AS & A Level · Practice questions · The normal distribution

Sketch, standardise, look up.

Six original Cambridge-style questions on the normal distribution, at Paper 5 standard. Show full details of each standardisation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch first, then \(z = \dfrac{x - \mu}{\sigma}\).

Sketch: mark \(\mu\), the value(s) and the shaded area.
Standardise: divide by \(\sigma\), not \(\sigma^2\); \(z\) to 3 d.p.
Look up: \(\Phi(z)\) is the area to the left; use \(1 - \Phi\) and symmetry for others.
Backwards: probability \(\to z\) from the table, with the sign from the sketch.
Two unknowns: two equations \(x = \mu + z\sigma\), solved together.

Recall

[3 marks]

\(X \sim \text{N}(70, 36)\). Find (a) \(\text{P}(X < 79)\), (b) \(\text{P}(X < 64)\).

\(\sigma = 6\). (a) \(z = \dfrac{79 - 70}{6} = 1.5\): \(\Phi(1.5) = 0.9332\). ✓
(b) \(z = \dfrac{64 - 70}{6} = -1\): \(1 - \Phi(1) = 1 - 0.8413 = 0.1587\). ✓

Calculation

[3 marks]

\(X \sim \text{N}(200, 12^2)\). Find \(\text{P}(185 < X < 220)\).

\(z_1 = \dfrac{185 - 200}{12} = -1.25\), \(z_2 = \dfrac{220 - 200}{12} = 1.667\). ✓
\(\Phi(1.667) - [1 - \Phi(1.25)] = 0.9522 - 0.1056 = 0.8466\). ✓
0.847 (3 s.f.). ✓

Calculation

[3 marks]

\(X \sim \text{N}(\mu, 8^2)\) and \(\text{P}(X > 60) = 0.25\). Find \(\mu\).

60 is above the mean, with \(\Phi(z) = 0.75\), so \(z = 0.674\). ✓
\(\dfrac{60 - \mu}{8} = 0.674\). ✓
\(\mu = 60 - 5.392 = 54.6\) (3 s.f.). ✓

Calculation

[3 marks]

The lengths of some rods are normally distributed with mean 25 cm. 90% of the rods are shorter than 28 cm. Find the standard deviation.

\(\Phi(z) = 0.9\) gives \(z = 1.282\). ✓
\(\dfrac{28 - 25}{\sigma} = 1.282\). ✓
\(\sigma = \dfrac{3}{1.282} = 2.34\) cm (3 s.f.). ✓

Analysis

[5 marks]

The masses of bags of flour are normally distributed. 10% of bags weigh less than 495 g and 5% weigh more than 520 g. Find the mean and standard deviation of the masses.

Lower tail 10%: \(z = -1.282\), so \(\mu - 1.282\sigma = 495\). ✓
Upper tail 5%: \(z = 1.645\), so \(\mu + 1.645\sigma = 520\). ✓
Subtract: \(2.927\sigma = 25\), so \(\sigma = 8.54\) g (3 s.f.). ✓
\(\mu = 495 + 1.282 \times 8.541 = 506\) g (3 s.f.). ✓

Analysis

[6 marks]

The heights of adults in a town are modelled by \(\text{N}(165, 6^2)\) cm.

(a) Find the probability that an adult is taller than 175 cm. (b) In a sample of 400 adults, how many would you expect to be taller than 175 cm? (c) Three adults are chosen at random. Find the probability that exactly one is taller than 175 cm. (d) Find the height exceeded by 20% of adults.

(a) \(z = \dfrac{10}{6} = 1.667\); \(1 - \Phi(1.667) = 1 - 0.9522 = 0.0478\). ✓
(b) \(400 \times 0.0478 = 19.1\), so about 19 adults. ✓
(c) Binomial with \(n = 3\), \(p = 0.0478\): \(3 \times 0.0478 \times 0.9522^2 = 0.130\) (3 s.f.). ✓
(d) \(\Phi(z) = 0.8\) gives \(z = 0.842\); height \(= 165 + 0.842 \times 6 = 170\) cm (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.