Maths › Statistics › The normal distribution
The normal distribution
Heights, masses, measurement errors: continuous quantities cluster around a mean and thin out symmetrically, and the normal curve is the standard model for that shape. Every question reduces to areas under one bell.
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.
WHAT YOU PROBABLY THINK
About half of a normal population lies more than one standard deviation from the mean.
The shape of natural variation
X ~ N(μ, σ²) says X is continuous, symmetric about its mean μ, and spread by its standard deviation σ. The curve's points of inflection sit at μ ± σ, which is how σ is read off a sketch. Roughly 68% of values fall within one σ of the mean, 95% within two, 99.7% within three: the rules of thumb that make answers checkable.
For a continuous variable, P(X = 40) exactly is zero; only intervals carry probability, so P(X < 40) and P(X ≤ 40) are the same thing. Every normal question is an area question, and a small sketch with the area shaded is the cheapest insurance in the paper.
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 to see the answer's size: 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.).
The rule-of-thumb check: beyond 1σ is about 16%, beyond 2σ about 2.5%, and 6.7% sits sensibly between.
The inverse normal runs the same machine backwards: given the area, find the value. For P(X < x) = 0.9 with the model above, the calculator returns x ≈ 55.1 g. Feed inverse-normal the area to the left; for “top 10%”, use 0.9, not 0.1.
Finding μ or σ
When μ or σ is unknown, the calculator cannot help until the problem is standardised: convert the known probability into a z-value with the inverse normal on N(0, 1), then solve the resulting equation.
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 forwards: 35 is then about 2.05 standard deviations above 30, and the area beyond 2.05σ is indeed about 2%.
Two unknown parameters need two given probabilities: standardise both, and two simultaneous equations in μ and σ fall out.
THE EXAM BIT
- Sketch and shade before calculating; the direction of the tail is where marks die.
- Inverse-normal takes the area to the left: convert “top 15%” to 0.85 first.
- Unknown-parameter questions must show the standardising step; calculator-only answers drop the method marks.
- Quote probabilities to 3 s.f. and z-values to at least 4, so the final answer survives rounding.
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 bell is symmetric about μ, with inflection points one σ out and 95% of it within two.
Standardise with Z = (X − μ)/σ whenever a parameter is unknown; the calculator handles everything else.
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.
CHECK YOUR PROGRESS
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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.
No animated video for this topic yet; these notes stand alone.