OCR A Level Mathematics A: The normal distribution and choosing a distribution (H240) – 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 follow the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, references 2.04e to 2.04h. All of this content is Stage 2, so it is A Level only. Give answers to 3 significant figures unless a question says otherwise, and show your working.

Links: OCR Mathematics A hub, printable checklist, study guide, revision notes and free diagnostics.

Questions

1. The scores on a reasoning test are modelled by X ~ N(72, 5²). (a) Find P(X < 65). [2] (b) Find P(68 < X < 79). [3]

2. A normal curve has points of inflection at x = 14 and x = 26. (a) Find the mean and the standard deviation. [2] (b) State the interval that contains about 95% of the values. [1] (c) State the approximate proportion of values between 14 and 26. [1]

3. X ~ N(55, 8²). (a) Find k such that P(X < k) = 0.92. [3] (b) Find m such that P(X > m) = 0.2. [2]

4. X ~ N(40, σ²) and P(X > 46) = 0.1. Find σ. [4]

5. X ~ N(μ, σ²), P(X < 12) = 0.15 and P(X > 20) = 0.25. Find μ and σ. [6]

6. The mass of a parcel handled by a depot is modelled by X ~ N(2.8, 0.35²) kg. (a) Find the probability that a parcel has mass over 3.5 kg. [2] (b) Of 600 parcels, how many would you expect to exceed 3.5 kg? [1] (c) The heaviest 5% of parcels are sent by a different route. Find the minimum mass of these parcels. [2]

7. Eight parcels from question 6 are chosen at random. Use the unrounded probability from question 6(a). (a) Find the probability that exactly 2 exceed 3.5 kg. [3] (b) State one assumption needed for your method in part (a). [1]

8. For each situation, name a suitable distribution and give a reason from the context. (a) The number out of 40 sown seeds that germinate, when each germinates with probability 0.9, independently. [2] (b) The heights of 800 adults, shown by a symmetrical bell-shaped histogram. [2] (c) The time customers wait at a counter, with mean 3 minutes and standard deviation 3 minutes, shown by a histogram with a long right tail. [2]

9. The waiting time T minutes at a clinic is modelled by T ~ N(18, 11²). (a) Find P(T < 0). [2] (b) What does your answer say about the model? [2]

10. X ~ B(150, 0.4). (a) Write down a normal distribution that could approximate X. [2] (b) Using this model, state the interval within which about 95% of values lie. [1] (c) Explain why a normal model suits this binomial distribution, and why P(X = 60) needs more care. [2]

11. The volume X ml of juice in a carton is modelled by X ~ N(μ, σ²). Of the cartons, 3% hold less than 245 ml and 8% hold more than 262 ml. (a) Find μ and σ. [4] (b) Find the proportion of cartons with less than 250 ml. [2] (c) A pack contains 12 cartons chosen at random. Find the probability that at least 2 hold less than 250 ml. [2] (d) The firm changes only the mean so that 1% of cartons hold less than 245 ml. Find the new mean. [2] Use unrounded values from part (a) in the later parts.

Answers

1. (a) z = (65 − 72)/5 = −1.4 [1]. P(X < 65) = 0.0808 [1] [2] (b) z = −0.8 and z = 1.4 [1]. P(68 < X < 79) = P(X < 79) − P(X < 68) [1]. = 0.707 [1] [3] Examiner insight: Write the distribution and the probability statement before the number, so your method is visible even if the calculator entry slips.

2. (a) μ is halfway between the inflection points: μ = (14 + 26)/2 = 20 [1]. Inflection points are μ ± σ, so 2σ = 12 and σ = 6 [1] [2] (b) μ ± 2σ gives 8 to 32 [1] [1] (c) The interval 14 to 26 is μ ± σ, so about two-thirds [1] [1] Examiner insight: Explain which fact you are using. A bare “20 and 6” gives the reader no way to follow your reasoning if one value is wrong.

3. (a) P(X < k) = 0.92, so z = 1.4051 [1]. k = 55 + 1.4051 × 8 [1]. k = 66.2 [1] [3] (b) P(X < m) = 0.8, so z = 0.8416 [1]. m = 55 + 0.8416 × 8 = 61.7 [1] [2] Examiner insight: Sketch the shaded area. Part (b) is an upper tail, and the most common slip is entering 0.2 where 0.8 belongs.

4. P(X < 46) = 0.9 [1]. z = 1.2816 [1]. (46 − 40)/σ = 1.2816 [1]. σ = 6/1.2816 = 4.68 [1] [4] Examiner insight: Show the equation with the z-value in it. Rounding z to 1.28 before dividing can change the third significant figure.

5. P(X < 12) = 0.15 gives z = −1.0364 [1]. 12 = μ − 1.0364σ [1]. P(X < 20) = 0.75 gives z = 0.6745 [1]. 20 = μ + 0.6745σ [1]. Subtract: 8 = 1.7109σ, so σ = 4.68 [1]. μ = 12 + 1.0364 × 4.6758 = 16.8 [1] [6] Examiner insight: Subtract the equations to find σ first, then substitute for μ. Keep unrounded z-values until the final line.

6. (a) z = (3.5 − 2.8)/0.35 = 2 [1]. P(X > 3.5) = 0.0228 [1] [2] (b) 600 × 0.02275 = 13.65, so about 14 parcels [1] [1] (c) P(X < w) = 0.95, so z = 1.6449 [1]. w = 2.8 + 1.6449 × 0.35 = 3.38 kg [1] [2] Examiner insight: Say that the answer in part (b) is an expected value, and round it in context. Part (c) is a boundary, so include the unit.

7. (a) Let Y be the number exceeding 3.5 kg, so Y ~ B(8, 0.02275) [1]. P(Y = 2) = 8C2 × (0.02275)² × (0.97725)⁶ [1]. = 0.0126 [1] [3] (b) The parcels are independent, with the same probability of exceeding 3.5 kg [1] [1] Examiner insight: An assumption must be about the model, in context. “The numbers are random” is too vague.

8. (a) Binomial B(40, 0.9) [1]. There is a fixed number of seeds, two outcomes and a constant, independent probability of success [1] [2] (b) Normal [1]. The data are continuous, symmetrical and bell-shaped, with one central peak [1] [2] (c) Normal is not suitable [1]. The histogram is skewed to the right, and a mean equal to the standard deviation would give a lot of probability below zero, which is impossible for a waiting time [1] [2] Examiner insight: Name the model and tie the reason to the wording of the context. A reason that fits any situation earns little.

9. (a) z = (0 − 18)/11 = −1.636 [1]. P(T < 0) = 0.0509 [1] [2] (b) The model gives about a 5% chance of a negative waiting time [1]. That is impossible, so the normal model is not a good fit for this variable [1] [2] Examiner insight: State the numerical result, then draw the conclusion about the model in context. A calculation with no comment misses the point of the question.

10. (a) μ = np = 60 [1]. σ² = npq = 150 × 0.4 × 0.6 = 36, so the model is N(60, 36) [1] [2] (b) σ = 6, so μ ± 2σ gives 48 to 72 [1] [1] (c) n is large and p is not close to 0 or 1, so the binomial distribution is roughly symmetrical and bell-shaped [1]. The normal variable is continuous, so P(X = 60) = 0, whereas the binomial has positive probability at 60. A continuity correction treats “X = 60” as the interval 59.5 to 60.5 [1] [2] Examiner insight: Give two linked ideas, one about why the shape suits a normal curve and one about discrete against continuous. Do not just restate the formulae.

11. (a) P(X < 245) = 0.03 gives z = −1.8808, so 245 = μ − 1.8808σ [1]. P(X < 262) = 0.92 gives z = 1.4051, so 262 = μ + 1.4051σ [1]. Subtract: 17 = 3.2859σ, so σ = 5.17 [1]. μ = 245 + 1.8808 × 5.1737 = 255 (254.7) [1] [4] (b) z = (250 − 254.73)/5.1737 = −0.914 [1]. P(X < 250) = 0.180 [1] [2] (c) Let Y be the number below 250 ml, so Y ~ B(12, 0.18026) [1]. P(Y ≥ 2) = 1 − P(Y ≤ 1) = 0.665 [1] [2] (d) P(X < 245) = 0.01 gives z = −2.3263 [1]. New mean = 245 + 2.3263 × 5.1737 = 257 ml [1] [2] Examiner insight: In a multi-part question, use the unrounded values from part (a) in later parts. Starting later parts from 255 and 5.17 can move the third significant figure.

Where marks are usually lost

• Entering the variance in the calculator where the standard deviation is needed.
• Using the area to the right as the input to the inverse normal function.
• Rounding z-values to two decimal places in a two-unknown problem.
• Skipping the sketch, then shading the wrong tail.
• Naming a model with no reason, or giving a reason that does not mention the context.
• Calling a distribution normal because the data are “random”, rather than continuous, symmetrical and bell-shaped.
• Stating a probability for a negative time and not commenting on the model.
• Confusing the two-thirds and 95% intervals, or reading σ as the distance from the mean to the edge of the curve.
• Treating an expected frequency as exact when the answer should be rounded in context.

Next steps

• Recap the ideas in the revision notes.
• Read the full explanations in the study guide.
• See the OCR Mathematics A hub and the printable checklist.
• Take one of the free 10-minute diagnostics to find gaps.
• Book a free trial class: /trial/.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, published by OCR. Section 2.04e to 2.04h of the detailed content.