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The normal distribution questions
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.
6 original questions · 21 marks · the the normal distribution notes · Statistics
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Describe the shape of a normal distribution in words: state its symmetry, and roughly what proportion of values fall within one and within two standard deviations of the mean.
Worked answer
A single symmetric peak at the mean, falling away identically on both sides, with the mean, median and mode all at the same place. About 68% of values lie within one standard deviation of the mean and about 95% within two. B1 for the symmetry about the mean, B1 for 68%, B1 for 95%. These two figures are worth memorising, because they catch a calculator answer that is wildly out before it costs anything.X ~ N(60, 52). Find P(X < 70) to 3 significant figures.
Worked answer
Standardise: z = (70 − 60)/5 = 2, so P(X < 70) = P(Z < 2) = 0.977. M1 for standardising, A1 for 0.977. Two standard deviations above the mean leaves only about 2.3% in the upper tail, so an answer just below 1 is what the sketch predicts. Write the standardising line out even when the calculator will take μ and σ directly, because the method mark is attached to it.Masses are modelled by X ~ N(60, 102) grams. Find P(65 < X < 75) to 3 significant figures.
Worked answer
Standardise both ends: z = (65 − 60)/10 = 0.5 and z = (75 − 60)/10 = 1.5. Then P(0.5 < Z < 1.5) = P(Z < 1.5) − P(Z < 0.5) = 0.9332 − 0.6915 = 0.242. M1 for standardising both ends, M1 for the subtraction, A1 for 0.242. Subtract the smaller cumulative probability from the larger; adding them, or subtracting the z values, are both worth nothing. Shade the strip on a sketch first and the subtraction writes itself.X ~ N(40, σ2) and P(X > 46) = 0.05. Find σ to 3 significant figures.
Worked answer
The top 5% of the standard normal starts at z = 1.6449. So (46 − 40)/σ = 1.6449, giving σ = 6/1.6449 = 3.65. B1 for z = 1.6449, M1 for the standardising equation, M1 for rearranging, A1 for 3.65. Check forwards: 46 is then 1.64 standard deviations above the mean, which cuts off 5%, as stated. Quote the z value before forming the equation, and note that z here is positive because 46 lies above the mean.X ~ N(μ, 42) and P(X < 50) = 0.975. Use z = 1.96 to find μ.
Worked answer
A probability of 0.975 below 50 puts 50 at z = +1.96, so (50 − μ)/4 = 1.96 and μ = 50 − 7.84 = 42.16. M1 for the standardising equation, M1 for rearranging, A1 for 42.16. The value 50 sits above the mean, not below it. Reading 0.975 as a lower-tail area is what keeps the sign of z honest, and a negative z here would give μ = 57.84, comfortably above a value that 97.5% of the distribution falls below.X ~ N(μ, σ2), with P(X < 20) = 0.10 and P(X > 35) = 0.05. Find μ and σ.
Worked answer
Turn each probability into a z value first. The bottom 10% ends at z = −1.2816, so (20 − μ)/σ = −1.2816. The top 5% begins at z = 1.6449, so (35 − μ)/σ = 1.6449. Rewrite them as 20 = μ − 1.2816σ and 35 = μ + 1.6449σ. Subtracting removes μ: 15 = 2.9265σ, so σ = 5.13. Then μ = 20 + 1.2816 × 5.1257 = 26.6. Check both statements before finishing: (35 − 26.57)/5.126 = 1.645, as required. B1 B1 for the two z values, M1 for the pair of simultaneous equations, M1 for the elimination, A1 for σ, A1 for μ. Sign is where this question is won and lost. The 20 is below the mean, so its z must be negative. Writing +1.2816 there gives σ = 41.3 and a negative mean, which is plainly impossible for a variable whose values sit around 20 to 35.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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