OxfordAQA International AS and A-level Mathematics: The normal distribution and sums of normal variables (9660) – Revision Notes

These revision notes condense section S2.4 (Normal distribution) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit S2 (Statistics); it is not examined at International AS. For full explanations and longer worked examples, read the study guide first.

When you have revised, try the practice questions. Tick off the topics on the printable checklist, see the course hub, and use the free 10-minute diagnostics to check what you remember.

Key facts

Item	Statement
Notation	X ~ N(μ, σ²). The second parameter is the variance
Shape	Bell-shaped, symmetric about x = μ, total area 1
Centre	Mean, median and mode all equal μ
Within μ ± σ	about 2/3 (0.6827)
Within μ ± 2σ	about 0.95 (0.9545)
Within μ ± 3σ	about 0.997 (0.9973)
Standardising	Z = (X − μ)/σ, with Z ~ N(0, 1)
Cumulative probability	Φ(z) = P(Z ≤ z)
Negative z	Φ(−z) = 1 − Φ(z)
Sum	X + Y ~ N(μ₁ + μ₂, σ₁² + σ₂²) for independent X, Y
Difference	X − Y ~ N(μ₁ − μ₂, σ₁² + σ₂²)
Combination	aX + bY ~ N(aμ₁ + bμ₂, a²σ₁² + b²σ₂²)

The sum and difference results are needed as results; proofs are not expected.

Method boxes

Finding a probability

1. Sketch the curve, mark μ and shade the required area.
2. Standardise each end point: z = (x − μ)/σ. Round to 2 d.p.
3. Convert to Φ values. Upper tail: 1 − Φ(z). Negative z: 1 − Φ(|z|). Between two values: subtract.
4. Check the answer against the sketch. A small shaded area needs a small answer.

Finding an unknown mean or standard deviation

1. Write the given probability as P(X < x) = p, converting an upper tail by 1 − p.
2. Use the table of percentage points to find z for p. If p < 0.5, z is negative.
3. Write (x − μ)/σ = z.
4. With two unknowns, write two equations and solve them simultaneously.

Sums and differences

1. Check the variables are independent, and note the context.
2. Find the new mean from the means with the coefficients and signs given.
3. Find the new variance by adding a² × variance for every term. Never subtract.
4. Standardise using the new standard deviation (square root of the new variance).

Percentage points used in these notes

P(Z < z)	0.75	0.80	0.90	0.95	0.975	0.99
z	0.6745	0.8416	1.2816	1.6449	1.9600	2.3263

These are supplied as data so you can follow the arithmetic. In a question, use the values you are given or look up.

Worked reminders

Both parameters unknown. X ~ N(μ, σ²) with P(X < 10) = 0.05 and P(X > 22) = 0.20.

(10 − μ)/σ = −1.6449     →  μ − 1.6449σ = 10
(22 − μ)/σ =  0.8416     →  μ + 0.8416σ = 22
Subtract: 2.4865σ = 12, so σ = 4.83 (3 s.f.)
μ = 10 + 1.6449 × 4.8261 = 17.9 (3 s.f.)

A linear combination. X ~ N(5, 1²) and Y ~ N(8, 2²) are independent. Find P(2X + Y > 21).

2X + Y ~ N(2 × 5 + 8, 2² × 1² + 2²) = N(18, 8)
z = (21 − 18)/√8 = 1.06 (2 d.p.)
P(2X + Y > 21) = 1 − Φ(1.06) = 1 − 0.8554 = 0.1446

A probability between two values. The delay in a train departure is D ~ N(72, 6²) seconds. Find P(65 < D < 80).

z for 65: (65 − 72)/6 = −1.17 (2 d.p.)
z for 80: (80 − 72)/6 =  1.33 (2 d.p.)
P = Φ(1.33) − Φ(−1.17)
  = Φ(1.33) − (1 − Φ(1.17))
  = 0.9082 − (1 − 0.8790) = 0.7872

Independent copies. Four independent lengths, each N(2.5, 0.3²) metres, are laid end to end. Find the probability that the total is under 9.6 metres.

Total ~ N(4 × 2.5, 4 × 0.3²) = N(10, 0.36)
sd = 0.6
z = (9.6 − 10)/0.6 = −0.67 (2 d.p.)
P = 1 − Φ(0.67) = 1 − 0.7486 = 0.2514

A difference in context. A shelf gap is G ~ N(30.5, 0.6²) cm and a book is B ~ N(29.8, 0.5²) cm, independent. The book fits when G − B > 0.

G − B ~ N(0.7, 0.36 + 0.25) = N(0.7, 0.61)
sd = √0.61 = 0.7810
z = (0 − 0.7)/0.7810 = −0.90 (2 d.p.)
P(G − B > 0) = Φ(0.90) = 0.8159

Symmetry shortcuts

These save time and reduce lookup errors, so check whether one applies before you start a calculation.

• P(Z > z) = P(Z < −z), so only positive z needs the table.
• P(−z < Z < z) = 2Φ(z) − 1. For z = 1.5 this is 2 × 0.9332 − 1 = 0.8664.
• If P(X < a) = p, then P(X > 2μ − a) = p too. The point reflected in the mean has the same tail.
• Percentage points for p < 0.5 are the negatives of those for 1 − p. For example, the point for 0.10 is −1.2816.

Wording clues

Phrase in the question	What to do
“Independent” stated in the question	A sum or difference result may be used
“Total mass of five packets”	Sum of five independent copies. Mean 5μ, variance 5σ²
“Five times the mass of one packet”	5X. Mean 5μ, variance 25σ²
“How much longer than” or “heavier than”	Look at X − Y and compare it with 0 or with the stated amount
“Lowest mark for the top 10%”	Percentage point for 0.90, then x = μ + zσ
“The middle 95%”	2.5% in each tail. Use the percentage point for 0.975
“Within 1.5 standard deviations of the mean”	Use P(−1.5 < Z < 1.5) = 2Φ(1.5) − 1

Must-know distinctions

Compare	Key difference
X ~ N(μ, σ²) and Z ~ N(0, 1)	Z has mean 0 and standard deviation 1. Convert with Z = (X − μ)/σ
X − Y and X + Y	Means subtract, but variances add in both
X₁ + X₂ + X₃ and 3X	Same mean. Variances are 3σ² and 9σ²
P(Z < −z) and P(Z < z)	Sum to 1, by symmetry
“Variance 16” and “standard deviation 16”	N(μ, 16) has σ = 4. N(μ, 16²) has σ = 16

Quick self-test

Let Y ~ N(35, 4²) for questions 1 to 5. Table values: Φ(1.5) = 0.9332 and Φ(1.75) = 0.9599.

1. For X ~ N(200, 16²), roughly what proportion of values lie between 184 and 216?
2. Find P(Y < 41).
3. Find P(Y > 29).
4. Find P(31 < Y < 42).
5. Find a such that P(Y < a) = 0.95.
6. W ~ N(μ, 2.5²) and P(W > 30) = 0.10. Find μ.
7. V ~ N(50, σ²) and P(V < 44) = 0.20. Find σ.
8. A ~ N(10, 3²) and B ~ N(6, 4²) are independent. State the distribution of A + B.
9. For the same A and B, find P(A − B > 0).
10. Compare the variance of 4A with the variance of A₁ + A₂ + A₃ + A₄, where the copies are independent.

Answers

1. 184 to 216 is μ ± σ, so about 2/3.
2. z = (41 − 35)/4 = 1.5, so P = Φ(1.5) = 0.9332.
3. z = (29 − 35)/4 = −1.5. P(Y > 29) = P(Z > −1.5) = Φ(1.5) = 0.9332.
4. z = −1.00 and z = 1.75. P = Φ(1.75) − (1 − Φ(1.00)) = 0.9599 − 0.1587 = 0.8012.
5. a = 35 + 1.6449 × 4 = 41.6 (3 s.f.).
6. P(W < 30) = 0.90, so (30 − μ)/2.5 = 1.2816. μ = 30 − 3.204 = 26.8 (3 s.f.).
7. (44 − 50)/σ = −0.8416, so σ = 6/0.8416 = 7.13 (3 s.f.).
8. A + B ~ N(16, 3² + 4²) = N(16, 25).
9. A − B ~ N(4, 25), so sd = 5. z = (0 − 4)/5 = −0.8. P = Φ(0.8) = 0.7881.
10. Var(4A) = 16 × 9 = 144. Var of the sum = 4 × 9 = 36. The first is four times the second.

Where marks are usually lost

• Writing N(35, 4) when the standard deviation is 4. The second number must be 16.
• Treating the variance as the standard deviation when standardising, or the reverse.
• Reading Φ(z) for an upper-tail question and forgetting 1 − Φ(z).
• Using a negative z in a table that has only positive values, with no symmetry step.
• Subtracting variances for a difference of two variables.
• Using 4A when the context has four independent observations, or the reverse.
• Giving the wrong sign to z when the unknown probability is below 0.5.
• Rounding z to one decimal place, or rounding too early in the middle of the working.
• Giving no sketch and no clear line showing which probability is being found.
• Omitting the statement that the variables are independent, where a sum or difference is being used.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section S2.4: Normal distribution, published by OxfordAQA.