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. Probabilities are areas under that curve, and standardising is what brings an unknown mean or standard deviation into the calculation.
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.
- Say when neither model suits the context.
COMMON MISCONCEPTION
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 you read σ off a sketch. Around 68% of values fall within one σ of the mean, 95% within two and 99.7% within three. Those three numbers make every answer checkable in about four seconds.
P(X = 40) is zero for a continuous variable. Only intervals carry probability, so P(X < 40) and P(X ≤ 40) mean the same thing and the difference between them never matters here. A normal probability is always an area, so a small shaded sketch before you calculate is worth the few seconds it takes.
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.).
Rule-of-thumb check: beyond 1σ is about 16%, beyond 2σ about 2.5%, and 6.7% sits sensibly between them.
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. Feed it the area to the left. Asked about the top 10%, you enter 0.9.
Finding μ or σ
When μ or σ is the unknown, the calculator cannot start until the problem is standardised. Turn the given probability into a z-value using 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 really 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
A binomial with a large n is a nuisance to sum by hand, and when p sits near 0.5 the bar chart is very nearly a bell already. Then the normal stands in for it.
Matching the mean and the variance is all the approximation does. There is one repair needed, because a discrete count is being replaced by a continuous curve. Widen every boundary by a half, which is the continuity correction. 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).
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 you use it. Large n, p close to 0.5. And say when neither model fits at all. A normal model allows any value from minus infinity to infinity, so it sits badly on a quantity that cannot be negative and has a long right tail, such as rainfall or waiting time.
ASSESSMENT FOCUS
- Sketch and shade before calculating. The direction of the tail is where marks are most often lost.
- Inverse normal takes the area to the left, so convert “the top 15%” into 0.85 before you type anything.
- Unknown-parameter questions must show the standardising line. A calculator-only answer drops the method marks even when the number is right.
- Quote probabilities to 3 s.f. and z-values to at least 4, so rounding does not affect the final answer.
- Approximation questions want the distribution named, the continuity correction visible, and the conditions stated. All three are separately marked.
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, and 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.
Or read them with their worked answers on the the normal distribution questions page.
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.
- Approximate a binomial by a normal when n is large and p is near 0.5, applying a continuity correction.
- Say when neither model suits the context.
Open the full revision checklist to see every objective in the course in one place.