The normal approximation to the binomial

9 minAS Level5.5.3

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By the end of this lesson

By the end you can state when a binomial distribution may be approximated by a normal one, find the approximating mean and variance, apply a continuity correction, and solve probability problems with it.

When a binomial starts to look normal

Recall that if $X \sim \mathrm{B}(n,p)$ , then $X$ counts successes in $n$ independent trials, each with success probability $p$ . Its mean and variance are $\mathrm{E}(X)=np, \qquad \mathrm{Var}(X)=npq,$ where $q=1-p$ .

For small $n$ the probability bar chart can be lopsided (statisticians say skewed: one tail is much longer than the other). As $n$ grows, with $p$ not too close to 0 or 1, the bars become a symmetric bell. Then a normal curve with the same mean and variance gives good estimates of binomial probabilities.

Working with the normal is much quicker than adding many binomial terms, which is why the approximation is useful.

Definition

Conditions for the normal approximation

The normal distribution may be used to approximate $X \sim \mathrm{B}(n,p)$ when $n$ is large enough that both $np>5 \qquad \text{and} \qquad nq>5,$ where $q=1-p$ . If either product is 5 or less, the distribution is too skewed and the approximation should not be used. ( $np$ is the expected number of successes and $nq$ the expected number of failures, so both need to be reasonably big.)

Under these conditions, take $Y \sim \mathrm{N}(np,\ npq)$ as the approximating normal variable, so the standard deviation is $\sqrt{npq}$ .

Why a continuity correction is needed

A binomial variable takes whole-number values: 8, 9, 10. A normal variable is continuous. To bridge the gap, give each whole number a bar of width 1, centred on the number.

The bar for 9 runs from 8.5 to 9.5. So a binomial event such as " $X \le 9$ " includes the whole bar for 9 and stops at its right-hand edge, 9.5: $\mathrm{P}(X \le 9) \approx \mathrm{P}(Y<9.5).$

This adjustment of half a unit is the continuity correction. The direction matters: include every bar that the event includes, and stop at the outer edge of the last one. Strict inequalities exclude the bar at $r$ : " $X<9$ " means $X \le 8$ , so it stops at 8.5.

Binomial event	Normal approximation
$\mathrm{P}(X \le r)$	$\mathrm{P}(Y<r+0.5)$
$\mathrm{P}(X<r)$	$\mathrm{P}(Y<r-0.5)$
$\mathrm{P}(X \ge r)$	$\mathrm{P}(Y>r-0.5)$
$\mathrm{P}(X>r)$	$\mathrm{P}(Y>r+0.5)$
$\mathrm{P}(X=r)$	$\mathrm{P}(r-0.5<Y<r+0.5)$
$\mathrm{P}(a \le X \le b)$	$\mathrm{P}(a-0.5<Y<b+0.5)$

Worked example: A cumulative probability with a continuity correction

The random variable $X \sim \mathrm{B}(50,0.3)$ . Use a normal approximation to estimate $\mathrm{P}(X \le 18)$ .

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