What is standardising?

For a normally distributed X with mean μ and standard deviation σ, the standardised value is

What is direction of the continuity correction?

P(X k) uses k + 0.5 in the numerator (extend the interval by half on the upper side). P(X k) uses k - 0.5 (extend the interval by half on the lower side).

What is approximation when conditions fail?

If n p or n(1 - p) is less than 5, the normal approximation is poor.

§-Syllabus dot point

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 Opus 4.812 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Watching the bell curve appear, stage by stage

The approximation is not a coincidence: as the number of trials grows, the binomial bars settle into the shape of a bell curve with the same mean and standard deviation. These panels overlay the matching normal $N(np,\, np(1 - p))$ on the bars of $B(n, 0.5)$ for growing $n$ .

Stage 1, a small $n$ is already close. With $n = 10$ the bars are wide and chunky, yet the bell curve $N(5, 2.5)$ already passes through their tops. The fit is rough in the tails but the centre is good.

Stage 2, more trials, narrower bars. At $n = 30$ (mean $15$ , standard deviation $2.74$ ) there are more, thinner bars and they hug the curve much more closely. This is the central limit theorem taking hold: summing more Bernoulli trials pulls the shape toward the normal.

Stage 3, a close fit you can compute with. By $n = 50$ (mean $25$ , standard deviation $3.54$ ) the bars and the curve are almost indistinguishable, so the area under the normal curve is a reliable stand-in for a sum of binomial probabilities. This is exactly when replacing a long pmf sum with one or two $z$ -lookups is justified.

When the approximation works

The approximation is good when:

$n$ is large.
$p$ 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. The diagram below zooms in on the bars near $k = 55$ : the bar for $X = 55$ stretches across the continuous interval $[54.5, 55.5]$ , so $P(X \le 55)$ keeps the whole of that bar and the boundary the normal curve should use is $55.5$ , not $55$ .

The direction is the part to get right: for an upper bound ("at most $k$ ", " $\le k$ ") push the boundary out to $k + 0.5$ ; for a lower bound ("at least $k$ ", " $\ge k$ ") push it out to $k - 0.5$ . Either way you are growing the interval to include the whole of the boundary bar.

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:

$n$ 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.

How exam questions ask about the normal approximation

These questions are heavily signposted, so the marks are in following the standard sequence without skipping a step:

"Using the normal approximation, estimate $P(\ldots)$ ": the explicit instruction to approximate. Quote $\mu = np$ and $\sigma = \sqrt{np(1 - p)}$ , apply the continuity correction, standardise, and look up the $z$ -value.
"Explain why the normal approximation is appropriate / valid": check and state both conditions, $np \ge 5$ and $n(1 - p) \ge 5$ . This is usually a 1 mark justification.
"A coin is tossed $100$ times ... probability of at least $60$ heads": a large- $n$ "at least" or "at most" probability that would need a huge pmf sum, the classic cue to approximate. Use the continuity-corrected boundary ( $59.5$ here for $X \ge 60$ ).
"State the mean and standard deviation of [the count]": just $np$ and $\sqrt{np(1 - p)}$ ; sometimes the lead-in to a later approximation part.
A given table value such as "use $P(Z \le 1.9) \approx 0.9713$ ": a strong hint about which $z$ -score you should be standardising to; if your $z$ does not match, recheck the continuity correction and the arithmetic.

Exam technique

Lay the answer out in fixed order: state $\mu = np$ and $\sigma = \sqrt{np(1 - p)}$ (note the square root, $\sigma$ not $\sigma^2$ ), confirm $np \ge 5$ and $n(1 - p) \ge 5$ , apply the continuity correction, then standardise and look up. Write the corrected boundary explicitly, $k + 0.5$ for an upper bound and $k - 0.5$ for a lower bound, because the wrong direction is the most common error and is easy for a marker to spot. Keep $z$ to two decimal places to match standard tables, and convert tail areas with $P(Z \ge z) = 1 - P(Z \le z)$ and symmetry $P(Z \le -z) = 1 - P(Z \le z)$ as needed. If the conditions fail (some $np < 5$ ), say the approximation is not valid and use the exact binomial instead, that observation is itself worth a mark.

Worked example

Approximate a cumulative probability with a continuity correction

For $X \sim B(100, 0.5)$ , approximate $P(X \le 55)$ . Both $np = 50 \ge 5$ and $n(1-p) = 50 \ge 5$ , so the normal approximation is valid. Compute the parameters of the approximating normal:

\mu = np = 50, \qquad \sigma^2 = np(1-p) = 25, \qquad \sigma = 5.

Because $X$ is discrete, the event $X \le 55$ corresponds to the continuous interval $X \le 55.5$ (the continuity correction extends the boundary by $0.5$ upward). Standardise and look up:

P(X \le 55) \approx P\!\left(Z \le \frac{55.5 - 50}{5}\right) = P(Z \le 1.1) \approx 0.8643.

Approximate a range probability using two z-scores

For $X \sim B(50, 0.4)$ , approximate $P(15 \le X \le 25)$ . Compute the parameters first, then apply continuity corrections at both ends of the range: extend the lower boundary down by $0.5$ and the upper boundary up by $0.5$ .

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

P(15 \le X \le 25) \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).

By symmetry of the standard normal: $P(-1.587 \le Z \le 1.587) \approx 1 - 2 \cdot P(Z > 1.587) \approx 1 - 2 \cdot 0.0563 \approx 0.887$ .

Approximate an exact-value probability with a continuity-correction interval

For $X \sim B(40, 0.5)$ , approximate $P(X = 20)$ . A single integer value $X = 20$ in the continuous picture corresponds to the interval $[19.5, 20.5]$ , so apply continuity corrections at both ends.

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

P(X = 20) \approx P\!\left(\frac{19.5 - 20}{3.162} \le Z \le \frac{20.5 - 20}{3.162}\right) = P(-0.158 \le Z \le 0.158) \approx 0.126.

The exact binomial gives $\binom{40}{20}(0.5)^{40} \approx 0.125$ , confirming the approximation is close.

Check the rule-of-thumb conditions before approximating

For $X \sim B(20, 0.1)$ , check whether the normal approximation is valid:

np = 20 \cdot 0.1 = 2 < 5.

The condition $np \ge 5$ fails, so the distribution is too skewed for the normal to approximate it reliably. Use the exact binomial pmf instead.

Apply the approximation to an "at least" probability

For $X \sim B(80, 0.3)$ , approximate $P(X \ge 30)$ . Both $np = 24 \ge 5$ and $n(1-p) = 56 \ge 5$ , so proceed. For an "at least" probability, extend the lower boundary down by $0.5$ :

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

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.

Final answer: $P(X \ge 30) \approx 0.0901$ .

Common traps

Skipping the continuity correction: Without it, the approximation systematically misses the true probability by a half-unit's worth.
Direction of the continuity correction: $P(X \le k)$ uses $k + 0.5$ in the numerator (extend the interval by half on the upper side). $P(X \ge k)$ uses $k - 0.5$ (extend the interval by half on the lower side).
Using the variance where the standard deviation is needed: $Z = (X - \mu)/\sigma$ uses $\sigma$ , not $\sigma^2$ .
Approximation when conditions fail: If $n p$ or $n(1 - p)$ is less than $5$ , the normal approximation is poor.
Forgetting which $z$ -table convention: Some tables give $P(0 \le Z \le z)$ , others give $P(Z \le z)$ . Convert as needed.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

HSC-style4 marksA fair coin is tossed

100

times. Using the normal approximation with a continuity correction, estimate the probability of getting at least

60

heads. Use

P(Z \le 1.9) \approx 0.9713

Show worked answer →

Let $X \sim B(100, 0.5)$ . Then $\mu = n p = 50$ and $\sigma^2 = n p (1 - p) = 25$ , so $\sigma = 5$ .

Both $n p = 50 \ge 5$ and $n(1 - p) = 50 \ge 5$ , so the approximation is valid.

For $P(X \ge 60)$ apply the continuity correction with $60 - 0.5 = 59.5$ :

P(X \ge 60) \approx P\!\left( Z \ge \frac{59.5 - 50}{5} \right) = P(Z \ge 1.9).

$P(Z \ge 1.9) = 1 - P(Z \le 1.9) \approx 1 - 0.9713 = 0.0287$ .

HSC-style3 marksA biased die shows a six with probability

\frac{1}{6}

. It is rolled

180

times. State the mean and standard deviation of the number of sixes, and explain why the normal approximation is appropriate.

Show worked answer →

Let $X \sim B(180, \tfrac{1}{6})$ count the sixes.

Mean: $\mu = n p = 180 \times \tfrac{1}{6} = 30$ .

Variance: $\sigma^2 = n p (1 - p) = 180 \times \tfrac{1}{6} \times \tfrac{5}{6} = 25$ , so $\sigma = 5$ .

The approximation is appropriate because $n p = 30 \ge 5$ and $n(1 - p) = 150 \ge 5$ , so the binomial is well approximated by $N(30, 25)$ .