Binomial to Normal Approximation

AS · S1 · 3 min

For large $n$ , binomial probabilities become tedious to add up, and the distribution's shape is close to a normal curve. The syllabus lets you replace $B(n, p)$ by a normal distribution with the same mean and variance, provided $np$ and $nq$ are both greater than $5$ , and provided you apply a continuity correction.

The approximation

Key result

If $X \sim B(n, p)$ with $np > 5$ and $n(1 - p) > 5$ , then approximately

X \sim N\big(np,\ np(1 - p)\big).

Because $X$ is discrete and the normal is continuous, each integer value $r$ is treated as the interval $r - 0.5$ to $r + 0.5$ (the continuity correction).

Binomial probability	Normal probability
$P(X = r)$	$P(r - 0.5 < Y < r + 0.5)$
$P(X \leq r)$	$P(Y < r + 0.5)$
$P(X < r)$	$P(Y < r - 0.5)$
$P(X \geq r)$	$P(Y > r - 0.5)$
$P(X > r)$	$P(Y > r + 0.5)$

Method

Check $np > 5$ and $nq > 5$ , and state the normal approximation with its mean and variance.
Rewrite the required probability with the continuity correction.
Standardise and use tables: $z = \dfrac{x - np}{\sqrt{npq}}$ .

At least a given number

A fair coin is tossed 100 times. Use a normal approximation to find the probability of at least 60 heads.

Solution

$X \sim B(100, 0.5)$ ; $np = 50 > 5$ , $nq = 50 > 5$ . $X \approx N(50, 25)$ .

$P(X \geq 60) \approx P(Y > 59.5) = P\left(Z > \dfrac{59.5 - 50}{5}\right) = P(Z > 1.9) = 1 - 0.9713 = 0.0287$ .

Between two values

$X \sim B(80, 0.3)$ . Find $P(20 \leq X \leq 30)$ approximately.

Solution

$np = 24$ , $npq = 16.8$ , $\sigma = 4.099$ . $X \approx N(24, 16.8)$ .

$P(19.5 < Y < 30.5) = P\left(\dfrac{19.5 - 24}{4.099} < Z < \dfrac{30.5 - 24}{4.099}\right) = P(-1.098 < Z < 1.586) = 0.9436 - (1 - 0.8639) = 0.808$ .

A single value

$X \sim B(150, 0.4)$ . Estimate $P(X = 60)$ .

Solution

$np = 60$ , $npq = 36$ , $\sigma = 6$ . $P(59.5 < Y < 60.5) = P(-0.083 < Z < 0.083) = 2(0.5331) - 1 = 0.0662$ .

Checking the conditions

Explain whether a normal approximation is appropriate for $B(40, 0.1)$ and for $B(400, 0.1)$ .

Solution

$B(40, 0.1)$ : $np = 4 < 5$ , not appropriate (the distribution is too skewed). $B(400, 0.1)$ : $np = 40$ , $nq = 360$ , both $> 5$ : appropriate, with $N(40, 36)$ .

Why it works

A binomial variable is a sum of $n$ independent $0/1$ variables, and sums of many independent variables are approximately normal. The condition $np > 5$ , $nq > 5$ ensures the distribution is not too lopsided for the symmetric normal curve to fit.

Watch out

The continuity correction is compulsory for full marks. $P(X \geq 60)$ becomes $P(Y > 59.5)$ , not $P(Y > 60)$ . Draw a number line if unsure which way the half goes: the region must include the integers you want.

Exam tip

Write the three ingredients: the check on $np$ and $nq$ , the approximating distribution $N(np, npq)$ , and the continuity-corrected inequality. Each typically carries a mark before any tables are used.

Practice

Question

$X \sim B(60, 0.25)$ . Approximate $P(X \leq 12)$ .
$X \sim B(200, 0.05)$ . Approximate $P(X > 15)$ .
A test has 50 true/false questions and a student guesses. Estimate the probability of scoring between 20 and 30 inclusive.
State why a normal approximation should not be used for $B(30, 0.9)$ .

Answers

$N(15, 11.25)$ ; $P(Y < 12.5) = P(Z < -0.745) = 0.228$ .
$N(10, 9.5)$ ; $P(Y > 15.5) = P(Z > 1.784) = 0.0372$ .
$N(25, 12.5)$ ; $P(19.5 < Y < 30.5) = P(-1.556 < Z < 1.556) = 0.880$ .
$nq = 3 < 5$ .

The approximation

Why it works

Practice

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