Normal Distribution

This lesson covers the normal distribution, the most important continuous probability distribution at A-Level. The normal distribution is used to model a wide range of natural phenomena and is central to statistical inference and hypothesis testing.

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Properties of the Normal Distribution

The normal distribution is a continuous probability distribution with a characteristic bell-shaped curve. If $X$X follows a normal distribution with mean $\mu$μ and variance $\sigma^2$σ2, we write:

$X \sim N(\mu, \sigma^2)$X∼N(μ,σ2)

Key properties:

• The curve is symmetrical about the mean $\mu$μ.
• The mean, median, and mode are all equal (at $\mu$μ).
• The total area under the curve is 1.
• Approximately 68% of data falls within $\mu \pm \sigma$μ±σ.
• Approximately 95% of data falls within $\mu \pm 2\sigma$μ±2σ.
• Approximately 99.7% of data falls within $\mu \pm 3\sigma$μ±3σ.

Region	Percentage of data
$\mu \pm \sigma$μ±σ	$\approx 68\%$≈68%
$\mu \pm 2\sigma$μ±2σ	$\approx 95\%$≈95%
$\mu \pm 3\sigma$μ±3σ	$\approx 99.7\%$≈99.7%

Exam Tip: The variance is $\sigma^2$σ2, not $\sigma$σ. When a question says $X \sim N(50, 16)$X∼N(50,16), the standard deviation is $\sigma = \sqrt{16} = 4$σ=16​=4. A common error is to confuse $\sigma$σ and $\sigma^2$σ2.

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The Standard Normal Distribution

The standard normal distribution has mean 0 and variance 1:

$Z \sim N(0, 1)$Z∼N(0,1)

Any normal variable $X$X can be standardised to $Z$Z using:

$Z = \frac{X - \mu}{\sigma}$Z=σX−μ​

This allows us to use standard normal tables to find probabilities for any normal distribution.

Finding Probabilities

To find $P(X < a)$P(X<a):

1. Standardise: $z = \frac{a - \mu}{\sigma}$z=σa−μ​.
2. Use the standard normal table or calculator to find $P(Z < z) = \Phi(z)$P(Z<z)=Φ(z).

Example: $X \sim N(100, 225)$X∼N(100,225), find $P(X < 115)$P(X<115).

$\sigma = \sqrt{225} = 15$σ=225​=15

$z = \frac{115 - 100}{15} = 1$z=15115−100​=1

$P(X < 115) = P(Z < 1) = \Phi(1) = 0.8413$P(X<115)=P(Z<1)=Φ(1)=0.8413

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Inverse Normal Problems

Given a probability, find the corresponding value of $x$x:

1. Find the $z$z-value from the table (or use the inverse normal function).
2. Convert back: $x = \mu + z\sigma$x=μ+zσ.

Example: $X \sim N(50, 9)$X∼N(50,9). Find $a$a such that $P(X < a) = 0.9$P(X<a)=0.9.

$\sigma = 3$σ=3

From tables: $z = 1.2816$z=1.2816 (since $\Phi(1.2816) = 0.9$Φ(1.2816)=0.9)

$a = 50 + 1.2816 \times 3 = 53.84$a=50+1.2816×3=53.84

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Finding Unknown Parameters

If the mean or standard deviation is unknown, set up equations using given probability information and solve simultaneously.

Example: $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2). Given $P(X < 20) = 0.2$P(X<20)=0.2 and $P(X > 50) = 0.1$P(X>50)=0.1:

$\frac{20 - \mu}{\sigma} = -0.8416 \quad \text{and} \quad \frac{50 - \mu}{\sigma} = 1.2816$σ20−μ​=−0.8416andσ50−μ​=1.2816

Solving these simultaneous equations gives $\mu$μ and $\sigma$σ.

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The Normal Distribution as an Approximation

The normal distribution can approximate the binomial distribution when $n$n is large and $p$p is not too close to 0 or 1. The rule of thumb is:

$np > 5 \quad \text{and} \quad n(1-p) > 5$np>5andn(1−p)>5

When approximating $B(n, p)$B(n,p) by $N(np, np(1-p))$N(np,np(1−p)), apply a continuity correction:

Binomial	Normal approximation
$P(X \leq r)$P(X≤r)	$P(Y < r + 0.5)$P(Y<r+0.5)
$P(X \geq r)$P(X≥r)	$P(Y > r - 0.5)$P(Y>r−0.5)
$P(X = r)$P(X=r)	$P(r - 0.5 < Y < r + 0.5)$P(r−0.5<Y<r+0.5)

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Summary

• $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2): symmetrical bell curve with mean $\mu$μ and variance $\sigma^2$σ2.
• Standardise using $Z = \frac{X - \mu}{\sigma}$Z=σX−μ​ to use $Z \sim N(0, 1)$Z∼N(0,1) tables.
• Use inverse normal to find values given probabilities.
• The normal can approximate the binomial with continuity correction when $np > 5$np>5 and $nq > 5$nq>5.

Exam Tip: Always show the standardisation step when finding normal probabilities. Write $P(X < a) = P\left(Z < \frac{a - \mu}{\sigma}\right)$P(X<a)=P(Z<σa−μ​) explicitly — this gains method marks even if the final answer is wrong.

A-Level Deep Dive: The Normal Distribution

Spec mapping

AQA 7357 specification, Paper 3 — Statistics, Section R (Year 2 content) covers the Normal distribution as a model; find probabilities using a Normal distribution; link to histograms, mean, standard deviation, points of inflection and the binomial distribution; 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). Section R extends Section O (probability) and Section Q (the binomial distribution from Year 1) and feeds directly into Section S (statistical hypothesis testing). Calculator-based standardisation and inverse-normal queries are the dominant exam tasks; the standard-normal table is no longer required for routine work but appears in the AQA formulae booklet for reference. The Central Limit Theorem is referenced informally — examined explicitly only in the context of "the sample mean is approximately normal".

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Worked example with full mark scheme

Question (8 marks):

A machine fills bottles with a nominal volume of 500 ml. The actual volume $X$X ml dispensed is modelled as $X \sim N(\mu, \sigma^2)$X∼N(μ,σ2).

(a) Given $\mu = 502$μ=502 and $\sigma = 1.5$σ=1.5, find $P(X < 500)$P(X<500). (2)

(b) Find the value of $x$x such that $P(X < x) = 0.95$P(X<x)=0.95. (2)

(c) A quality engineer wants $P(X < 500) = 0.02$P(X<500)=0.02 and $P(X > 504) = 0.05$P(X>504)=0.05. Find $\mu$μ and $\sigma$σ. (4)

Solution with mark scheme:

(a) Step 1 — standardise.

$Z = \dfrac{X - \mu}{\sigma} = \dfrac{500 - 502}{1.5} = -\dfrac{4}{3} \approx -1.333$Z=σX−μ​=1.5500−502​=−34​≈−1.333

M1 — correct standardisation with the right values substituted. A common slip is to invert the fraction or to use $\sigma^2 = 2.25$σ2=2.25 in the denominator instead of $\sigma = 1.5$σ=1.5. The standardisation formula uses $\sigma$σ, not $\sigma^2$σ2.

Step 2 — read probability.

$P(X < 500) = P(Z < -1.333) \approx 0.0912$P(X<500)=P(Z<−1.333)≈0.0912.

A1 — answer to 3 s.f. (calculator output). Acceptable range typically $0.091$0.091–$0.092$0.092.

(b) Step 1 — inverse normal.

We need $x$x with $\Phi\!\left(\dfrac{x - 502}{1.5}\right) = 0.95$Φ(1.5x−502​)=0.95. From inverse normal, $z_{0.95} \approx 1.6449$z0.95​≈1.6449.

M1 — identification of the correct $z$z-value (positive because the cumulative probability exceeds 0.5 — sketch the normal curve and shade the left tail of area 0.95 to confirm sign).

Step 2 — un-standardise.

$x = \mu + z\sigma = 502 + 1.6449 \times 1.5 \approx 504.47$x=μ+zσ=502+1.6449×1.5≈504.47.

A1 — $x \approx 504$x≈504 ml (3 s.f.).

(c) Step 1 — translate each probability into a $z$z-equation.

From $P(X < 500) = 0.02$P(X<500)=0.02:

$\dfrac{500 - \mu}{\sigma} = z_{0.02} \approx -2.0537$σ500−μ​=z0.02​≈−2.0537

From $P(X > 504) = 0.05 \Leftrightarrow P(X < 504) = 0.95$P(X>504)=0.05⇔P(X<504)=0.95:

$\dfrac{504 - \mu}{\sigma} = z_{0.95} \approx 1.6449$σ504−μ​=z0.95​≈1.6449

M1 — both $z$z-values correctly identified, with correct sign on the lower tail ($z_{0.02}$z0.02​ is negative — students who write $+2.0537$+2.0537 here lose both marks).

Step 2 — solve simultaneously.

$500 - \mu = -2.0537\sigma$500−μ=−2.0537σ and $504 - \mu = 1.6449\sigma$504−μ=1.6449σ. Subtracting the first from the second:

$4 = (1.6449 - (-2.0537))\sigma = 3.6986\sigma \implies \sigma \approx 1.0815$4=(1.6449−(−2.0537))σ=3.6986σ⟹σ≈1.0815

M1 — eliminating $\mu$μ to find $\sigma$σ.

Step 3 — back-substitute.

$\mu = 500 + 2.0537 \times 1.0815 \approx 502.22$μ=500+2.0537×1.0815≈502.22.

A1 — $\mu \approx 502$μ≈502 ml.

A1 — $\sigma \approx 1.08$σ≈1.08 ml, with both stated to a sensible accuracy and units carried through.

Total: 8 marks (M4 A4).

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Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): The lengths $L$L mm of components produced by a process are modelled as $L \sim N(48, 1.6^2)$L∼N(48,1.62).

(a) Find $P(46 < L < 49)$P(46<L<49). (2)

(b) A component is acceptable if its length is within 2 mm of the mean. Find the probability that a component is acceptable. (2)

(c) Of $200$200 components, find the expected number that are not acceptable. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — recognise the need for $P(L < 49) - P(L < 46)$P(L<49)−P(L<46) via the calculator's "between" mode or two cumulative queries.
• A1 (AO1.1b) — answer $\approx 0.628$≈0.628.

(b)

• M1 (AO1.2) — interpret "within 2 mm of the mean" as $P(46 < L < 50) = P(|L - 48| < 2)$P(46<L<50)=P(∣L−48∣<2).
• A1 (AO1.1b) — answer $\approx 0.789$≈0.789.

(c)

• M1 (AO3.1a) — $P(\text{not acceptable}) = 1 - 0.789 = 0.211$P(not acceptable)=1−0.789=0.211, multiplied by $200$200.
• A1 (AO2.2b) — expected number $\approx 42$≈42 (rounded sensibly; "42.2 components" is unphysical and would be marked as not interpreted).

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Section R questions are AO1-heavy at routine level but pivot to AO3 when the context demands modelling judgement (the rounding step in part (c)).

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Synoptic links

Connects to:

1. Section Q — the binomial distribution: when $n$n is large and $p$p is not too close to $0$0 or $1$1, $\text{Bin}(n, p) \approx N(np, np(1-p))$Bin(n,p)≈N(np,np(1−p)). The continuity correction $P(X \leq k) \approx P(Y < k + 0.5)$P(X≤k)≈P(Y<k+0.5) (with $Y$Y continuous-normal) is essential. AQA tests this approximation in Section R when the binomial calculation would otherwise be tedious.

2. Section S — hypothesis testing for the population mean: the test statistic $Z = \dfrac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}$Z=σ/n​Xˉ−μ0​​ is normal under $H_0$H0​ when the population is normal (or large $n$n via CLT). Critical values $\pm 1.96$±1.96 (two-tailed, $5\%$5%) and $\pm 1.6449$±1.6449 (one-tailed) come straight from inverse normal.

3. Central Limit Theorem (informal Year 2 content): for any distribution with finite variance, the sample mean $\bar{X}_n$Xˉn​ approaches normality as $n \to \infty$n→∞. This is why the normal distribution shows up everywhere — sums of independent effects tend toward it.

4. Section P — conditional probability: combining $P(A \cap B)/P(B)$P(A∩B)/P(B) with normal probabilities (e.g., "given that $X > 50$X>50, find $P(X > 55)$P(X>55)") tests both topics. The calculation reduces to a ratio of two upper-tail probabilities.

5. Statistical inference (university-level extension): confidence intervals $\bar{X} \pm 1.96 \, \sigma/\sqrt{n}$Xˉ±1.96σ/n​ depend on the same $z$z-values as hypothesis testing. The same standardisation underpins both procedures.

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Mark-scheme literacy

Section R questions split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Standardising correctly, using the calculator's normal CDF and inverse-normal modes, simplifying to required accuracy
AO2 (reasoning / interpretation)	25–35%	Interpreting context (e.g. "within 2 of the mean"), choosing between standard-normal and $N(\mu, \sigma^2)$N(μ,σ2) form, justifying the use of normal as a model
AO3 (problem-solving / modelling)	10–25%	Solving for $\mu$μ and $\sigma$σ simultaneously, rounding sensibly, commenting on model fit

Examiner-rewarded phrasing: "standardising using $Z = (X - \mu)/\sigma$Z=(X−μ)/σ"; "from inverse normal, $z = 1.6449$z=1.6449 to 4 d.p."; "expected number $= np = 200 \times 0.211 \approx 42$=np=200×0.211≈42 components". Phrases that lose marks: writing "$z = 1.6449 \implies x = 1.6449$z=1.6449⟹x=1.6449" (forgetting to un-standardise); using $\sigma^2$σ2 where $\sigma$σ is needed in standardisation; quoting probabilities to 1 s.f. when 3 s.f. is expected.

A specific AQA pattern: "give your answer to an appropriate degree of accuracy" is itself an AO2 cue — three significant figures for probabilities, two for expected counts of physical objects, integer for "number of components".

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Grade-band model answers

3-mark question

Question: $X \sim N(50, 16)$X∼N(50,16). Find $P(X > 58)$P(X>58).

Grade C response (~150 words):

$\sigma = \sqrt{16} = 4$σ=16​=4. Standardise: $Z = (58 - 50)/4 = 2$Z=(58−50)/4=2.

$P(X > 58) = P(Z > 2) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228$P(X>58)=P(Z>2)=1−Φ(2)≈1−0.9772=0.0228.

Examiner commentary: Full marks (3/3). The candidate correctly extracts $\sigma = 4$σ=4 from the variance $\sigma^2 = 16$σ2=16, standardises cleanly, and converts the upper-tail query to $1 - \Phi(z)$1−Φ(z). The answer is given to 3 s.f., which matches AQA's expectation for normal probabilities. Many candidates lose the M1 here by treating the second parameter as $\sigma$σ, computing $Z = (58 - 50)/16 = 0.5$Z=(58−50)/16=0.5, and producing $P(Z > 0.5) \approx 0.309$P(Z>0.5)≈0.309. That is the most common single error on Section R.

Grade A response (~210 words):*

The model is $X \sim N(50, 16)$X∼N(50,16), so $\mu = 50$μ=50 and $\sigma^2 = 16$σ2=16, giving $\sigma = 4$σ=4.

To find $P(X > 58)$P(X>58), standardise to the standard normal:

$Z = \dfrac{X - \mu}{\sigma} = \dfrac{58 - 50}{4} = 2$Z=σX−μ​=458−50​=2

So $P(X > 58) = P(Z > 2)$P(X>58)=P(Z>2). Using the symmetry and $\Phi(2) \approx 0.9772$Φ(2)≈0.9772: