for any sample size. The centre stays at $\mu$ , but the variance is divided by $n$ . The standard deviation of $\bar{X}$ is $\dfrac{\sigma}{\sqrt{n}}$ (often called the standard error of the mean). Bigger samples give means that cluster more tightly around $\mu$ .

The total of the sample, $X_1 + X_2 + \dots + X_n$ , is normal too, with mean $n\mu$ and variance $n\sigma^2$ .

The central limit theorem

What if the population isn’t normal? The central limit theorem says:

For a random sample of size $n$ from any population with mean $\mu$ and variance $\sigma^2$ , the distribution of $\bar{X}$ approaches $N\left(\mu, \dfrac{\sigma^2}{n}\right)$ as $n$ gets large.

How large is “large” depends on the population: for a roughly symmetric population, quite small samples are fine, while for a very skewed one you need more. In IB examinations, $n \gt 30$ is considered sufficient. The same applies to the sample total, which is approximately $N(n\mu, n\sigma^2)$ .

The figure shows this for a skewed population (waiting times with mean $4$ minutes and variance $8$ ). The mean of $5$ values is less skewed; the mean of $30$ values is almost a perfect bell curve, centred at $4$ and much narrower.

A skewed population (mean

4

, variance

8

) and the exact distributions of

\bar{X}

for

n = 5

and

n = 30

. As

n

grows,

\bar{X}

becomes normal and narrower.

Which result do I use?

Population	Sample size	Distribution of the sample mean
Normal	any $n$	exactly $N\left(\mu, \dfrac{\sigma^2}{n}\right)$
Not normal (or unknown)	$n \gt 30$	approximately $N\left(\mu, \dfrac{\sigma^2}{n}\right)$ by the CLT
Not normal (or unknown)	small $n$	can’t assume normal

Worked examples

Example 1: A carton of eggs

The mass of an egg is $N(60, 4^2)$ grams, and the mass of an empty carton is $N(30, 2^2)$ grams, all independent. A carton holds $12$ eggs. Find the probability that a full carton has a mass of more than $770$ g.

Solution. Let $T = C + E_1 + E_2 + \dots + E_{12}$ (twelve different eggs, so a sum, not $12E$ ).

E(T) = 30 + 12(60) = 750, \qquad \mathrm{Var}(T) = 2^2 + 12(4^2) = 4 + 192 = 196

A sum of independent normal variables is normal, so $T \sim N(750, 196)$ , with standard deviation $\sqrt{196} = 14$ .

GDC normal cdf with lower bound $770$ , upper bound a very large number, $\mu = 750$ , $\sigma = 14$ :

P(T \gt 770) = 0.0766 \text{ (3 s.f.)}

Example 2: Who jumps further?

In the long jump, Aisha’s distances are $A \sim N(5.2, 0.3^2)$ metres and Bea’s are $B \sim N(5.0, 0.4^2)$ metres, independently. Find the probability that Bea jumps further than Aisha on a given attempt.

Solution. $P(B \gt A) = P(A - B \lt 0)$ . Let $D = A - B$ .

E(D) = 5.2 - 5.0 = 0.2, \qquad \mathrm{Var}(D) = 0.3^2 + 0.4^2 = 0.25

So $D \sim N(0.2, 0.25)$ , with standard deviation $0.5$ .

P(D \lt 0) = 0.345 \text{ (3 s.f.)}

Bea wins about a third of the time, even though Aisha’s mean is higher. (Check: $P(D \gt 0) = 0.655$ , and the two add to $1$ .)

Example 3: One bottle or the mean of nine?

The volume of juice in a bottle is $X \sim N(250, 12^2)$ ml.

(a) Find the probability that one bottle contains less than $245$ ml.
(b) A random sample of $9$ bottles is taken. Find the probability that their mean volume is less than $245$ ml.

Solution.

(a) $P(X \lt 245) = 0.338$ (3 s.f.), using $\mu = 250$ , $\sigma = 12$ .

(b) The population is normal, so $\bar{X} \sim N\left(250, \dfrac{12^2}{9}\right) = N(250, 16)$ , with standard deviation $\dfrac{12}{\sqrt{9}} = 4$ .

P(\bar{X} \lt 245) = 0.106 \text{ (3 s.f.)}

A single bottle is quite likely to be $5$ ml short, but the average of nine bottles being $5$ ml short is much less likely, because the short and full bottles balance out.

Example 4: Using the CLT for a skewed population

The daily screen time of teenagers in a city is skewed to the right, with mean $3.2$ hours and standard deviation $1.5$ hours. A random sample of $40$ teenagers is taken.

(a) Find the probability that the sample mean is more than $3.5$ hours.
(b) Find the probability that the total screen time of the $40$ teenagers is more than $135$ hours.

Solution.

(a) The population isn’t normal, but $n = 40 \gt 30$ , so by the central limit theorem

\bar{X} \approx N\left(3.2,\ \frac{1.5^2}{40}\right), \qquad \text{standard deviation } \frac{1.5}{\sqrt{40}} = 0.23717\ldots

P(\bar{X} \gt 3.5) \approx 0.103 \text{ (3 s.f.)}

(b) The total $S = X_1 + \dots + X_{40}$ is approximately normal with mean $40(3.2) = 128$ and variance $40(1.5^2) = 90$ , so its standard deviation is $\sqrt{90} = 9.4868\ldots$

P(S \gt 135) \approx 0.230 \text{ (3 s.f.)}

Check: $S \gt 135$ is the same as $\bar{X} \gt \dfrac{135}{40} = 3.375$ , and the normal cdf for $\bar{X}$ above $3.375$ gives the same $0.230$ .

Common mistakes

Using σ instead of σ/√n for the sample mean. Questions about one value use $\sigma$ . Questions about a mean of $n$ values use $\dfrac{\sigma}{\sqrt{n}}$ . Underline the word “mean” when you see it.

Typing the variance into the normal cdf. Most GDCs ask for the standard deviation. If $\bar{X} \sim N(250, 16)$ , enter $\sigma = 4$ , not $16$ .

Writing nX for a total. The total of $12$ eggs is $E_1 + \dots + E_{12}$ , with variance $12\sigma^2$ . Using $12E$ gives variance $144\sigma^2$ , which is far too big.

Subtracting variances for a difference. $\mathrm{Var}(A - B) = \mathrm{Var}(A) + \mathrm{Var}(B)$ , always (for independent variables).

Using the CLT for small samples. With a skewed population and $n = 8$ , you can’t assume $\bar{X}$ is normal. The CLT needs a large sample ( $n \gt 30$ in exams), unless the population itself is normal.

Thinking the CLT makes the population normal. The CLT is about the distribution of the sample mean (or total). The individual values keep whatever shape the population has.

Practice

1. (Warm-up) $X \sim N(80, 36)$ . A random sample of $16$ values is taken. State the distribution of $\bar{X}$ , including its standard deviation.

Solution

\bar{X} \sim N\left(80, \frac{36}{16}\right) = N(80, 2.25)

The standard deviation is $\sqrt{2.25} = 1.5$ (or $\dfrac{6}{\sqrt{16}} = 1.5$ ).

2. (Warm-up) In each case, can you assume that $\bar{X}$ is (at least approximately) normal? Give a reason.

(a) A sample of $5$ from a normal population.
(b) A sample of $50$ from a skewed population.
(c) A sample of $8$ from a skewed population.

Solution

(a) Yes, exactly normal: the mean of a sample from a normal population is normal for any $n$ .

(b) Yes, approximately normal by the central limit theorem, since $n = 50 \gt 30$ .

3. (Warm-up) $X \sim N(20, 9)$ and $Y \sim N(15, 16)$ are independent. State the distributions of:

(a) $X + Y$
(b) $X - Y$
(c) $2X - Y$

Solution

(a) $X + Y \sim N(35, 25)$

(b) $X - Y \sim N(5, 25)$

4. (Core) The masses of adults using an elevator are normally distributed with mean $75$ kg and standard deviation $12$ kg. The elevator’s safe load is $1000$ kg. Find the probability that $12$ randomly chosen adults have a total mass of more than $1000$ kg.

Solution

$T = X_1 + \dots + X_{12}$ has mean $12(75) = 900$ and variance $12(12^2) = 1728$ , so $T \sim N(900, 1728)$ with standard deviation $\sqrt{1728} = 41.569\ldots$

P(T \gt 1000) = 0.00807 \text{ (3 s.f.)}

5. (Core) A coffee machine pours $F \sim N(250, 8^2)$ ml into a cup. The capacity of a cup is $C \sim N(260, 3^2)$ ml, independently. Find the probability that a cup overflows.

Solution

The cup overflows when $F \gt C$ , that is, $F - C \gt 0$ .

E(F - C) = 250 - 260 = -10, \qquad \mathrm{Var}(F - C) = 64 + 9 = 73

So $F - C \sim N(-10, 73)$ , with standard deviation $\sqrt{73} = 8.544\ldots$

P(F - C \gt 0) = 0.121 \text{ (3 s.f.)}

6. (Core) A fair six-sided die has mean score $3.5$ and variance $\dfrac{35}{12}$ . It is rolled $50$ times. Find the approximate probability that the mean score is between $3.3$ and $3.7$ .

Solution

The scores are not normal, but $n = 50 \gt 30$ , so by the CLT

\bar{X} \approx N\left(3.5,\ \frac{35/12}{50}\right), \qquad \text{standard deviation } \sqrt{\frac{35}{600}} = 0.24152\ldots

P(3.3 \lt \bar{X} \lt 3.7) \approx 0.592 \text{ (3 s.f.)}

7. (Core) The masses of chocolate bars are normally distributed with standard deviation $2.5$ g. A sample of $n$ bars is taken. Find the smallest $n$ for which the probability that the sample mean is within $0.5$ g of the population mean is at least $0.95$ .

Solution

$\bar{X} \sim N\left(\mu, \dfrac{2.5^2}{n}\right)$ . We need $P(|\bar{X} - \mu| \lt 0.5) \ge 0.95$ . By symmetry, $0.5$ must be at least $1.95996\ldots$ standard deviations (the inverse normal of $0.975$ ):

1.95996\ldots \times \frac{2.5}{\sqrt{n}} \le 0.5 \quad\Rightarrow\quad \sqrt{n} \ge 9.7998\ldots \quad\Rightarrow\quad n \ge 96.04\ldots

The smallest sample size is $n = 97$ .

8. (Challenge) The lengths of wooden planks are $L \sim N(2.4, 0.02^2)$ metres, independently.

(a) Three planks are laid end to end. Find the probability that their total length is more than $7.25$ m.
(b) Find the probability that the total length of two planks is more than $0.05$ m longer than twice the length of a third plank.

Solution

(a) The total has mean $3(2.4) = 7.2$ and variance $3(0.02^2) = 0.0012$ , so standard deviation $\sqrt{0.0012} = 0.034641\ldots$

P(\text{total} \gt 7.25) = 0.0745 \text{ (3 s.f.)}

(b) Let $D = L_1 + L_2 - 2L_3$ . Then $E(D) = 2.4 + 2.4 - 2(2.4) = 0$ and

\mathrm{Var}(D) = 0.02^2 + 0.02^2 + 2^2(0.02^2) = 6(0.0004) = 0.0024

So $D \sim N(0, 0.0024)$ , with standard deviation $0.048989\ldots$

P(D \gt 0.05) = 0.154 \text{ (3 s.f.)}

Notice that $2L_3$ is one plank doubled, so its variance is $4(0.02^2)$ , while $L_1 + L_2$ is two different planks.

9. (Challenge) The amount spent by a customer at a hardware store has mean $42 and standard deviation $18, and the distribution is skewed. On one day, $64$ customers visit.

(a) Find the probability that the total amount spent is more than $2900.
(b) Find the amount, to the nearest dollar, that the total exceeds with probability $0.9$ .

Solution

(a) $n = 64 \gt 30$ , so by the CLT the total $S$ is approximately normal, with mean $64 \times 42 = 2688$ dollars and standard deviation $18\sqrt{64} = 144$ dollars.

P(S \gt 2900) \approx 0.0705 \text{ (3 s.f.)}

(b) We need $k$ with $P(S \gt k) = 0.9$ , so $P(S \lt k) = 0.1$ . Inverse normal with area $0.1$ , $\mu = 2688$ , $\sigma = 144$ :

k = 2503.45\ldots

The total exceeds about $2503 with probability $0.9$ .