Maths › Statistics › The normal distribution
The normal distribution
The normal distribution models continuous data that are approximately symmetric and bell-shaped. Calculate probabilities as areas under the curve, standardise values, and determine unknown parameters from stated probabilities.
Builds on The binomial distribution.
IN THIS TOPIC
- Use the shape and symmetry of X ~ N(μ, σ²), including the points of inflection at μ ± σ.
- Find probabilities and inverse-normal values with a calculator.
- Standardise with Z = (X − μ)/σ to find an unknown μ or σ from given probabilities.
- Approximate a binomial by a normal when n is large and p is near 0.5, applying a continuity correction.
- State when neither model suits the context.
COMMON MISCONCEPTION
About half of a normal population lies more than one standard deviation from the mean.
About 68% of a normal population lies within one standard deviation of the mean, so about 32% lies further out, roughly a third rather than a half. The matching figures are 95% within two standard deviations and 99.7% within three, for a distribution that is symmetric and bell-shaped about μ. Use them to check the size of a calculated probability.
The normal curve and when it applies
X ~ N(μ, σ²) says X is continuous, symmetric about its mean μ, with spread measured by its standard deviation σ. The curve's points of inflection are at μ ± σ, which allows σ to be read from a sketch. About 68% of values fall within one σ of the mean, 95% within two and 99.7% within three; use these figures to check the size of a calculated probability.
The normal distribution is not a default model for every continuous quantity. It applies to data that are approximately symmetric and bell-shaped; continuous distributions can also be skewed, bounded, multimodal or heavy-tailed, and a normal model then gives poor probabilities.
P(X = 40) is zero for a continuous variable: only intervals have non-zero probability, so P(X < 40) and P(X ≤ 40) are equal. A normal probability is an area under the curve, so sketch and shade the required region before calculating.
Areas from the calculator
WORKED EXAMPLE
A straightforward tail
Masses are modelled by X ~ N(50, 4²) in grams. Find the probability a mass exceeds 56 g.
Standardise first to see how big the answer should be: z = (56 − 50)/4 = 1.5, so this is the area beyond one and a half standard deviations.
The calculator gives P(X > 56) = 0.0668 (3 s.f.).
Check: the area beyond 1σ is about 16% and beyond 2σ about 2.5%, and 6.7% lies between those values, as expected for z = 1.5.
The inverse normal reverses that process, taking an area and returning a value. For P(X < x) = 0.9 with the model above, the calculator returns x ≈ 55.1 g. It takes the area to the left, so for the top 10% enter 0.9.
Finding μ or σ
When μ or σ is unknown, standardise first. Convert the given probability into a z-value using the inverse normal on N(0, 1), then solve the resulting equation for the unknown parameter.
WORKED EXAMPLE
An unknown standard deviation
X ~ N(30, σ²) and P(X > 35) = 0.02. Find σ.
P(Z > z) = 0.02 gives z = 2.0537 from the standard normal.
So (35 − 30)/σ = 2.0537, giving σ = 5/2.0537 = 2.43 (3 s.f.).
Check by working forwards: 35 is then about 2.05 standard deviations above 30, and the area beyond 2.05σ is about 2%.
Two unknown parameters need two given probabilities. Standardise both to obtain a pair of simultaneous equations in μ and σ.
Approximating a binomial
Summing a binomial with large n by hand is laborious, and when p is close to 0.5 the distribution is approximately symmetric and bell-shaped, so a normal distribution approximates it.
The approximation matches the mean and the variance. Because a discrete count is being replaced by a continuous variable, apply the continuity correction, extending each boundary by a half: P(X ≤ 60) becomes P(Y < 60.5), P(X ≥ 60) becomes P(Y > 59.5), and P(X = 60) becomes P(59.5 < Y < 60.5).
Describe the drawing
The twenty-one bars of B(20, 0.5) with the matching normal curve for N(10, 5) drawn over them, the bars up to twelve coloured and a dashed boundary line standing at twelve and a half: widened by the half-unit continuity correction the normal area is 0.868, agreeing with the exact binomial 0.868, while cutting at twelve would give only 0.814.
WORKED EXAMPLE
Using the approximation
X ~ B(150, 0.45). Use a suitable approximation to estimate P(X ≤ 60).
np = 67.5 and np(1 − p) = 37.125, so Y ~ N(67.5, 37.125) with σ = 6.09.
With the continuity correction, P(X ≤ 60) ≈ P(Y < 60.5), and z = (60.5 − 67.5)/6.09 = −1.149.
That gives 0.125 (3 s.f.). The exact binomial value is 0.1251, so the approximation agrees to 3 significant figures here.
State the conditions when using it: n large and p close to 0.5. State also when neither model is appropriate. A normal model assigns positive probability to every interval of values, negative ones included, so it is unsuitable for a quantity that cannot be negative and has a long right tail, such as rainfall or waiting time.
ASSESSMENT FOCUS
- Sketch and shade the required region before calculating, to fix which tail is wanted.
- The inverse normal takes the area to the left, so convert “the top 15%” into 0.85 before entering it.
- For an unknown parameter, show the standardising line Z = (X − μ)/σ before solving.
- Quote probabilities to 3 significant figures and z-values to at least 4, so that rounding does not affect the final answer.
- When using an approximation, name the approximating distribution, show the continuity correction and state the conditions under which it is used.
CHECK YOURSELF
X ~ N(μ, 6²) and P(X < 82) = 0.975. Use z = 1.96 to find μ.
Show a hint
Standardise, then solve for the one unknown.
Show the answer
(82 − μ)/6 = 1.96, so μ = 82 − 6 × 1.96 = 70.24, about 70.2.
Check: 82 sits nearly two standard deviations above 70.2, matching an area of 0.975.
The normal curve is symmetric about μ, with points of inflection at μ ± σ and about 95% of the distribution within 2σ of the mean.
Use a normal model only for data that are approximately symmetric and bell-shaped, and standardise with Z = (X − μ)/σ whenever a parameter is unknown.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the normal distribution questions page.
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- Use the shape and symmetry of X ~ N(μ, σ²), including the points of inflection at μ ± σ.
- Find probabilities and inverse-normal values with a calculator.
- Standardise with Z = (X − μ)/σ to find an unknown μ or σ from given probabilities.
- Approximate a binomial by a normal when n is large and p is near 0.5, applying a continuity correction.
- State when neither model suits the context.
Open the full revision checklist to see every objective in the course in one place.