OxfordAQA International AS and A-level Mathematics: The normal distribution and sums of normal variables (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section S2.4 (Normal distribution) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit S2 (Statistics); it is not examined at International AS.

Read the study guide or the revision notes first. Use the course hub, the printable checklist and the free 10-minute diagnostics alongside.

Round z-values to 2 decimal places and give probabilities to 4 decimal places unless told otherwise. Where a percentage point is needed, it is given in the question.

Questions

1. The lifetime X of a type of sensor is modelled by X ~ N(72, 8²) months.

(a) State P(X > 72). [1] (b) Write down the approximate proportion of sensors with a lifetime between 64 and 80 months. [1] (c) Write down the approximate proportion of sensors with a lifetime of more than 88 months. [1]

2. The journey time W of a commuter is modelled by W ~ N(36, 5²) minutes.

(a) Find P(W < 42). [2] (b) Find P(30 < W < 42). [3]

3. The marks M in a test are modelled by N(62, 11²). The top 10% of marks receive a distinction. Given that P(Z < 1.2816) = 0.90, find the lowest whole-number mark that receives a distinction. [4]

4. The mass T of a tablet is modelled by N(μ, 4.5²) milligrams. A process is set so that 2.5% of tablets have a mass below 95 mg. Given that P(Z < 1.96) = 0.975, find μ to one decimal place. [4]

5. Y ~ N(85, σ²) and P(Y > 100) = 0.05. Given that P(Z < 1.6449) = 0.95, find σ to 3 significant figures. [4]

6. X ~ N(μ, σ²), P(X < 30) = 0.10 and P(X > 46) = 0.25. Given that P(Z < 1.2816) = 0.90 and P(Z < 0.6745) = 0.75, find μ and σ to 3 significant figures. [6]

7. The mass of a passenger’s hold bag is X ~ N(23, 3²) kg, and the mass of the same passenger’s cabin bag is Y ~ N(17, 2²) kg. The two are independent. Find the probability that the two bags have a total mass of more than 45 kg. [5]

8. A rod has length R ~ N(50.4, 0.3²) mm and a slot has width S ~ N(50.9, 0.4²) mm, independent. The rod fits if the slot is at least 0.2 mm wider than the rod. Find the probability that a randomly chosen rod fits a randomly chosen slot. [5]

9. The mass of a packet of rice is X ~ N(8, 1.2²) hundred grams.

(a) A box holds 5 independent packets. Find the probability that the total mass in a box is more than 43 hundred grams. [3] (b) Find P(5X > 43). [3] (c) Explain why the answers to (a) and (b) are different. [1]

10. The length P of a pipe is modelled by N(μ, σ²) metres. 15% of pipes are shorter than 1.96 m and 20% are longer than 2.04 m. Given that P(Z < 1.0364) = 0.85 and P(Z < 0.8416) = 0.80:

(a) Find μ to 4 significant figures and σ to 3 significant figures. [6] (b) Two pipes chosen independently are joined end to end. Using μ = 2.004 and σ = 0.0426, find the probability that the joined length is between 4.00 m and 4.10 m. [4]

11. X ~ N(20, 4²) and Y ~ N(15, 3²) are independent. A student writes: “X − Y ~ N(20 − 15, 4² − 3²) = N(5, 7)”.

(a) Identify the mistake and state the correct distribution of X − Y. [2] (b) Find P(X < Y). [2]

Answers

1. (a) The distribution is symmetric about the mean 72, so P(X > 72) = 0.5 [1] (b) 64 to 80 is μ ± σ, so about 2/3 [1] (c) 88 = μ + 2σ. About 0.95 lies within 2σ, so the upper tail is about 0.025 [1] Examiner insight: Rewrite each end point as μ plus or minus a number of σ before quoting a proportion. A bare answer with no link to μ ± σ is harder to credit if it is wrong.

2. (a) z = (42 − 36)/5 = 1.2 [1], so P(W < 42) = Φ(1.20) = 0.8849 [1] (b) z = (30 − 36)/5 = −1.2 [1] P(W < 30) = 1 − 0.8849 = 0.1151 [1] P(30 < W < 42) = 0.8849 − 0.1151 = 0.7698 [1] Examiner insight: For an interval, show both standardised values. Using symmetry here (an answer of 2Φ(1.2) − 1) is a quicker route, but show the line.

3. Distinction means P(M > x) = 0.10, so P(M < x) = 0.90 [1] (x − 62)/11 = 1.2816 [1] x = 62 + 14.098 = 76.1 [1] Marks are whole numbers and a mark of 76 is below 76.1, so the lowest whole mark is 77 [1] Examiner insight: Read the last line of the question. The context asks for a whole-number mark, so stopping at 76.1 leaves the final step undone.

4. P(T < 95) = 0.025, so 95 is below the mean. By symmetry z = −1.96 [1] (95 − μ)/4.5 = −1.96 [1] 95 − μ = −8.82 [1] μ = 103.8 mg [1] Examiner insight: Show how the negative sign arises. A positive 1.96 would give a mean below 95 mg, which cannot satisfy a lower tail of 2.5%.

5. P(Y < 100) = 0.95 [1] (100 − 85)/σ = 1.6449 [1] σ = 15/1.6449 [1] σ = 9.12 [1] Examiner insight: Convert the upper tail to a lower tail before choosing the percentage point. Using 0.05 instead of 0.95 gives the wrong sign.

6. P(X < 30) = 0.10, so (30 − μ)/σ = −1.2816 [1] P(X < 46) = 0.75, so (46 − μ)/σ = 0.6745 [1] μ − 1.2816σ = 30 and μ + 0.6745σ = 46 [1] Subtract: 1.9561σ = 16 [1] σ = 16/1.9561 = 8.18 [1] μ = 30 + 1.2816 × 8.1795 = 40.5 [1] Examiner insight: Keep unrounded values in the working until the end. A rounded σ used in the substitution can shift the third significant figure of μ.

7. X + Y is normal [1] with mean 23 + 17 = 40 [1] Variance = 3² + 2² = 13, so sd = 3.6056 [1] z = (45 − 40)/3.6056 = 1.39 [1] P(X + Y > 45) = 1 − Φ(1.39) = 1 − 0.9177 = 0.0823 [1] Examiner insight: Write the full distribution N(40, 13) and state the independence. Marks are best earned by showing the variance as a sum of two squares.

8. Let D = S − R. D is normal with mean 50.9 − 50.4 = 0.5 [1] Variance = 0.4² + 0.3² = 0.25, so sd = 0.5 [1] Required: P(D ≥ 0.2). z = (0.2 − 0.5)/0.5 = −0.6 [1] P = P(Z > −0.6) = Φ(0.6) [1] = 0.7257 [1] Examiner insight: Translate the context into an inequality in one variable first. The variances add even though the means are subtracted.

9. (a) Total ~ N(5 × 8, 5 × 1.2²) = N(40, 7.2) [1] z = (43 − 40)/√7.2 = 1.12 [1] P = 1 − Φ(1.12) = 1 − 0.8686 = 0.1314 [1] (b) 5X ~ N(40, 5² × 1.44) = N(40, 36) [1] z = (43 − 40)/6 = 0.5 [1] P = 1 − 0.6915 = 0.3085 [1] (c) The five packets vary independently and partly cancel, so the total has variance 7.2. Scaling one packet by 5 gives variance 36, so the spread is larger [1] Examiner insight: Both parts have mean 40, so the explanation must refer to the variances (7.2 and 36) and not to the means.

10. (a) P(P < 1.96) = 0.15, so (1.96 − μ)/σ = −1.0364 [1] P(P > 2.04) = 0.20, so P(P < 2.04) = 0.80 and (2.04 − μ)/σ = 0.8416 [1] μ − 1.0364σ = 1.96 and μ + 0.8416σ = 2.04 [1] Subtract: 1.878σ = 0.08 [1] σ = 0.0426 [1] μ = 1.96 + 1.0364 × 0.042599 = 2.004 [1] (b) Joined length ~ N(2 × 2.004, 2 × 0.0426²) = N(4.008, 0.00363) [1] sd = 0.0602; z limits (4.00 − 4.008)/0.0602 = −0.13 and (4.10 − 4.008)/0.0602 = 1.53 [1] P = Φ(1.53) − Φ(−0.13) = 0.9370 − (1 − 0.5517) [1] = 0.4887 [1] Examiner insight: In part (b), use the figures the question tells you to use. Two independent pipes mean variance doubles, so the sd is not 2 × 0.0426.

11. (a) Variances must be added, not subtracted, for a difference of independent variables [1]. The correct distribution is X − Y ~ N(5, 25) [1] (b) P(X < Y) = P(X − Y < 0). z = (0 − 5)/5 = −1 [1] P = 1 − Φ(1) = 1 − 0.8413 = 0.1587 [1] Examiner insight: A response that only says “the formula is wrong” gains less than one that states variances add and gives the corrected distribution.

Where marks are usually lost

• Writing the standard deviation where the variance should be, as in N(35, 4) for a standard deviation of 4.
• Not subtracting from 1 for an upper-tail probability.
• Looking up a negative z directly, or forgetting the symmetry step.
• Choosing the percentage point for the tail probability (0.05) instead of the cumulative one (0.95).
• Wrong sign for z when the stated probability is below 0.5.
• Subtracting variances for a difference of two variables.
• Treating independent copies as a multiple, 5X, or the reverse.
• Giving μ or σ to too few significant figures, or rounding midway through simultaneous equations.
• Stopping at an unrounded value when the context asks for a whole number.

Next steps

• Revisit the revision notes for the methods boxes.
• Reread the study guide for any question you missed.
• Check coverage on the course hub and the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section S2.4: Normal distribution, published by OxfordAQA.