Home / A-Level Maths 9709 / The normal distribution / The normal distribution
Cambridge International AS & A Level · Topic 5.5
A-Level Maths 9709 / Topic 5.5 / AS & A Level

The bell, standardised.

Heights, masses and measurement errors are modelled by a continuous, symmetrical bell-shaped curve. Every normal distribution turns into the same standard one, \(Z\), so one table of \(\Phi(z)\) answers every question, forwards and backwards.

Topic 5.5All the normal distribution lessons →
The key idea

If \(X \sim \text{N}(\mu, \sigma^2)\), then \(Z = \dfrac{X - \mu}{\sigma} \sim \text{N}(0, 1)\) and \(\text{P}(X < x) = \Phi\!\left(\dfrac{x - \mu}{\sigma}\right)\). Use symmetry, \(\Phi(-z) = 1 - \Phi(z)\), for negative \(z\), and always sketch the curve.

384246505462 x 56 (z = 1.5) 0.9332 X ~ N(50, 4²) P(X < 56) = Φ(1.5)
Fig. 1: X ~ N(50, 4²) drawn to scale. The value 56 is 1.5 standard deviations above the mean, and the shaded area to its left is Φ(1.5) = 0.9332
Section 01

One curve for all.

A normal distribution is fixed by its mean \(\mu\) and variance \(\sigma^2\), written \(\text{N}(\mu, \sigma^2)\). It is symmetrical about \(\mu\), and about 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean. Standardising counts how many standard deviations a value is from the mean.

Worked example Less than
\(X \sim \text{N}(50, 4^2)\). Find \(\text{P}(X < 56)\).
  1. \(z = \dfrac{56 - 50}{4} = 1.5\).
  2. \(\text{P}(X < 56) = \Phi(1.5)\), read from the table (Fig. 1).
Answer: 0.9332
Worked example Greater than
\(X \sim \text{N}(50, 4^2)\). Find \(\text{P}(X > 56)\).
  1. \(z = 1.5\) as before; the area to the right is the rest of the curve.
  2. \(1 - \Phi(1.5) = 1 - 0.9332\).
Answer: 0.0668
Examiner trap

Dividing by the variance. \(\text{N}(50, 16)\) means \(\sigma^2 = 16\), so \(\sigma = 4\); \(\dfrac{56 - 50}{16}\) is wrong.

Section 02

Using symmetry.

The table only gives areas to the left of a positive \(z\). Sketch the curve, shade the area you want, and rewrite it using the symmetry of the curve and the fact that the total area is 1.

Worked example A negative z
\(X \sim \text{N}(50, 4^2)\). Find \(\text{P}(X > 45)\).
  1. \(z = \dfrac{45 - 50}{4} = -1.25\).
  2. The area to the right of \(-1.25\) equals, by symmetry, the area to the left of \(+1.25\).
  3. \(\Phi(1.25)\).
Answer: 0.8944
Worked example Between two values
\(X \sim \text{N}(50, 4^2)\). Find \(\text{P}(46 < X < 53)\).
  1. \(z_1 = \dfrac{46 - 50}{4} = -1\) and \(z_2 = \dfrac{53 - 50}{4} = 0.75\).
  2. \(\text{P} = \Phi(0.75) - \Phi(-1) = \Phi(0.75) - [1 - \Phi(1)]\).
  3. \(= 0.7734 - (1 - 0.8413) = 0.7734 - 0.1587\).
Answer: 0.6147, or 0.615 (3 s.f.)
Your turn
\(X \sim \text{N}(120, 15^2)\). Find \(\text{P}(X > 130)\).
Examiner trap

Writing \(\Phi(0.75) - \Phi(1)\) for \(\text{P}(46 < X < 53)\). The left end has \(z = -1\), not \(+1\); a sketch shows the area must be bigger than \(\Phi(0.75) - 0.5\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

\(X \sim \text{N}(40, 25)\). What is the standardised value of \(x = 48\)?

Quick check+10 XP

Given \(\Phi(1.2) = 0.8849\), find \(\Phi(-1.2)\).

Quick check+10 XP

About what percentage of a normal distribution lies within 2 standard deviations of the mean?

Quick check+10 XP

\(X \sim \text{N}(10, 9)\). What is the standard deviation of \(X\)?

Section 03

Working backwards.

When the probability is given and a value is wanted, use the table in reverse (or the critical values table: 0.674, 0.842, 1.036, 1.282, 1.645, 1.960, 2.326, 2.576). Decide from a sketch whether \(z\) is positive or negative BEFORE writing the equation \(\dfrac{x - \mu}{\sigma} = z\).

Worked example Finding a value
\(X \sim \text{N}(50, 4^2)\). Find \(k\) such that \(\text{P}(X > k) = 0.1\).
  1. The top 10% is above \(k\), so \(\Phi(z) = 0.9\) and \(z = 1.282\) (positive: \(k\) is above the mean).
  2. \(\dfrac{k - 50}{4} = 1.282\).
  3. \(k = 50 + 4 \times 1.282 = 55.128\).
Answer: \(k = 55.1\) (3 s.f.)
Worked example Finding \(\sigma\)
\(X \sim \text{N}(30, \sigma^2)\) and \(\text{P}(X < 35) = 0.8\). Find \(\sigma\).
  1. \(\Phi(z) = 0.8\) gives \(z = 0.842\).
  2. \(\dfrac{35 - 30}{\sigma} = 0.842\).
  3. \(\sigma = \dfrac{5}{0.842} = 5.938\).
Answer: \(\sigma = 5.94\) (3 s.f.)
Examiner trap

Using \(z = 0.1\) or \(z = 0.8\) in the equation. Those are probabilities; you must convert them to \(z\)-values with the table first.

Section 04

Two unknowns, two equations.

If neither \(\mu\) nor \(\sigma\) is known, each given probability produces one equation \(x = \mu + z\sigma\). Solve them simultaneously, taking great care with the sign of each \(z\).

Worked example Mean and standard deviation
\(X\) is normally distributed. \(\text{P}(X < 20) = 0.15\) and \(\text{P}(X > 40) = 0.05\). Find \(\mu\) and \(\sigma\).
  1. 20 is in the lower tail: \(\Phi(z) = 0.85\) gives 1.036, so \(z = -1.036\) and \(\mu - 1.036\sigma = 20\).
  2. 40 is in the upper tail: \(\Phi(z) = 0.95\) gives \(z = 1.645\), so \(\mu + 1.645\sigma = 40\).
  3. Subtract: \(2.681\sigma = 20\), so \(\sigma = 7.460\).
  4. \(\mu = 20 + 1.036 \times 7.460 = 27.73\).
Answer: \(\mu = 27.7\), \(\sigma = 7.46\) (3 s.f.)
Worked example Expected numbers
The heights of 400 adults are modelled by \(\text{N}(165, 6^2)\) cm. How many would you expect to be taller than 175 cm?
  1. \(z = \dfrac{175 - 165}{6} = 1.667\).
  2. \(\text{P}(X > 175) = 1 - \Phi(1.667) = 1 - 0.9522 = 0.0478\).
  3. \(400 \times 0.0478 = 19.1\).
Answer: About 19 adults
Examiner trap

Giving both \(z\)-values the same sign. A value below the mean always has a negative \(z\); a quick sketch with 20 to the left of \(\mu\) and 40 to the right prevents this.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

\(X \sim \text{N}(120, 15^2)\). Find \(\text{P}(X > 130)\).

Exam style+20 XP

\(X \sim \text{N}(50, 16)\). Find \(\text{P}(46 < X < 53)\).

Exam style+20 XP

\(X \sim \text{N}(30, \sigma^2)\) and \(\text{P}(X < 35) = 0.8\). Find \(\sigma\).

Exam style+20 XP

\(X\) is normally distributed with \(\text{P}(X < 20) = 0.15\) and \(\text{P}(X > 40) = 0.05\). Find \(\mu\).

Skill unlocked

Normal distribution, sorted.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Standardising, symmetry, probabilities between two values, finding a value, finding \(\mu\) or \(\sigma\), both unknowns, and a context with expected numbers.