The Normal Distribution

This lesson covers the normal distribution as required by the Edexcel A-Level Mathematics specification (9MA0), Paper 3 Section A -- Statistics. You need to understand the properties of the normal distribution, standardise using the Z formula, and use the standard normal distribution to find probabilities.

---

What is the Normal Distribution?

The normal distribution is a continuous probability distribution described by a symmetric, bell-shaped curve. It models many real-world variables such as heights, weights, IQ scores, and measurement errors.

Notation

X ~ N(mu, sigma²) where mu is the mean and sigma² is the variance.

Note: the second parameter is the variance, not the standard deviation.

Example

Heights of adult women: X ~ N(164, 49) (mean 164 cm, SD 7 cm, since 7² = 49).

---

Properties of the Normal Distribution

1. Bell-shaped and symmetrical about the mean mu.
2. Mean = median = mode (all equal to mu).
3. Asymptotic to the x-axis -- extends infinitely but never touches it.
4. Total area under the curve = 1.
5. Completely defined by mu (location) and sigma² (spread).

The 68-95-99.7 Rule

• 68% of values lie within 1 SD of the mean (mu - sigma to mu + sigma).
• 95% of values lie within 2 SDs (mu - 2sigma to mu + 2sigma).
• 99.7% of values lie within 3 SDs (mu - 3sigma to mu + 3sigma).

Effect of changing parameters

• Changing mu shifts the curve left or right without changing its shape.
• Increasing sigma makes the curve wider and flatter.
• Decreasing sigma makes the curve taller and narrower.

---

The Standard Normal Distribution

Z ~ N(0, 1) -- the special case with mu = 0 and sigma = 1.

Any normal distribution can be converted to this using standardisation.

---

Standardising (the Z-Score)

Z = (X - mu) / sigma

The z-score tells you how many standard deviations x is from the mean.

Example

X ~ N(50, 16), so mu = 50, sigma = 4.

P(X < 56): Z = (56 - 50)/4 = 1.5. P(Z < 1.5) = 0.9332.

---

Using the Standard Normal Table

The table gives P(Z < z) -- the area to the left of z.

• P(Z > a) = 1 - P(Z < a)
• P(a < Z < b) = P(Z < b) - P(Z < a)
• P(Z < -a) = 1 - P(Z < a) (by symmetry)

Examples

P(Z > 0.84) = 1 - 0.7995 = 0.2005

P(-1.2 < Z < 0.8) = 0.7881 - 0.1151 = 0.6730

---

Inverse Normal (Finding Values Given Probabilities)

1. Find z such that P(Z < z) = given probability (from table or calculator).
2. Convert back: x = mu + z x sigma

Example

X ~ N(100, 225), sigma = 15. Find x such that P(X < x) = 0.9.

z = 1.2816. x = 100 + 1.2816 x 15 = 119.2.

---

Finding mu and sigma

Set up equations using Z = (x - mu) / sigma and the given probabilities.

Example: Finding sigma

X ~ N(80, sigma²). P(X > 90) = 0.1 --> P(X < 90) = 0.9 --> z = 1.2816.

(90 - 80)/sigma = 1.2816 --> sigma = 10/1.2816 = 7.80.

Example: Finding mu

X ~ N(mu, 25), sigma = 5. P(X < 68) = 0.3 --> z = -0.5244.

(68 - mu)/5 = -0.5244 --> mu = 70.6.

Exam Tip: If the probability is > 0.5, z is positive (above the mean). If < 0.5, z is negative.

---

Summary

• X ~ N(mu, sigma²) is symmetric and bell-shaped with mean = median = mode = mu.
• 68-95-99.7 rule: 68% within 1 SD, 95% within 2 SD, 99.7% within 3 SD.
• Standardise: Z = (X - mu) / sigma. Z ~ N(0, 1).
• Inverse normal: find z from table, then x = mu + z x sigma.
• To find unknown mu or sigma, set up Z = (x - mu)/sigma and solve.

A-Level Deep Dive: The Normal Distribution

Spec mapping

Edexcel 9MA0-03 specification section 7 — Statistical distributions (continuous), Paper 3 Statistics covers the Normal distribution as a model; find probabilities using the Normal distribution; link to histograms, mean, standard deviation, points of inflection; 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). Although examined principally on Paper 3, the Normal distribution underpins section 8 (Statistical hypothesis testing — testing for the mean of a Normal distribution with known variance) and connects synoptically to section 5 (probability) and section 6 (statistical distributions, including the binomial). The Edexcel formula booklet provides the standardisation $Z = (X - \mu)/\sigma$Z=(X−μ)/σ but candidates are expected to use calculator distribution functions for direct computation of $P(X < x)$P(X<x) and inverse normal lookups.

---

Worked example with full mark scheme

Question (8 marks):

The random variable $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2). It is given that $P(X < 50) = 0.2$P(X<50)=0.2 and $P(X < 70) = 0.9$P(X<70)=0.9.

(a) Find the values of $\mu$μ and $\sigma$σ. (5)

(b) Hence find $P(45 < X < 65)$P(45<X<65). (3)

Solution with mark scheme:

(a) Step 1 — standardise both probability statements.

If $P(X < 50) = 0.2$P(X<50)=0.2, then by symmetry of the standard normal, $P(Z < z_1) = 0.2$P(Z<z1​)=0.2 gives a negative $z_1$z1​. From the inverse normal: $z_1 = -0.8416$z1​=−0.8416 (to 4 d.p.).

If $P(X < 70) = 0.9$P(X<70)=0.9, then $P(Z < z_2) = 0.9$P(Z<z2​)=0.9 gives $z_2 = 1.2816$z2​=1.2816.

M1 — using the inverse normal correctly to extract two $z$z-values, with the correct sign for each. Common error: ignoring the sign of $z_1$z1​ because $P(X < 50) < 0.5$P(X<50)<0.5 implies the value lies below the mean.

A1 — both $z$z-values correct to at least 3 d.p.

Step 2 — set up two equations from $Z = (X - \mu)/\sigma$Z=(X−μ)/σ.

$\dfrac{50 - \mu}{\sigma} = -0.8416 \quad \text{and} \quad \dfrac{70 - \mu}{\sigma} = 1.2816$σ50−μ​=−0.8416andσ70−μ​=1.2816

M1 — both equations correctly formed.

Step 3 — solve simultaneously.

Subtracting equation 1 from equation 2:

$\dfrac{70 - 50}{\sigma} = 1.2816 - (-0.8416) = 2.1232$σ70−50​=1.2816−(−0.8416)=2.1232

$\sigma = \dfrac{20}{2.1232} \approx 9.42$σ=2.123220​≈9.42

A1 — $\sigma \approx 9.42$σ≈9.42 (accept 9.41–9.43).

Step 4 — back-substitute to find $\mu$μ.

Using $50 - \mu = -0.8416 \cdot 9.42 \approx -7.93$50−μ=−0.8416⋅9.42≈−7.93, so $\mu \approx 57.93$μ≈57.93.

A1 — $\mu \approx 57.9$μ≈57.9 (accept 57.8–58.0).

(b) Step 1 — standardise both bounds with the values from (a).

$Z_1 = (45 - 57.93)/9.42 \approx -1.373$Z1​=(45−57.93)/9.42≈−1.373 and $Z_2 = (65 - 57.93)/9.42 \approx 0.751$Z2​=(65−57.93)/9.42≈0.751.

M1 — correct standardisation of both bounds using the candidate's own $\mu, \sigma$μ,σ (follow-through from (a)).

Step 2 — compute the probability.

$P(45 < X < 65) = \Phi(0.751) - \Phi(-1.373) = 0.7737 - 0.0848 \approx 0.689$P(45<X<65)=Φ(0.751)−Φ(−1.373)=0.7737−0.0848≈0.689

M1 — using the difference $\Phi(z_2) - \Phi(z_1)$Φ(z2​)−Φ(z1​) (or, equivalently, direct calculator computation $P(45 < X < 65)$P(45<X<65) with the candidate's parameters).

A1 — final answer $\approx 0.689$≈0.689 (accept 0.685–0.692).

Total: 8 marks (M3 A5).

---

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): The lengths, in cm, of cucumbers grown commercially follow a normal distribution with mean $25$25 and standard deviation $3$3. A cucumber is classed as "premium" if its length exceeds $28$28 cm.

(a) Find the probability that a randomly chosen cucumber is premium. (2)

(b) A grower picks 5 cucumbers at random. Find the probability that exactly 2 of them are premium. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — standardising $Z = (28 - 25)/3 = 1$Z=(28−25)/3=1, or directly using calculator $P(X > 28)$P(X>28) with $X \sim N(25, 9)$X∼N(25,9).
• A1 (AO1.1b) — $P(X > 28) \approx 0.1587$P(X>28)≈0.1587.

(b)

• M1 (AO3.3) — recognising that "exactly 2 of 5" is binomial $Y \sim B(5, p)$Y∼B(5,p) with $p$p from part (a).
• M1 (AO1.1b) — applying $P(Y = 2) = \binom{5}{2} p^2 (1 - p)^3$P(Y=2)=(25​)p2(1−p)3 with $p = 0.1587$p=0.1587.
• A1 (AO1.1b) — $\binom{5}{2} (0.1587)^2 (0.8413)^3 \approx 10 \cdot 0.02519 \cdot 0.5957 \approx 0.150$(25​)(0.1587)2(0.8413)3≈10⋅0.02519⋅0.5957≈0.150.
• A1 (AO2.4) — interpreting answer in context: "approximately 0.15, so about a 15% chance of exactly two premium cucumbers in a sample of five".

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a classic Paper 3 synoptic item: the normal model supplies the success probability that feeds a binomial calculation. AO3 sits in the recognition that "exactly 2 of 5" demands a binomial frame, not a continuation of normal arithmetic.

---

Synoptic links

Connects to:

1. Section 5 — Probability: the Normal is a continuous probability distribution, so axiomatic probability rules ($0 \leq P \leq 1$0≤P≤1, $P(\text{complement}) = 1 - P$P(complement)=1−P, addition for disjoint events) apply throughout. The translation $P(X > a) = 1 - P(X < a)$P(X>a)=1−P(X<a) is the most-used identity.

2. Section 6 — Binomial distribution: for large $n$n and $p$p not too close to 0 or 1, $B(n, p)$B(n,p) is approximated by $N(np, np(1-p))$N(np,np(1−p)). Outside the 9MA0 specification (it is no longer an explicit examinable approximation in current Edexcel), but the conceptual link — that sums of independent Bernoullis tend to normal — is foundational.

3. Section 8 — Hypothesis testing for the mean of a normal distribution: $H_0: \mu = \mu_0$H0​:μ=μ0​ versus $H_1: \mu \neq \mu_0$H1​:μ=μ0​ using the test statistic $Z = (\bar{X} - \mu_0)/(\sigma/\sqrt{n}) \sim N(0,1)$Z=(Xˉ−μ0​)/(σ/n​)∼N(0,1) assumes a normal sampling distribution. Without confidence in normal-distribution probability calculations, hypothesis-test conclusions cannot be reached.

4. Central limit theorem (CLT): the sample mean $\bar{X}$Xˉ of $n$n independent identically distributed observations approaches $N(\mu, \sigma^2/n)$N(μ,σ2/n) as $n \to \infty$n→∞, regardless of the underlying distribution. This is why the normal distribution is so prevalent in applied statistics — it appears as a limit even when the data themselves are not normally distributed.

5. Statistical inference (Year 2): confidence intervals for the mean — $\bar{x} \pm z_{\alpha/2} \cdot \sigma/\sqrt{n}$xˉ±zα/2​⋅σ/n​ — depend on inverse normal lookups for the critical $z$z-values ($z_{0.025} = 1.96$z0.025​=1.96 for a 95% CI). The same standardisation that turns $X$X into $Z$Z on a single observation turns $\bar{X}$Xˉ into $Z$Z on a sample mean.

---

Mark-scheme literacy

Normal distribution questions on 9MA0-03 split AO marks between procedural fluency and contextual interpretation:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–70%	Standardising correctly, using calculator distribution functions, applying inverse normal, computing $P(a < X < b)$P(a<X<b) as a difference
AO2 (reasoning / interpretation)	20–30%	Sketching the curve and shading the relevant region, interpreting answers in context, justifying why the normal model is or is not appropriate
AO3 (problem-solving)	10–20%	Setting up simultaneous equations from two probability statements; recognising when a binomial-of-normal-tail synoptic structure applies

Examiner-rewarded phrasing: "since $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2), standardising gives $Z \sim N(0, 1)$Z∼N(0,1)"; "by the symmetry of the normal curve, $P(Z < -z) = P(Z > z) = 1 - \Phi(z)$P(Z<−z)=P(Z>z)=1−Φ(z)"; "in context, this means approximately 16% of cucumbers are classified premium". Phrases that lose marks: writing $\sigma^2$σ2 where $\sigma$σ is required (or vice versa) inside the standardisation formula; quoting $z$z-values without specifying their direction (positive/negative); reporting probabilities greater than 1 or less than 0 without recognising the impossibility.

A specific Edexcel pattern to watch: when a question gives $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2) with the variance written out, candidates frequently divide by $\sigma^2$σ2 instead of $\sigma$σ in the standardisation. Read the parameters precisely — the second argument is the variance, but the standardisation uses the standard deviation $\sigma = \sqrt{\sigma^2}$σ=σ2​.

---

Grade-band model answers

3-mark question

Question: $X \sim N(50, 16)$X∼N(50,16). Find $P(X < 56)$P(X<56).

Grade C response (~180 words):

Standardise: $Z = (56 - 50)/\sqrt{16} = 6/4 = 1.5$Z=(56−50)/16​=6/4=1.5.

So $P(X < 56) = P(Z < 1.5) = \Phi(1.5)$P(X<56)=P(Z<1.5)=Φ(1.5). Using a calculator (or table), $\Phi(1.5) \approx 0.9332$Φ(1.5)≈0.9332.

Therefore $P(X < 56) \approx 0.933$P(X<56)≈0.933.

Examiner commentary: Full marks (3/3). The candidate correctly identifies that $\sigma^2 = 16$σ2=16 implies $\sigma = 4$σ=4 (not 16), standardises cleanly, and reads off the normal CDF to three significant figures — Edexcel's default reporting precision. The answer is presented with three decimal places, which is the convention for normal probabilities in 9MA0. Many candidates lose a mark here by dividing by 16 instead of 4, producing $z = 0.375$z=0.375 and an incorrect probability around 0.646. The discriminating step at this level is recognising the variance/standard-deviation distinction.

Grade A response (~225 words):*

Since $X \sim N(50, 16)$X∼N(50,16), the variance is $\sigma^2 = 16$σ2=16, so $\sigma = 4$σ=4.

Standardise: