IB DP Mathematics: Applications and Interpretation – Random variables, the central limit theorem, confidence intervals and the Poisson distribution (HL) 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the HL random variables unit of IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 4.14–4.17, which are all AHL content (HL only). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

The guide requires technology on all three HL papers, so every question is labelled “(calculator allowed)”. Give answers exactly or to 3 s.f.

Learn the methods in the study guide and the revision notes. The IB DP Maths AI course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed, HL only) E(X) = 7 and Var(X) = 2.5. Find E(4 − 3X) and Var(4 − 3X). [2]

2. (calculator allowed, HL only) The masses (kg) of six parcels taken at random are

12.4, 13.1, 11.8, 12.9, 13.5, 12.3

Find unbiased estimates of the population mean and the population variance. [3]

3. (calculator allowed, HL only) X ~ Po(3.6).

(a) Find P(X = 2). [1] (b) Find P(X ≥ 5). [2] (c) Write down the standard deviation of X. [1]

4. (calculator allowed, HL only) For each situation, state whether a binomial, Poisson or normal model is most appropriate, with a reason.

(a) The number of typing errors on a page, where errors occur at random at an average of 0.8 per page. [1] (b) The number of faulty bulbs in a pack of 15, each faulty with probability 0.05, independently. [1] (c) The mass of a bag of flour filled by a machine. [1]

5. (calculator allowed, HL only) The mass of an empty jar is E ~ N(180, 5²) grams. The mass of jam put in a jar is J ~ N(454, 8²) grams. E and J are independent.

(a) Find the probability that a filled jar has total mass more than 650 g. [3] (b) Find the probability that the mass of jam in a jar is more than 2.5 times the mass of the empty jar. [3]

6. (calculator allowed, HL only) Gym visit times X ~ N(72, 9²) minutes. A random sample of 12 visits is taken.

(a) State the distribution of X̄, the sample mean, giving a reason. [2] (b) Find P(70 < X̄ < 75). [2] (c) Find the smallest sample size n for which P(X̄ > 74) < 0.05. [3]

7. (calculator allowed, HL only) Clinic waiting times have mean 4.2 minutes, standard deviation 3.1 minutes and a strongly skewed distribution. A random sample of 50 is taken.

(a) Explain why X̄ is approximately normally distributed. [1] (b) Write down the approximate distribution of X̄. [2] (c) Find the probability that the sample mean exceeds 4.8 minutes. [2]

8. (calculator allowed, HL only) Battery life is normally distributed with standard deviation 1.8 hours. A sample of 20 batteries has mean life 31.4 hours.

(a) Find a 95% confidence interval for the mean battery life. [2] (b) The maker claims a mean life of 33 hours. Comment. [2]

9. (calculator allowed, HL only) Reaction times (ms) are normally distributed. A sample of ten gives

248, 262, 255, 271, 244, 259, 266, 250, 257, 263

(a) Find unbiased estimates of the population mean and variance. [2] (b) Find a 90% confidence interval for the population mean. [1] (c) Explain why you used the distribution you chose in (b). [1] (d) Interpret your interval in context. [1] (e) State what would happen to the width of the interval if the confidence level were raised to 99%. [1]

10. (calculator allowed, HL only) Complaints to Shop A and Shop B follow independent Poisson distributions with means 1.5 per day and 0.9 per day respectively.

(a) Find the probability that Shop A receives exactly 2 complaints on a given day. [1] (b) Find the probability that the two shops together receive at least 4 complaints on a given day. [3] (c) Find the probability that Shop A receives more than 10 complaints in a 5-day week. [2] (d) On one day the shops receive 3 complaints in total. Find the probability that all 3 were at Shop A. [3] (e) Over 30 days, find the probability that there are at least 3 days on which neither shop receives a complaint. [3]

11. (calculator allowed, HL only) A lift has a safe load of 1000 kg. Adult masses are independent and N(78, 11²) kg.

(a) Twelve adults enter the lift. Find the probability that their total mass exceeds 1000 kg. [3] (b) Explain why modelling the total mass as 12X, where X is one adult’s mass, is wrong. [1] (c) Find the largest number of adults for which the probability that the total mass exceeds 1000 kg is less than 0.01. [3] (d) For the twelve adults, find the probability that their mean mass exceeds 80 kg. [2]

Answers

1. E(4 − 3X) = 4 − 3(7) = −17 [1]; Var(4 − 3X) = (−3)² × 2.5 = 22.5 [1] [2] Examiner insight: The −3 is squared and the 4 does not affect the variance; a negative variance earns nothing.

2. x̄ = 76.0/6 = 12.7 (3 s.f.) [1]. From the GDC, sₙ₋₁ = 0.61536… [1], so s²ₙ₋₁ = 0.379 (3 s.f.) [1] [3] Examiner insight: An unbiased estimate of variance needs sₙ₋₁ squared; quoting σx² = 0.316 loses the final accuracy mark.

3. (a) P(X = 2) = 0.177 (3 s.f.) [1] (b) P(X ≥ 5) = 1 − P(X ≤ 4) [1] = 0.294 (3 s.f.) [1] (c) Var(X) = 3.6, so SD = √3.6 = 1.90 (3 s.f.) [1] Examiner insight: Writing 1 − P(X ≤ 5) earns no method mark because it is the wrong complement, even if the GDC use is correct.

4. (a) Poisson: errors occur independently at a uniform average rate, with no fixed maximum [1] (b) Binomial: a fixed number (15) of independent trials with constant p = 0.05 [1] (c) Normal: mass is a continuous measurement [1] Examiner insight: Each mark needs the model and a reason from the context; a name alone scores nothing.

5. (a) T = E + J; E(T) = 180 + 454 = 634 [1]; Var(T) = 25 + 64 = 89 [1]; P(T > 650) = 0.0449 (3 s.f.) [1] (b) Let D = J − 2.5E, so the event is D > 0 [1]. E(D) = 454 − 450 = 4; Var(D) = 64 + 2.5² × 25 = 220.25 [1]. P(D > 0) = 0.606 (3 s.f.) [1] Examiner insight: In (b) the coefficient 2.5 must be squared (6.25 × 25); using 2.5 × 25 is a method error, so later marks are lost.

6. (a) The population is normal, so X̄ is normal for any n [1]. X̄ ~ N(72, 6.75), since 81/12 = 6.75 [1] (b) SD(X̄) = 9/√12 = 2.598… [1]; P(70 < X̄ < 75) = 0.655 (3 s.f.) [1] (c) Need (74 − 72)/(9/√n) > 1.6449 [1], so √n > 7.402 and n > 54.8 [1]. Smallest n = 55 [1] Examiner insight: In (c), rounding 54.8 down to 54 loses the final mark; n = 54 does not satisfy the inequality.

7. (a) n = 50 > 30, so by the central limit theorem X̄ is approximately normal [1] (b) Mean 4.2 [1]; variance 3.1²/50 = 0.1922, so X̄ ≈ N(4.2, 0.1922) [1] (c) SD(X̄) = 3.1/√50 = 0.4384… [1]; P(X̄ > 4.8) = 0.0856 (3 s.f.) [1] Examiner insight: Name the central limit theorem and quote n > 30; “the sample is large” alone may not earn the reasoning mark.

8. (a) σ is known, so use a Z-interval with σ = 1.8, x̄ = 31.4, n = 20 [1]: [30.6, 32.2] (3 s.f.) [1] (b) 33 lies above the interval [1], so the sample does not support the claim; the mean appears lower than 33 hours [1] Examiner insight: “The claim is wrong” with no reference to the interval earns neither mark; compare 33 with the interval, then conclude in context.

9. (a) x̄ = 257.5 [1]; s²ₙ₋₁ = 8.4492…² = 71.4 (3 s.f.) [1] (b) T-interval at 90%: [253, 262] (3 s.f.), that is [252.6, 262.4] to 1 d.p. [1] (c) The population standard deviation is unknown, so the t-distribution is used [1] (d) We are 90% confident that the population mean reaction time lies between 253 ms and 262 ms [1] (e) The interval would become wider [1] Examiner insight: A Z-interval built from sₙ₋₁ loses the mark in (b) and the reasoning mark in (c); σ unknown means t, whatever n is.

10. (a) A ~ Po(1.5): P(A = 2) = 0.251 (3 s.f.) [1] (b) T = A + B ~ Po(2.4) [1]; P(T ≥ 4) = 1 − P(T ≤ 3) [1] = 0.221 (3 s.f.) [1] (c) W ~ Po(5 × 1.5) = Po(7.5) [1]; P(W > 10) = 1 − P(W ≤ 10) = 0.138 (3 s.f.) [1] (d) P(A = 3 and B = 0)/P(T = 3) [1] = (0.12551… × 0.40657…)/0.20901… = 0.051029…/0.20901… [1] = 0.244 (3 s.f.) [1] (e) P(no complaints on a day) = P(T = 0) = e⁻²·⁴ = 0.090718… [1]. Y, the number of such days, ~ B(30, 0.090718) [1]. P(Y ≥ 3) = 1 − P(Y ≤ 2) = 0.520 (3 s.f.) [1] Examiner insight: In (e) state the binomial model with its parameters; a bare GDC value earns at most the final mark.

11. (a) T = X₁ + … + X₁₂; E(T) = 12 × 78 = 936 [1]; Var(T) = 12 × 11² = 1452 [1]; P(T > 1000) = 0.0465 (3 s.f.) [1] (b) 12X is one adult’s mass multiplied by 12 (Var = 17 424); the twelve masses are independent, so Var(T) = 12Var(X), not 12²Var(X) [1] (c) n = 11: T ~ N(858, 1331), P(T > 1000) = 0.0000497 < 0.01 [1]. n = 12: P = 0.0465 > 0.01 [1]. Largest number is 11 [1] (d) X̄ ~ N(78, 121/12), SD = 11/√12 = 3.175… [1]; P(X̄ > 80) = 0.264 (3 s.f.) [1] Examiner insight: To justify “largest” in (c), show the probability for both n = 11 and n = 12.

Where marks are usually lost

• Subtracting variances for a difference such as J − 2.5E.
• Using 12X for the total of 12 separate people.
• Quoting the GDC’s σx instead of Sx when an unbiased estimate of σ² is asked for.
• Entering σ²/n instead of σ/√n into normalcdf.
• Stating “X̄ is normal” with no reason.
• Using a Z-interval with sₙ₋₁; when σ is unknown the guide requires t for any sample size.
• Interpreting a confidence interval without context.
• Not rescaling the Poisson mean (1.5 per day becomes 7.5 per 5-day week).
• Using the wrong complement for “at least” or “more than” with poissoncdf.
• Rounding e⁻²·⁴ to 0.09 before using it.

Next steps

• Review weak spots in the revision notes.
• Re-read worked examples in the study guide.
• See the IB DP Maths AI course hub.
• Tick this unit off the printable syllabus checklist.
• Paper advice: AI exam preparation.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections AHL 4.14, 4.15, 4.16 and 4.17.