NSWMaths Extension 1Syllabus dot point

When and how do we use the normal distribution to approximate binomial probabilities?

Use the normal approximation $X \sim N(n p, n p (1 - p))$ to approximate binomial probabilities for large $n$

A focused answer to the HSC Maths Extension 1 dot point on the normal approximation of the binomial. The rule of thumb $n p \ge 5$ and $n(1 - p) \ge 5$ , continuity correction, standardising and computing approximate probabilities, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to recognise when the binomial distribution can be approximated by a normal distribution, write down the approximating normal $N(n p, n p (1 - p))$ , apply the continuity correction, and compute approximate probabilities using z-scores.

The answer

The result

If $X \sim B(n, p)$ with $n$ "large" (rule of thumb: both $n p \ge 5$ and $n (1 - p) \ge 5$ ), then

X \approx N(\mu, \sigma^2) \quad \text{with} \quad \mu = n p, \quad \sigma^2 = n p (1 - p).

This is a consequence of the central limit theorem: the binomial is a sum of $n$ independent identically distributed Bernoulli trials, and sums of many i.i.d. random variables tend to a normal distribution.

When the approximation works

The approximation is good when:

IMATH_13 is large.
IMATH_14 is not too close to $0$ or $1$ (which makes the distribution very skewed).

The HSC rule of thumb is $n p \ge 5$ and $n (1 - p) \ge 5$ . With $n = 100$ , this works for $0.05 \le p \le 0.95$ . With $n = 20$ , it works for roughly $0.25 \le p \le 0.75$ .

When $p$ is very small and $n$ is large, the Poisson approximation is more appropriate, but that is beyond HSC Extension 1.

Continuity correction

The binomial is discrete; the normal is continuous. To improve the approximation, adjust the boundary by $\pm 0.5$ .

For $X \sim B(n, p)$ and $k$ an integer,

P(X \le k) \approx P\!\left( Z \le \frac{k + 0.5 - \mu}{\sigma} \right),

P(X \ge k) \approx P\!\left( Z \ge \frac{k - 0.5 - \mu}{\sigma} \right),

P(X = k) \approx P\!\left( \frac{k - 0.5 - \mu}{\sigma} \le Z \le \frac{k + 0.5 - \mu}{\sigma} \right).

The half-unit shift accounts for the fact that the binomial $X = k$ corresponds to the interval $[k - 0.5, k + 0.5]$ in the continuous picture.

Standardising

For a normally distributed $X$ with mean $\mu$ and standard deviation $\sigma$ , the standardised value is

Z = \frac{X - \mu}{\sigma}.

Look up the probability in a standard normal table (or use $\arcsin$ -style estimates for HSC if a table is not provided).

When to use the approximation

For HSC Extension 1, use it when:

IMATH_34 is large enough that summing pmf values is tedious.
The question asks for a numerical answer (not exact).
The question explicitly says "using the normal approximation".

Otherwise, compute the binomial directly.

Worked example

Standard approximation

$X \sim B(100, 0.5)$ . Approximate $P(X \le 55)$ .

$\mu = 50$ , $\sigma^2 = 25$ , $\sigma = 5$ .

Continuity correction: $P(X \le 55) \approx P(Z \le \frac{55.5 - 50}{5}) = P(Z \le 1.1) \approx 0.8643$ .

Range probability

$X \sim B(50, 0.4)$ . Approximate $P(15 \le X \le 25)$ .

$\mu = 20$ , $\sigma^2 = 12$ , $\sigma \approx 3.464$ .

Continuity correction: $P(14.5 \le X \le 25.5) \approx P\!\left( \frac{14.5 - 20}{3.464} \le Z \le \frac{25.5 - 20}{3.464} \right) = P(-1.587 \le Z \le 1.587)$ .

$\approx 1 - 2 \cdot P(Z > 1.587) \approx 1 - 2 \cdot 0.0563 \approx 0.887$ .

Probability of an exact value

$X \sim B(40, 0.5)$ . Approximate $P(X = 20)$ .

$\mu = 20$ , $\sigma \approx 3.162$ .

$P(X = 20) \approx P(19.5 \le X \le 20.5) \approx P(-0.158 \le Z \le 0.158) \approx 0.126$ .

(Exact binomial: $\binom{40}{20} (0.5)^{40} \approx 0.125$ . Close.)

Check the conditions

$X \sim B(20, 0.1)$ . Can we use the normal approximation?

$n p = 2 < 5$ . The condition fails. The approximation is not reliable here.

(Use the exact binomial pmf instead.)

At least

$X \sim B(80, 0.3)$ . Approximate $P(X \ge 30)$ .

$\mu = 24$ , $\sigma^2 = 16.8$ , $\sigma \approx 4.099$ .

Continuity correction: $P(X \ge 30) \approx P\!\left( Z \ge \frac{29.5 - 24}{4.099} \right) = P(Z \ge 1.342) \approx 1 - 0.9099 = 0.0901$ .