The normal distribution

The normal distribution is a continuous probability distribution with a symmetrical, bell-shaped curve. It is described by two numbers: the mean μ, which sits at the centre, and the standard deviation σ, which sets how spread out the curve is. It is written X ~ N(μ, σ²). About 68% of values lie within one standard deviation of the mean, 95% within two and 99.7% within three.

What is the normal distribution?

Measure the heights of a thousand adults, or the masses of a thousand apples from one tree, and draw a histogram. Most values cluster round the middle, fewer lie further out, and the shape is roughly symmetrical. The normal distribution is the smooth curve that models data like this. AQA's A-level Maths specification asks students to understand and use it as a model, and to find probabilities with it, linking it to histograms, the mean, standard deviation and points of inflection.

Here x is the number of standard deviations from the mean, and y is the probability density, drawn ten times taller so the shape is clear. The curve is highest at the mean and falls away symmetrically. It changes from curving down to curving up at one standard deviation either side: those are its points of inflection.

Properties of the normal distribution

• Symmetrical about the mean, so the mean, median and mode are all equal.
• Bell-shaped, with points of inflection at μ − σ and μ + σ.
• The total area under the curve is 1, because it covers every possible value.
• The tails never touch the x-axis, though they get extremely close.
• Continuous: probabilities are areas, so P(X = a) for one exact value is 0.

The notation X ~ N(μ, σ²)

The second number is the variance, σ², not the standard deviation. So X ~ N(60, 144) has a mean of 60 and a standard deviation of √144 = 12. Mixing these up is one of the most common errors.

The 68-95-99.7 rule

Example: adult heights in a group are modelled as N(170, 8²), in centimetres. Then about 68% are between 162 and 178 cm, about 95% are between 154 and 186 cm, and almost all, 99.7%, are between 146 and 194 cm. Values more than three standard deviations from the mean are very unusual.

The standard normal distribution and z-scores

The standard normal distribution, Z ~ N(0, 1), has mean 0 and standard deviation 1. Any normal variable can be turned into it by working out a z-score: how many standard deviations a value is from the mean.

If test scores are N(60, 12²), a score of 78 has z = (78 − 60) ÷ 12 = 1.5: it is one and a half standard deviations above the mean. A score of 45 has z = −1.25. Probabilities for Z are written with Φ, so Φ(1.5) means P(Z < 1.5).

Finding probabilities

In A-level exams you use the normal distribution function on your calculator: enter the lower bound, the upper bound, μ and σ. Tables of Φ(z) do the same job. Always sketch the curve and shade the area you want first.

Try it

Probability Calculator

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Test scores are X ~ N(60, 12²). These probabilities are to 4 decimal places:

For a value above the mean, the answer to P(X < x) is more than 0.5; for a value below, less. That quick check catches a lot of mistakes.

Working backwards: the inverse normal

Sometimes the probability is given and you need the value. What score do 90% of students get below? Use the inverse normal function: P(Z < z) = 0.9 gives z = 1.2816, so x = 60 + 1.2816 × 12 = 75.4.

Finding an unknown mean or standard deviation

The masses of bags are normally distributed with σ = 8 g, and 20% weigh less than 50 g. Find the mean. P(Z < z) = 0.2 gives z = −0.8416. Then (50 − μ) ÷ 8 = −0.8416, so μ = 50 + 0.8416 × 8 = 56.7 g.

Try a normal distribution question, then press Mark to check your answer in the course.

The normal approximation to the binomial

A binomial distribution B(n, p) with large n and p not too close to 0 or 1 has a bell shape, so it can be approximated by a normal distribution with the same mean and variance:

A common rule of thumb is that np and n(1 − p) should both be more than 5. Because the binomial is discrete and the normal is continuous, use a continuity correction: treat each whole number as the interval half a unit either side.

Example: X ~ B(100, 0.4). The mean is 100 × 0.4 = 40 and the variance is 100 × 0.4 × 0.6 = 24. To estimate P(X ≤ 45), use P(Y < 45.5) for Y ~ N(40, 24): z = (45.5 − 40) ÷ √24 = 1.123, giving 0.8692. The exact binomial answer is 0.8689, very close. Your calculator can find binomial probabilities exactly, so the approximation matters most when it is asked for or when no calculator function fits.

When is the normal distribution a good model?

AQA expects students to recognise when the normal model may not be appropriate. It suits data that is continuous, roughly symmetrical and bell-shaped, such as heights, masses and measurement errors. It is a poor model for:

• Skewed data, such as incomes or waiting times.
• Data with a hard limit near the mean, such as times that can't be negative but cluster near zero.
• Small discrete counts, such as the number of goals in a match.

A histogram of the data is the quickest check: if it isn't roughly symmetrical and bell-shaped, a normal model won't fit well. The standard deviation calculator finds μ and σ from data, and the histogram calculator shows the shape.

Common mistakes

• Using the variance as the standard deviation. In N(60, 144), σ = 12, not 144.
• Forgetting 1 −. Calculators and tables usually give the area to the left. For P(X > a), subtract from 1 or set the bounds the right way.
• Skipping the sketch. A ten-second sketch shows whether the answer should be above or below 0.5.
• Leaving out the continuity correction when approximating a binomial.
• Giving P(X = a) a value. For a continuous distribution it is always 0.

To practise statistics questions under time pressure, try today's A-level Maths: Statistics short timed paper. Each paper mixes topics. The Probability course builds up to the normal distribution from the basics.

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