OCR A Level Mathematics A: The normal distribution and choosing a distribution (H240) – Revision Notes

These notes condense references 2.04e to 2.04h of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. All four are Stage 2 content, so they are A Level only. For full explanations and longer worked examples, start with the study guide.

Links: OCR Mathematics A hub, printable checklist, practice questions, statistics revision notes and free diagnostics.

Key facts: 2.04e and 2.04g

Item	Fact
Notation	X ~ N(μ, σ²): mean μ, variance σ², standard deviation σ
Shape	Bell-shaped, symmetrical about x = μ, total area 1
Probability	Area under the curve. A single point has zero area, so P(X = a) = 0 and the two inequalities P(X < a), P(X ≤ a) agree
Points of inflection	x = μ − σ and x = μ + σ
About two-thirds	of values lie in μ ± σ
About 95%	of values lie in μ ± 2σ
Almost all	of values lie in μ ± 3σ
Not required	The equation of the normal curve (excluded)

Effect of the parameters. Changing μ moves the curve left or right without changing its shape. A larger σ gives a wider, lower curve. The area is always 1.

Link to histograms. If a histogram of a large continuous sample is symmetrical and bell-shaped, a normal model may fit. Estimate μ from the sample mean and σ from the sample standard deviation.

Sketching a normal curve

1. Draw a bell shape with its peak above x = μ and label μ on the axis.
2. Mark x = μ − σ and x = μ + σ where the curve changes from curving upwards to curving downwards, or the reverse.
3. Let the tails approach the axis without touching or crossing it.
4. Shade the area that represents the probability asked for, and label the boundary values.

Two sketches on the same axes with different σ should cross, with the narrower curve taller. If they share μ, they cross twice and are symmetrical about the same line.

Key facts: 2.04f

Item	Fact
Standard normal	Z ~ N(0, 1)
Standardising	Z = (X − μ)/σ, so X = μ + zσ
Calculator inputs	Bounds, μ, and σ (the standard deviation, not the variance)
Inverse normal	Input the area to the left of x; the output is x
Expected frequency	probability × number of items

Method: probability

1. Write the distribution, with σ² given as a square.
2. Sketch the curve and shade the required area.
3. Use the calculator, or standardise first, and write the probability statement.
4. Give 3 significant figures unless told otherwise.

Method: find x for a given probability

1. Shade the area. If it is an upper tail p, the area to the left is 1 − p.
2. Find z from the inverse normal function for that area to the left.
3. Convert: x = μ + zσ.
4. Check the sign: a left-tail area below 0.5 gives a negative z.

Method: find unknown μ and/or σ

1. Turn each probability into the area to the left, then into a z-value.
2. Write each as μ + zσ = x (or (x − μ)/σ = z).
3. One unknown: solve directly. Two unknowns: subtract the equations to find σ first, then μ.
4. Keep unrounded z-values until the end.

Worked reminder

X ~ N(64, 3²). Find P(X > 70) and the value exceeded by 80% of the distribution.

z = (70 − 64)/3 = 2, so P(X > 70) = 0.0228 (3 s.f.)
80% exceed t, so P(X < t) = 0.2, z = −0.8416
t = 64 − 0.8416 × 3 = 61.5 (3 s.f.)

Useful z-values to recognise

Area to the left	0.75	0.80	0.85	0.90	0.95	0.975	0.99
z	0.6745	0.8416	1.0364	1.2816	1.6449	1.9600	2.3263

These come from the inverse normal function on your calculator. By symmetry, an area of 0.25 on the left has z = −0.6745. Recognising them lets you check an answer for sense. For example, a 90% point is a little over one standard deviation above the mean.

Worked reminder: two unknowns

Y ~ N(μ, σ²) with P(Y < 18) = 0.25 and P(Y < 30) = 0.9. Find μ and σ.

18 = μ − 0.6745σ
30 = μ + 1.2816σ
Subtract: 12 = 1.9561σ, so σ = 6.13 (3 s.f.)
μ = 18 + 0.6745 × 6.1348 = 22.1 (3 s.f.)

Worked reminder: histogram facts

A bell-shaped histogram of 500 values has mean 52 and standard deviation 4. About two-thirds, roughly 333 values, lie between 48 and 56. About 95% lie between 44 and 60, and almost all lie between 40 and 64. The curve fitted to it has points of inflection at x = 48 and x = 56.

2.04h: Choosing a model

Context	Model	Reason to quote
Count of successes in a fixed number of independent trials, constant p	Binomial B(n, p)	The conditions hold in context
Continuous measurement, symmetrical, one peak	Normal N(μ, σ²)	Bell-shaped histogram
Large n in a binomial	Binomial, or normal as an approximation	See below
Skewed data with a hard lower limit	Neither standard model fits well	Normal is symmetrical and unbounded

Normal model limitations. It gives non-zero probability to impossible values (negative lengths or times). It cannot represent skew. Real measurements are recorded to a finite accuracy.

Binomial model limitations. The probability may change between trials. Trials may depend on each other, for example when selecting without replacement from a small group. The number of trials may not be fixed.

Binomial approximated by normal. If X ~ B(n, p) with n large, X is approximately N(np, npq) where q = 1 − p. The specification excludes questions that explicitly require calculations using this approximation. The continuity correction (treat “X ≤ r” as “X < r + 0.5”) is the idea behind it, because a whole-number count is being modelled by a continuous variable.

Must-know distinctions

• Variance and standard deviation. In N(25, 9), σ = 3. In N(25, 3²), σ = 3. Calculators need 3.
• Left and right areas. Inverse normal gives the point with the stated area to its left.
• Continuous and discrete. A normal variable has P(X = a) = 0. A binomial variable has positive probability at each whole number.
• Stating a model and justifying it. A bare model name is weaker than a name tied to the context.

Quick self-test

Let X ~ N(40, 5²) for questions 1 to 6.

1. Find P(X < 46).
2. Find P(X > 35).
3. Find P(34 < X < 47).
4. Find k such that P(X < k) = 0.9.
5. A second variable Y ~ N(40, σ²) has P(Y < 30) = 0.2. Find σ.
6. A third variable W ~ N(μ, 5²) has P(W > 50) = 0.1. Find μ.
7. A normal curve has points of inflection at x = 12 and x = 20. Find μ and σ, and the interval containing about 95% of values.
8. For T ~ N(100, 15²), find P(T < 130).
9. X ~ B(200, 0.3). Write down the normal model that could approximate X.
10. Give one reason why a normal model may be unsuitable for waiting times at an emergency department.

Answers

1. z = 1.2, so P = 0.885.
2. z = −1, so P = 0.841.
3. P(X < 47) − P(X < 34) = 0.804.
4. z = 1.2816, so k = 40 + 1.2816 × 5 = 46.4.
5. P(Y < 30) = 0.2 gives z = −0.8416, so (30 − 40)/σ = −0.8416 and σ = 11.9.
6. P(W < 50) = 0.9 gives z = 1.2816, so 50 = μ + 1.2816 × 5 and μ = 43.6.
7. μ = (12 + 20)/2 = 16, 2σ = 8 so σ = 4. About 95% lie between 8 and 24.
8. z = 2, so P = 0.977.
9. np = 60 and npq = 200 × 0.3 × 0.7 = 42, so N(60, 42).
10. Waiting times tend to be skewed to the right with a lower limit of zero, but a normal curve is symmetrical and extends below zero.

Where marks are usually lost

• Typing σ² into the calculator where σ is needed.
• Giving a probability for the wrong tail because no sketch was drawn.
• Using the area to the right in the inverse normal function.
• Rounding z-values early, which shifts a two-unknown answer in the third significant figure.
• Giving “it is normal” as the reason for choosing a normal model, with no reference to continuity, symmetry or the shape of the data.
• Writing a limitation that is generic instead of tied to the context in the question.
• Forgetting to state the distribution and the probability statement before the number.
• Treating P(X = a) as non-zero for a normal variable.
• Subtracting the variance instead of the standard deviation when finding the points of inflection from a given N(μ, σ²).
• Stating the mean as np but forgetting that the variance is npq, not np.
• Quoting the 95% fact as “within one standard deviation”, or mixing up the two-thirds and 95% intervals.

Next, test yourself with the practice questions and the free diagnostics.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, published by OCR. Section 2.04e to 2.04h of the detailed content.