AQA A-Level Mathematics: N: Statistical distributions (7357) – 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 N: Statistical distributions (N1 to N3) of the AQA A-level Mathematics (7357) specification, version 1.3, for A-level exams from June 2018 onwards. Section N is Paper 3 content in the specification. A calculator is required in every 7357 paper, so every question here assumes you have one with binomial, Normal and inverse Normal functions. Questions 9 to 11 are extended problems that combine several outcomes. Give answers to 3 significant figures unless a question says otherwise.

Learn the methods first in the Statistical distributions study guide and the Statistical distributions revision notes. The 7357 course hub and the printable checklist show where this topic fits, and the free 10-minute diagnostics help you find weak spots.

Questions

1. A discrete random variable X has probability function P(X = x) = c/x for x = 1, 2, 3, 6.

(a) Show that c = 1/2. [2] (b) Find P(X < 6). [1]

2. A quiz has 25 multiple-choice questions, each with four options. Jo guesses every answer at random. X is the number Jo gets right.

(a) State the distribution of X, and give two conditions needed for it, in context. [3] (b) Find the probability that Jo gets at least 10 right. [2]

3. X ~ B(15, 0.35). Find:

(a) P(X = 5) [2] (b) P(X < 4) [2] (c) P(4 ≤ X ≤ 8) [2]

4. A seed company states that 85% of its seeds germinate. A gardener plants 40 seeds.

(a) Find the probability that at least 36 germinate. [3] (b) Find the probability that fewer than 32 germinate. [2] (c) The seeds are all planted in one tray on a cold windowsill. Explain why a binomial model may not be appropriate. [1]

5. The heights of adult men in a town are modelled by X ~ N(170, 8²), in cm.

(a) Find P(X > 180). [2] (b) Find P(160 < X < 175). [2] (c) Find the height exceeded by 5% of the men. [2]

6. The curve of a Normal distribution has points of inflection at x = 38 and x = 52.

(a) Find the mean and standard deviation of the distribution. [2] (b) Find the probability that a value from this distribution is greater than 50. [2]

7. The time, T minutes, that customers spend in a shop is Normally distributed. 20% of customers spend less than 40 minutes and 10% spend more than 70 minutes. Find the mean and standard deviation of T. [6]

8. A random sample of 400 leaves from a tree has mean length 24.6 mm and standard deviation 1.8 mm. A histogram of the lengths is symmetrical and bell-shaped.

(a) Using a Normal model, estimate how many of the 400 leaves are longer than 27 mm. [3] (b) In the sample, 38 leaves were longer than 27 mm. Comment on the model. [1]

9. For each situation, state whether a binomial model, a Normal model or neither is appropriate, with a reason.

(a) The number of hearts when 5 cards are dealt from one standard pack of 52 cards. [2] (b) The masses of eggs laid at a large farm, shown by a symmetrical histogram. [2] (c) The number of times a dice is rolled until the first six appears. [2] (d) Delivery delays have mean 5 minutes and standard deviation 4 minutes. Find P(X < 0) under N(5, 4²), and use it to comment on a Normal model for the delays. [3]

10. The lengths of bolts made by a machine are modelled by X ~ N(50, 0.4²), in mm. A bolt is rejected if its length is less than 49.3 mm or more than 50.6 mm.

(a) Show that the probability a bolt is rejected is 0.107, to 3 significant figures. [3] (b) Bolts are packed in boxes of 30. Find the probability that a box contains more than 3 rejected bolts. [3] (c) State one assumption needed for your model in (b). [1] (d) The mean stays at 50 mm. Find the standard deviation needed so that only 1% of bolts are longer than 50.6 mm. [3]

11. In a town, 45% of adults hold a library card. A random sample of 150 adults is taken. X is the number with a library card.

(a) Find P(X ≤ 60). [1] (b) Explain why X can be approximated by a Normal distribution, and state its mean and variance. [3] (c) Use the Normal approximation, with a continuity correction, to estimate P(X ≤ 60). [3] (d) Compare your answers to (a) and (c). [1]

Answers

1. (a) c(1 + 1/2 + 1/3 + 1/6) = 1 [1], so 2c = 1 and c = 1/2 [1]. (b) P(X < 6) = (1/2)(1 + 1/2 + 1/3) = 11/12 [1]. Examiner insight: (a) is a “show that”, so the sum of the fractions equal to 1 must appear; writing only “c = 1/2” earns nothing.

2. (a) X ~ B(25, 0.25) [1]. Each guess is independent of the others [1]. The probability of a correct guess is 1/4 for every question, as each has four options [1]. (b) P(X ≥ 10) = 1 − P(X ≤ 9) [1] = 0.0713 [1]. Examiner insight: conditions copied from a textbook without the quiz context usually earn at most one of the two marks.

3. (a) ¹⁵C₅ (0.35)⁵ (0.65)¹⁰ [1] = 0.212 [1]. (b) P(X < 4) = P(X ≤ 3) [1] = 0.173 [1]. (c) P(X ≤ 8) − P(X ≤ 3) [1] = 0.785 [1]. Examiner insight: the method mark is for the correct rewrite; P(X ≤ 4) for (b) loses both marks even with a correct calculator entry.

4. (a) X ~ B(40, 0.85) [1]. P(X ≥ 36) = 1 − P(X ≤ 35) [1] = 0.263 [1]. (b) P(X < 32) = P(X ≤ 31) [1] = 0.135 [1]. (c) Seeds in the same tray share conditions, so if it is cold, germination of one seed is linked to others and they are not independent (or p may be lower than 0.85) [1]. Examiner insight: stating the distribution earns the first mark in (a); a correct probability with no model shown risks losing it.

5. (a) P(X > 180) with σ = 8 [1] = 0.106 [1]. (b) P(160 < X < 175) [1] = 0.628 [1]. (c) P(X < h) = 0.95, inverse Normal [1], h = 183.2 = 183 cm [1]. Examiner insight: in (c), using 0.05 instead of 0.95 gives a value below the mean; a quick check that the answer is above 170 catches this.

6. (a) Inflection points are at μ ± σ, so μ = 45 [1] and σ = 7 [1]. (b) X ~ N(45, 7²) [1], P(X > 50) = 0.238 [1]. Examiner insight: σ is half the distance between the points of inflection; quoting σ = 14 loses the second mark in (a) and the follow-on accuracy in (b).

7. P(T < 40) = 0.2 gives z = −0.8416 [1], so (40 − μ)/σ = −0.8416 [1]. P(T > 70) = 0.1 gives z = 1.2816 [1], so (70 − μ)/σ = 1.2816 [1]. Subtracting: 30 = 2.1232σ, so σ = 14.1 [1], and μ = 51.9 [1]. Examiner insight: the z for 0.2 must be negative; using +0.8416 makes the equations inconsistent and loses the accuracy marks that follow.

8. (a) X ~ N(24.6, 1.8²) [1]. P(X > 27) = 0.0912 [1]. 400 × 0.0912 = 36.5, so about 36 or 37 leaves [1]. (b) 38 is close to the expected 36.5, so the Normal model fits well in this tail [1]. Examiner insight: an expected frequency need not be a whole number, but you lose the final mark if you leave the answer as a probability.

9. (a) Neither binomial nor Normal [1]: cards are dealt without replacement, so the probability of a heart changes and the deals are not independent [1]. (b) Normal [1]: mass is continuous and the data are symmetrical and bell-shaped [1]. (c) Neither [1]: the number of trials is not fixed, so it is not binomial, and the variable is discrete [1]. (d) P(X < 0) = 0.106 [1]. A delay cannot be negative [1], yet the model gives about 11% of deliveries a negative delay, so a Normal model is not appropriate (the data are likely skewed) [1]. Examiner insight: the reason mark needs the context; “p is not constant” without saying why (cards are not replaced) is unlikely to earn it.

10. (a) P(X < 49.3) = 0.0401 [1]. P(X > 50.6) = 0.0668 [1]. Sum = 0.10687 = 0.107 [1]. (b) R ~ B(30, 0.107) [1]. P(R > 3) = 1 − P(R ≤ 3) [1] = 0.401 [1]. (c) Bolts are rejected independently of each other [1]. (d) P(X > 50.6) = 0.01 gives z = 2.3263 [1]. 0.6/σ = 2.3263 [1], σ = 0.258 mm [1]. Examiner insight: (a) is a “show that”, so give both tail probabilities and an answer to more than 3 s.f. before rounding; quoting 0.107 alone earns nothing.

11. (a) 0.125 [1]. (b) n is large and p is close to 0.5 (np = 67.5 and n(1 − p) = 82.5 are both large) [1]. Mean = 150 × 0.45 = 67.5 [1]. Variance = 150 × 0.45 × 0.55 = 37.125 [1]. (c) P(X ≤ 60) ≈ P(Y < 60.5) [1]. z = (60.5 − 67.5)/√37.125 = −1.149 [1], so P = 0.125 [1]. (d) The approximation agrees with the exact value to 3 s.f. [1]. Examiner insight: in (c), using 60 instead of 60.5 loses the continuity-correction mark and gives 0.109, which then fails the accuracy mark too.

Where marks are usually lost

• Putting the variance into the Normal function: N(170, 8²) needs σ = 8, and N(40, 9) needs σ = 3.
• Rewriting “at least” or “more than” one step out for a binomial.
• Stating binomial conditions in general terms with no link to the context.
• Missing the sign of z when the given probability is below the mean.
• Rounding z to 2 decimal places before multiplying by σ in a “find σ” question.
• Using “show that” answers as the only working, so no method can be credited.
• Judging a Normal model without checking for impossible values such as negative times.
• Leaving an expected frequency as a probability.
• Forgetting the continuity correction when using the Normal link to the binomial.

Next steps

• Recap with the Statistical distributions revision notes.
• Re-read the Statistical distributions study guide for any method you missed.
• Review sampling in the Statistical sampling practice questions.
• See every topic on the 7357 course hub.
• Tick off outcomes on the printable checklist.
• Try one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams June 2018 onwards, published by AQA. Section 3.15, N: Statistical distributions.