A-Level Maths the normal distribution: standardising, inverse problems and approximations

Where the normal distribution sits in each specification

The key ideas

X ~ N(μ, σ^2) has mean μ and variance σ^2, so the standard deviation is σ. Note that the second parameter is the variance: N(165, 64) has σ = 8. The curve is symmetrical about μ, so P(X < μ) = 0.5 and P(X > μ + a) = P(X < μ - a).

Standardise by converting to Z ~ N(0, 1) with z = (x - μ)/σ. Tables give Φ(z) = P(Z < z). For P(Z > z) use 1 - Φ(z), and for negative z use symmetry: Φ(-z) = 1 - Φ(z). Always sketch the curve and shade the region you need.

For inverse problems, find the z-value that gives the probability, then solve z = (x - μ)/σ for the unknown. With two unknowns, two probabilities give two equations in μ and σ to solve simultaneously. When approximating a binomial B(n, p) by N(np, np(1 - p)), apply a continuity correction: P(X ≥ 30) becomes P(Y > 29.5), because each whole number covers the interval half a unit either side.

Worked example 1: heights are N(165, 8^2). Find the probability a person is taller than 175 cm

1. Standardise: z = (175 - 165)/8 = 1.25.
2. P(X > 175) = P(Z > 1.25) = 1 - Φ(1.25) = 1 - 0.8944.
3. Answer: 0.1056, about 10.6%. A sketch with the tail to the right of 175 confirms a small probability.

Worked example 2: X ~ N(μ, σ^2) with P(X < 20) = 0.1 and P(X > 35) = 0.2. Find μ and σ

1. P(Z < z) = 0.1 gives z = -1.282, so (20 - μ)/σ = -1.282, that is μ - 1.282σ = 20.
2. P(Z > z) = 0.2 means Φ(z) = 0.8, so z = 0.842, giving (35 - μ)/σ = 0.842, that is μ + 0.842σ = 35.
3. Subtract the first equation from the second: 2.124σ = 15, so σ = 7.06.
4. Then μ = 20 + 1.282 × 7.06 = 29.1 (3 s.f.).

Worked example 3 (9709 P5, Edexcel S2): X ~ B(60, 0.4). Use a normal approximation to find P(X ≥ 30)

1. Check the conditions: np = 24 and nq = 36 are both greater than 5.
2. Mean 24, variance np(1 - p) = 60 × 0.4 × 0.6 = 14.4, so Y ~ N(24, 14.4) and σ = 3.795.
3. Continuity correction: P(X ≥ 30) ≈ P(Y > 29.5).
4. z = (29.5 - 24)/3.795 = 1.449.
5. P(Z > 1.449) = 1 - 0.9264 = 0.0736. The exact binomial value is 0.0746, so the approximation is close.

Common mistakes that cost marks

• Dividing by the variance instead of the standard deviation: in N(165, 64), divide by 8, not 64.
• Using the wrong tail, such as Φ(1.25) when P(Z > 1.25) is needed.
• Getting the sign of z wrong in inverse problems when the value is below the mean.
• Missing or wrong continuity corrections, such as P(X ≥ 30) to P(Y > 30.5).
• Not stating the conditions for an approximation when asked.
• Giving only a calculator answer with no standardisation shown. Cambridge 9709 asks for full working when standardising.

Exam technique and how a tutor helps

Write P(X > 175) = P(Z > (175 - 165)/8) = P(Z > 1.25) on one line before giving the number. This line carries method marks. Keep z-values to at least 3 decimal places from the tables or 4 significant figures from the calculator, and give probabilities to 3 or 4 significant figures.

Draw a quick sketch for every question. It takes seconds and prevents the most common error, which is choosing the wrong side of the curve.

In one-to-one lessons a tutor drills the standardising step and the choice of tail with short mixed questions, then moves to two-unknown problems and approximations, which are where most students lose time. Working on the shared whiteboard lets the tutor see each sketch and correct it before it turns into a wrong answer.

Self-check: can you do these?

• X ~ N(50, 16). Find P(X < 46). (Answer: Φ(-1) = 0.1587)
• X ~ N(100, 15^2). Find the value exceeded by 5% of the population. (Answer: 100 + 1.645 × 15 = 124.7)
• Write P(X ≤ 12) with a continuity correction for a normal approximation. (Answer: P(Y < 12.5))
• In N(70, 9), what is σ? (Answer: 3)
• What are the 9709 conditions for approximating B(n, p) by a normal distribution? (Answer: np > 5 and nq > 5)