Central limit theorem

Theorem taught in Functions of random variables and limit theorems, part of Probability

Central limit theorem

The central limit theorem at work. From Functions of random variables and limit theorems.
Type Theorem
Taught in Functions of random variables and limit theorems
Part Probability
Worked examples 4
Questions using it 4

If the X_i are independent and identically distributed with mean \mu and variance \sigma^2\in(0,\infty), then for every x

\lim_{n\to\infty}\mathbb{P}\Bigl(\frac{S_n-n\mu}{\sigma\sqrt{n}}\le x\Bigr)=\Phi(x),

where \Phi is the standard normal distribution function.

Examples

Normal approximation to a binomial

Worked example from Functions of random variables and limit theorems

Let X\sim\operatorname{Bin}(20,1/2).

Question. Approximate \mathbb{P}(X\le8) by the central limit theorem, with and without a continuity correction, and compare with the exact value.

Solution. Here np=10 and np(1-p)=5. The exact probability is 0.2517. The plain approximation gives

\mathbb{P}(X\le8)\approx\Phi\Bigl(\frac{8-10}{\sqrt5}\Bigr)=\Phi(-0.894)=0.186,

which is poor. X is integer-valued, so \mathbb{P}(X\le8)=\mathbb{P}(X\le8.5), and the normal curve should be read at the midpoint between 8 and 9:

\mathbb{P}(X\le8)\approx\Phi\Bigl(\frac{8.5-10}{\sqrt5}\Bigr)=\Phi(-0.671)=0.251,

almost exact. A usual rule of thumb asks for both np and n(1-p) to be above 5.

A normal approximation in the tail

Question page: A normal approximation in the tail

Let X\sim\operatorname{Bin}(100,1/2). Approximate \mathbb{P}(X\ge 60) by a normal probability. Use \Phi(2)\approx 0.97725.

The approximating normal has mean 50 and variance 25.

\mathbb{P}(Y\ge 60)=\mathbb{P}\bigl(Z\ge (60-50)/5\bigr)=\mathbb{P}(Z\ge 2)=1-\Phi(2)\approx 0.0228.

A sieve for blueberries

Question page: A sieve for blueberries

Blueberry diameters are \operatorname{N}(1, 0.1^2), in centimetres. A sieve of diameter d classifies a berry as large when its diameter exceeds d.

  1. Choose d so that 30% of berries are large. Use the standard normal quantile z_{0.7}\approx 0.524, that is the point with \Phi(z_{0.7})=0.7.

  2. With that choice of d, approximate the probability that fewer than 280 of 1000 independent berries are large.

  1. A berry is large with probability \mathbb{P}(D>d)=1-\Phi((d-1)/0.1). Setting this to 0.3 requires (d-1)/0.1=z_{0.7}=0.524, so d=1.0524 cm.

  2. The number of large berries is N\sim\operatorname{Bin}(1000,0.3), with mean 300 and variance 210. By the central limit theorem with a continuity correction,

\mathbb{P}(N<280)=\mathbb{P}(N\le279)\approx\Phi\Bigl(\frac{279.5-300}{\sqrt{210}}\Bigr)=\Phi(-1.41)=0.079.

The exact binomial value is 0.078.

A lift and a weight limit

Question page: A lift and a weight limit

A lift carries at most 650 kg. A person’s weight is \operatorname{N}(75, 10^2), and weights are independent. Let Z_n be the total weight of n people.

  1. Find \mathbb{P}(Z_8\ge 650). Use \Phi(1.77)\approx 0.962.

  2. Find the largest n such that \mathbb{P}(Z_n\ge 650)\le 0.01.

  1. Z_8\sim\operatorname{N}(600, 800).

\mathbb{P}(Z_8\ge 650)=1-\Phi\bigl((650-600)/\sqrt{800}\bigr)=1-\Phi(1.768)\approx 1-0.962=0.038.

  1. Z_n\sim\operatorname{N}(75n, 100n), so \mathbb{P}(Z_n\ge 650)=1-\Phi\bigl((650-75n)/(10\sqrt{n})\bigr). At n=8 the probability is about 0.038. At n=7 the standardised point is (650-525)/(10\sqrt{7})\approx 4.72, and 1-\Phi(4.72) is far below 0.01. The largest such n is 7.

Figures

The distribution function of \operatorname{Bin}(20,1/2), a staircase, against the \operatorname{N}(10,5) distribution function. The normal curve passes through the middle of each step, which is why the continuity correction reads it half a unit to the right. From Functions of random variables and limit theorems.

Used in

See also

Cumulative distribution function · Transformation rule · Multivariate normal distribution · Affine images of normal vectors · Normal linear model · Law of large numbers

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