Normal Approximation to the Binomial

This lesson covers the normal approximation to the binomial distribution as required by the Edexcel A-Level Mathematics specification (9MA0), Paper 3 Section A -- Statistics. You need to understand when and why this approximation is used, how to apply it with a continuity correction, and how to calculate probabilities.

Why Approximate?

When n is large, calculating binomial probabilities involves large factorials and summing many terms. The normal distribution provides a simpler approximation.

When to Use the Normal Approximation

The approximation is appropriate when:

n is large
np > 5 and n(1 - p) > 5

These conditions ensure the binomial is roughly symmetric and bell-shaped.

Exam Tip: Always check np > 5 and n(1 - p) > 5 and state your conclusion.

Setting Up the Approximation

If X ~ B(n, p) and conditions are met:

X is approximately N(mu, sigma²) where:

mu = np
sigma² = np(1 - p)
sigma = sqrt(np(1 - p))

Example

X ~ B(100, 0.4): np = 40 > 5, n(1-p) = 60 > 5. So X approximately N(40, 24).

The Continuity Correction

The binomial is discrete, the normal is continuous. We adjust by +/- 0.5:

Binomial probability	Normal approximation
P(X = r)	P(r - 0.5 < Y < r + 0.5)
P(X ≤ r)	P(Y < r + 0.5)
P(X < r)	P(Y < r - 0.5)
P(X ≥ r)	P(Y > r - 0.5)
P(X > r)	P(Y > r + 0.5)

Think of each integer r as occupying the interval (r - 0.5, r + 0.5).

Exam Tip: The continuity correction is essential. Forgetting it is a very common error.

Worked Example 1

X ~ B(80, 0.3). Find P(X ≤ 20).

mu = 24, sigma² = 16.8, sigma = 4.099.

P(X ≤ 20) --> P(Y < 20.5)

Z = (20.5 - 24)/4.099 = -0.854

P(Z < -0.854) = 1 - 0.803 = 0.197

Worked Example 2

X ~ B(200, 0.45). Find P(X ≥ 100).

mu = 90, sigma = 7.036.

P(X ≥ 100) --> P(Y > 99.5)

Z = (99.5 - 90)/7.036 = 1.350

P(Z > 1.350) = 1 - 0.9115 = 0.0885

Worked Example 3: Exact Value

X ~ B(50, 0.5). Find P(X = 25).

mu = 25, sigma = 3.536.

P(X = 25) --> P(24.5 < Y < 25.5)

Z1 = -0.141, Z2 = 0.141

P = 0.5557 - 0.4443 = 0.111 (exact binomial gives 0.1123 -- good approximation).

Common Errors to Avoid

Forgetting the continuity correction.
Using the wrong direction for the correction.
Confusing sigma and sigma² in N(mu, sigma²).
Not checking conditions (np > 5 and n(1-p) > 5).
Not stating that the approximation and continuity correction are being used.

Summary

When X ~ B(n, p) with np > 5 and n(1-p) > 5, approximate by N(np, np(1-p)).
Apply continuity correction: ≤ r becomes < r + 0.5; ≥ r becomes > r - 0.5; etc.
Standardise and use normal tables.
Always check conditions and state the continuity correction for full marks.

A-Level Deep Dive: Normal Approximation to the Binomial

Spec mapping

Edexcel 9MA0-03 specification, Statistics section 4 — Statistical distributions, sub-strand 4.4 and Section 7 — The normal distribution, sub-strand 7.5 covers select an appropriate probability distribution for a context, with appropriate reasoning, including recognising when the binomial or normal model may not be appropriate (refer to the official specification document for exact wording). The normal approximation to the binomial is examinable on 9MA0-03 Paper 3 (Statistics and Mechanics), sitting at the join between the binomial distribution (Year 1) and the normal distribution (Year 2). It is also synoptic with Section 5 (Statistical hypothesis testing) because larger $n$ tests are routinely conducted using the normal approximation. The Edexcel formula booklet lists the normal distribution but does not state the approximation rule — students must recall $X \sim B(n,p) \approx N(np, np(1-p))$ and the continuity correction $\pm 0.5$ from memory.

Worked example with full mark scheme

Question (8 marks):

The random variable $X \sim B(100, 0.4)$ .

(a) Explain why $X$ may be approximated by a normal distribution, stating the parameters of the approximating distribution. (3)

(b) Use a normal approximation, with continuity correction, to estimate $P(X \leq 45)$ . (4)

Solution with mark scheme:

(a) Step 1 — justify the approximation.

Since $n = 100$ is large and $p = 0.4$ is not close to $0$ or $1$ , both $np = 40$ and $n(1-p) = 60$ exceed $10$ , so the binomial distribution is approximately symmetric and the normal approximation is appropriate.

B1 — stating that $n$ is large with $p$ not extreme, or equivalently that $np > 5$ and $n(1-p) > 5$ . Reasons in words score; numerical values without justification do not.

Step 2 — state the parameters.

$\mu = np = 100 \times 0.4 = 40$ and $\sigma^2 = np(1-p) = 100 \times 0.4 \times 0.6 = 24$ .

B1 — correct mean.

B1 — correct variance $24$ (or standard deviation $\sigma = \sqrt{24} \approx 4.899$ ). So $Y \sim N(40, 24)$ .

(b) Step 1 — apply the continuity correction.

$P(X \leq 45) \approx P(Y < 45.5)$ , where $Y \sim N(40, 24)$ .

M1 — correct continuity correction in the right direction (upper bound $+0.5$ because $X \leq 45$ is inclusive on the upper end).

Step 2 — standardise.

$Z = \dfrac{45.5 - 40}{\sqrt{24}} = \dfrac{5.5}{4.8990} \approx 1.1227$

M1 — correct standardisation using $\mu = 40$ and $\sigma = \sqrt{24}$ .

Step 3 — read tables / use calculator.

$P(Z < 1.1227) \approx 0.8692$ .

A1 — correct probability value to at least 3 significant figures.

A1 — final stated probability $P(X \leq 45) \approx 0.869$ in context.

The binomial distribution is discrete whereas the normal distribution is continuous, so any conversion via continuity correction is inherently an approximation; agreement is best when $n$ is large.

B1 — any one valid reason (discreteness vs continuity; finite $n$ ; tail behaviour differs slightly).

Total: 8 marks (B3 M2 A2 B1).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A factory produces components, of which historically $30\%$ are classified as premium grade. A random sample of $200$ components is taken; let $X$ denote the number that are premium grade.

(a) Write down the exact distribution of $X$ and justify the use of a normal approximation. (2)

(b) Using a suitable approximation with continuity correction, estimate $P(55 \leq X \leq 70)$ . (4)

Mark scheme decomposition by AO:

(a)

B1 (AO1.2) — $X \sim B(200, 0.3)$ stated.
B1 (AO2.4) — justification: $n = 200$ large, $np = 60 > 10$ and $n(1-p) = 140 > 10$ , so normal approximation appropriate; approximating distribution $Y \sim N(60, 42)$ .

(b)

M1 (AO1.1b) — correct continuity correction: $P(55 \leq X \leq 70) \approx P(54.5 < Y < 70.5)$ .
M1 (AO1.1b) — correct standardisation: lower $z = (54.5 - 60)/\sqrt{42} \approx -0.8487$ , upper $z = (70.5 - 60)/\sqrt{42} \approx 1.6202$ .
A1 (AO1.1b) — $\Phi(1.6202) - \Phi(-0.8487) \approx 0.9474 - 0.1981 = 0.7493$ .
A1 (AO2.5) — final answer $P(55 \leq X \leq 70) \approx 0.749$ in context with appropriate accuracy.

Total: 6 marks split AO1 = 4, AO2 = 2. This is balanced between procedural fluency (standardisation) and reasoning (interpreting the inclusive interval and correcting both endpoints in the right direction).

Synoptic links

Connects to:

Section 4 — The binomial distribution: the approximation begins from $X \sim B(n,p)$ . Confidence with the binomial pmf and its mean $np$ and variance $np(1-p)$ is prerequisite. Without these formulas memorised, the approximating normal cannot be parameterised.
Section 7 — The normal distribution: the entire machinery of standardisation $Z = (Y - \mu)/\sigma$ and looking up $\Phi(z)$ is reused unchanged. The approximation is essentially a parameter mapping from $(n,p)$ to $(\mu, \sigma^2)$ ; everything afterwards is normal-distribution practice.
Central limit theorem (informal at A-Level, formal at university): the binomial is a sum of $n$ independent Bernoulli trials, so the CLT says the standardised sum tends to a standard normal. The approximation is the CLT applied at a fixed (large) $n$ rather than as a limit.
Section 5 — Statistical hypothesis testing: for tests on a binomial proportion with large $n$ , the test statistic is constructed using the normal approximation. Critical regions and $p$ -values rely on the same standardisation step used here.
Modelling cycle (AO3): choosing the approximation is a modelling decision — students must justify why the binomial is acceptable, why the normal approximation is appropriate, and what the continuity correction compensates for. This is precisely the AO3 reasoning Edexcel rewards.

Mark-scheme literacy

Normal-approximation questions on 9MA0-03 spread AO marks more evenly than pure procedural questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Computing $\mu = np$ , $\sigma^2 = np(1-p)$ , applying standardisation, reading normal-table values
AO2 (reasoning / interpretation)	25–35%	Justifying use of approximation, choosing correct continuity correction direction, interpreting result in context
AO3 (problem-solving / modelling)	10–20%	Selecting between exact binomial and normal approximation, stating limitations, contextual conclusions

Examiner-rewarded phrasing: "since $n$ is large and $p$ is not close to $0$ or $1$ "; "applying a continuity correction because the binomial is discrete and the normal is continuous"; "let $Y \sim N(np, np(1-p))$ ". Phrases that lose marks: writing " $X \sim N(np, np(1-p))$ " — the original variable is still binomial, only the approximating variable is normal; omitting the continuity correction "to save time"; using $\sigma^2 = np$ (confusing with Poisson).

A specific Edexcel pattern to watch: when the question says "use a suitable approximation", students must state which approximation is being used. Silently swapping in a normal model and computing a probability scores the procedural marks but loses the AO2 marks for justification.

Grade-band model answers

3-mark question

Question: $X \sim B(80, 0.5)$ . Use a normal approximation with continuity correction to estimate $P(X \geq 45)$ .

Grade C response (~190 words):

$X \sim B(80, 0.5)$ so $\mu = 80 \times 0.5 = 40$ and $\sigma^2 = 80 \times 0.5 \times 0.5 = 20$ .

So $Y \sim N(40, 20)$ .

$P(X \geq 45) \approx P(Y > 44.5)$ .

$z = (44.5 - 40)/\sqrt{20} = 4.5/4.472 \approx 1.006$ .

$P(Z > 1.006) = 1 - 0.8427 = 0.1573$ .

So $P(X \geq 45) \approx 0.157$ .

Examiner commentary: Full marks (3/3). The candidate identifies the parameters correctly, applies the continuity correction in the right direction ( $X \geq 45$ becomes $Y > 44.5$ because $45$ is included in the original event so we shift down to $44.5$ ), and standardises accurately. The final probability is given to three significant figures, matching examiner expectations. Working is brief but every step is verifiable. A typical Grade C answer for a procedural question — efficient and correct.

Grade A response (~240 words):*

The exact distribution is $X \sim B(80, 0.5)$ . Since $n = 80$ is large and $p = 0.5$ , both $np = 40$ and $n(1-p) = 40$ comfortably exceed $10$ , so the normal approximation is appropriate.

Approximating: $Y \sim N(\mu, \sigma^2)$ with $\mu = np = 40$ and $\sigma^2 = np(1-p) = 20$ , giving $\sigma = \sqrt{20} = 2\sqrt{5}$ .

Apply the continuity correction. Because the event $X \geq 45$ includes the value $45$ , the corresponding continuous event is $Y > 44.5$ (we shift the boundary down by $0.5$ to capture the discrete probability mass at $X = 45$ ):

$P(X \geq 45) \approx P(Y > 44.5)$

Standardise:

$Z = \dfrac{44.5 - 40}{2\sqrt{5}} = \dfrac{4.5}{2\sqrt{5}} \approx 1.0062$

Normal Approximation to the Binomial

Normal Approximation to the Binomial

Why Approximate?

When to Use the Normal Approximation

Setting Up the Approximation

Example

The Continuity Correction

Worked Example 1

Worked Example 2

Worked Example 3: Exact Value

Common Errors to Avoid

Summary

A-Level Deep Dive: Normal Approximation to the Binomial

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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