Pearson Edexcel International Advanced Level Mathematics: The Normal distribution (YMA01) – Revision Notes

These revision notes condense the Normal distribution content from the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Specification Issue 3, April 2019. The content is section 6 of Statistics 1, numbered 5.1 in the specification, and S1 is an optional unit for IAS and IAL Mathematics and Further Mathematics. For full explanations and longer worked examples, read the study guide.

Other pages: practice questions, the Edexcel A Level Mathematics hub, the printable checklist and the free diagnostics.

Key facts

Item	Result
Notation	X ~ N(μ, σ²). The second value is the variance
Parameters	Mean μ, variance σ², standard deviation σ
Shape	A bell, mirror-symmetrical about the line x = μ
Mean, median, mode	All equal μ
Total area	1, and 0.5 on each side of μ
Exact values	A single point has zero probability, so strict and non-strict inequalities give the same answer
Standardising	Z = (X − μ)/σ, with Z ~ N(0, 1)
Cumulative function	Φ(z) = P(Z ≤ z)
Not required	Probability density function; derivations; interpolation

Reading probabilities from Φ

You want	Use
P(Z < z), z > 0	Φ(z)
P(Z > z), z > 0	1 − Φ(z)
P(Z < −z)	1 − Φ(z)
P(Z > −z)	Φ(z)
P(−z < Z < z)	2Φ(z) − 1
P(a < Z < b)	Φ(b) − Φ(a)

Method boxes

Finding a probability

1. Sketch the bell and shade the region.
2. Standardise each end: z = (x − μ)/σ.
3. Rewrite using Φ and symmetry so that every z is positive.
4. Read the table and combine.

Finding a value from a probability (inverse)

1. Sketch and shade.
2. Turn the given probability into Φ(z) = p with p above 0.5.
3. Read z backwards from the table. If the shaded left tail is below 0.5, the z is negative.
4. Use x = μ + zσ.

Finding μ and σ

1. Write each statement as P(X < something) = a probability.
2. Find the z for each, with the correct sign.
3. Write (x₁ − μ)/σ = z₁ and (x₂ − μ)/σ = z₂, then rearrange to x = μ + zσ.
4. Subtract to eliminate μ, solve for σ, then substitute for μ.
5. Check that σ is positive and that μ lies between x₁ and x₂ when they sit on opposite sides.

Symmetry shortcut: if P(X < a) = P(X > b), then μ = (a + b)/2.

Small worked reminders

The data for these reminders: Φ(1.2) = 0.8849, Φ(1.4) = 0.9192, Φ(z) = 0.75 gives z = 0.6745, Φ(z) = 0.90 gives z = 1.2816, Φ(z) = 0.70 gives z = 0.5244.

Probability. X ~ N(40, 25), so σ = 5.

P(X < 46) = Φ((46 − 40)/5) = Φ(1.2) = 0.8849
P(X > 33) = P(Z > −1.4) = Φ(1.4) = 0.9192
P(34 < X < 47) = Φ(1.4) + Φ(1.2) − 1 = 0.8041

Inverse. X ~ N(80, 12²). Find k with P(X > k) = 0.25.

P(X < k) = 0.75, so z = 0.6745
k = 80 + 0.6745 × 12 = 88.094 = 88.1

Two unknowns. X ~ N(μ, σ²), P(X < 20) = 0.30, P(X < 35) = 0.90.

P(X < 20) = 0.30: z = −0.5244, so μ − 0.5244σ = 20
P(X < 35) = 0.90: z = 1.2816,  so μ + 1.2816σ = 35
1.806σ = 15, so σ = 8.31 (3 s.f.)
μ = 35 − 1.2816 × 8.3056... = 24.4 (3 s.f.)

One unknown, σ. T ~ N(60, σ²) and P(T < 52) = 0.0668. Data: Φ(1.5) = 0.9332.

0.0668 = 1 − 0.9332, so the z-value is −1.5
(52 − 60)/σ = −1.5
σ = 16/3 = 5.33 (3 s.f.)

One unknown, μ. Y ~ N(μ, 6²) and P(Y > 50) = 0.0228. Data: Φ(2) = 0.9772.

P(Y < 50) = 0.9772, so (50 − μ)/6 = 2
μ = 38

Expected number. In a batch of 500 items with M ~ N(310, 20²), P(M > 340) = 1 − Φ(1.5) = 0.0668. Expected number = 500 × 0.0668 = 33.4, so about 33.

Symmetry with one more fact. P(X < 18) = P(X > 30) = 0.0668 gives μ = 24. With Φ(1.5) = 0.9332 the z-value for 18 is −1.5, so (18 − 24)/σ = −1.5 and σ = 4.

Sketch checklist

A sketch takes ten seconds and catches most sign errors. Before you read the table, check three things:

• The shaded region is on the correct side of μ, and the probability you expect is more or less than 0.5.
• Every z you carry forward has the sign that matches whether the x value is below or above μ.
• The final answer sits on the same side of μ as the tail you shaded. A value for “the top 10%” must be above the mean.

For a two-sided interval symmetrical about the mean, such as P(μ − a < X < μ + a), the answer is 2Φ(a/σ) − 1. Run it in reverse for an unknown a: if the probability is 0.90, then Φ(a/σ) = 0.95, and a/σ = 1.6449.

Wording that points to each skill

The question says	The skill
“Find P(X > 35)”	Standardise, then use 1 − Φ
“The top 5% of values exceed k”	Inverse, with Φ(z) = 0.95
“Show that σ = 4”	Standardise with a known z and show each line
“Find the values of μ and σ”	Two equations, solved simultaneously
“Estimate how many in a sample of 400”	Probability multiplied by the size
“Sketch the distribution”	Bell shape, label μ and the axis, symmetrical

Must-know distinctions

• Variance and standard deviation: N(μ, σ²) holds the variance, but you divide by σ.
• P(Z > z) and Φ(z): they add to 1.
• Forward and inverse: forward gives a probability from a value, inverse gives a value from a probability.
• One unknown and two unknowns: one statement gives one equation, two unknowns need two statements.
• Positive and negative z: the value is below the mean exactly when z is negative.

Quick self-test

Use these values where you need them: Φ(0.5) = 0.6915, Φ(0.8) = 0.7881, Φ(1) = 0.8413, Φ(2) = 0.9772, and Φ(z) = 0.95 gives z = 1.6449.

1. State the standard deviation of X ~ N(7, 16).
2. State the value of P(X > μ) for any Normal X.
3. Z ~ N(0, 1). Write P(Z < −0.5) using Φ, then evaluate it.
4. X ~ N(50, 10²). Find P(X < 60).
5. X ~ N(50, 10²). Find P(X > 42).
6. X ~ N(100, 5²). Find P(95 < X < 105).
7. X ~ N(30, 4²). Find k such that P(X < k) = 0.95.
8. P(X < 14) = P(X > 26) for a Normal X. State μ.
9. X ~ N(μ, 3²) and P(X < 20) = 0.8413. Find μ.
10. X ~ N(45, σ²) and P(X > 53) = 0.0228. Find σ.

Answers

1. 4, the square root of 16.
2. 0.5, by symmetry about the mean.
3. P(Z < −0.5) = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085.
4. z = 1, so Φ(1) = 0.8413.
5. z = −0.8, so P = Φ(0.8) = 0.7881.
6. z = ±1, so 2Φ(1) − 1 = 0.6826.
7. z = 1.6449, so k = 30 + 1.6449 × 4 = 36.5796, which is 36.6 (3 s.f.).
8. μ = (14 + 26)/2 = 20.
9. Φ(z) = 0.8413 gives z = 1, so (20 − μ)/3 = 1 and μ = 17.
10. P(X < 53) = 0.9772, so z = 2. (53 − 45)/σ = 2 gives σ = 4.

Setting out a full answer

Examiner advice on layout is the same for every Normal question, and following it also protects you from slips:

1. Name the variable and its distribution at the top, such as “X ~ N(40, 5²)”, and say clearly which number is σ.
2. Write the probability statement in symbols, for example P(34 < X < 47).
3. Show the standardised values before you read any table.
4. Show the Φ expression with its subtractions, then the number.
5. Finish with a sentence in context, such as “the probability that a mango weighs less than 325 g is 0.773”, and the units where there are any.

For inverse and parameter questions, write the z-value as a labelled line (for example “z = 1.2816”) before forming the equation. When two statements are used, label the equations (1) and (2) so that subtracting them is easy to follow.

Revision plan

• Day 1: forward probabilities in all six table cases, using sketches.
• Day 2: inverse questions, both tails, giving x to 3 significant figures.
• Day 3: one-unknown questions for μ and for σ.
• Day 4: two-unknown simultaneous questions and the symmetry shortcut.
• Day 5: a mixed set, then the practice questions under timed conditions.

Where marks are usually lost

• Dividing by the variance instead of the standard deviation.
• Leaving a negative z unconverted, or reading a table entry that does not match the question.
• Reporting P(X > a) as Φ(z) with no subtraction from 1.
• Using the wrong tail in an inverse question, so the answer lies on the wrong side of μ.
• Forgetting the minus sign on z when the probability given is below 0.5.
• Rounding z or σ early in a simultaneous-equation question, which shifts the final answers.
• Skipping the standardising line. Showing (x − μ)/σ makes your method visible even when a later table read-off slips.
• Giving a probability with too few figures. Unless the question says otherwise, 3 significant figures is a safe default, and table-based answers are usually quoted to 4 decimal places.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics – Specification, Issue 3, April 2019, first teaching September 2018, published by Pearson Education Limited.