OxfordAQA International AS and A-level Mathematics: Mean and variance of continuous variables and the exponential distribution (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 follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section S2.2 (second part: mean and variance of continuous random variables and of sums of independent variables) and section S2.3 (exponential distribution). The specification places this content in International A-level Unit S2 (Statistics); it is not examined at International AS. Read the study guide first if the topic is new. Course pages: hub and checklist.

Questions

1. The random variable X has pdf f(x) = x³/4 for 0 ≤ x ≤ 2, and f(x) = 0 otherwise.

(a) Find E(X). [2] (b) Find Var(X). [3]

2. The random variable X has pdf f(x) = 2/x² for 1 ≤ x ≤ 2, and f(x) = 0 otherwise.

(a) Show that E(X) = 2 ln 2. [3] (b) Find the exact value of E(6X⁻¹). [3] (c) Find Var(X), giving your answer to 3 significant figures. [3]

3. The random variable X has E(X) = 7.5 and Var(X) = 2.25. Find E(4 − 2X), Var(4 − 2X) and the standard deviation of 4 − 2X. [3]

4. The times, in minutes, taken by two independent stages of a process are X and Y. E(X) = 20, Var(X) = 6.25, E(Y) = 8 and Var(Y) = 2.25.

(a) Find the mean and variance of X + Y. [2] (b) Find the mean and variance of X − Y. [2] (c) Explain why your two variances are equal. [1]

5. A sensor gives readings R with mean 2.4 and variance 0.09. Five independent readings R₁, …, R₅ are added to give S. A second quantity is T = 5R, using a single reading.

(a) Find E(S) and Var(S). [2] (b) Find E(T) and Var(T), and say why the variances differ. [2]

6. Faults occur at random along a long cable at an average rate of 0.25 per kilometre. The distance in kilometres between consecutive faults is D.

(a) State the distribution of D and write down its mean. [2] (b) Find P(D > 6). [2] (c) Find P(1 < D < 5). [3]

7. The time X minutes between customers entering a shop has an exponential distribution with mean 12.

(a) Find P(X < 8). [3] (b) Find the value t such that P(X > t) = 0.05. [2]

8. The random variable X has cumulative distribution function F(x) = 1 − e^(−0.2x) for x ≥ 0.

(a) Find the median of X. [2] (b) Write down E(X) and the standard deviation of X. [1] (c) Find P(X > 7). [2]

9. The random variable X has an exponential distribution with parameter 0.25. Find E(2X + 3), E(5X²) and Var(4X − 5). [5]

10. The random variable X has pdf f(x) = 3x(8 − x)/256 for 0 ≤ x ≤ 8, and f(x) = 0 otherwise.

(a) Use integration to find E(X). [3] (b) Find Var(X). [4] (c) Find E(3X² − 10). [2] (d) Find Var(5 − 2X). [2]

11. Customers arrive at a stall at random, at an average rate of 1.5 per hour. The time in hours between consecutive arrivals is T.

(a) Find P(T > 1). [2] (b) Find the mean and the standard deviation of T, in minutes. [2] (c) The number of arrivals in a 2-hour period is N. Find P(N ≥ 2). [3] (d) T₁ and T₂ are independent gaps with the distribution of T. Find the mean and the standard deviation of T₁ − T₂, in hours. [3] (e) Find Var(2T₁) and Var(T₁ + T₂), and comment on the difference. [2]

Answers

1. (a) E(X) = ∫ x · x³/4 dx from 0 to 2 [1] = [x⁵/20] = 32/20 = 8/5 [1] (b) E(X²) = ∫ x⁵/4 dx = [x⁶/24] from 0 to 2 = 64/24 = 8/3 [1]; Var(X) = 8/3 − (8/5)² [1] = 8/75 (0.107) [1] [5] Examiner insight: Subtracting μ² rather than μ at the last step is a common slip, so write the formula before substituting.

2. (a) E(X) = ∫ x · 2/x² dx from 1 to 2 = ∫ 2/x dx [1] = [2 ln x] [1] = 2 ln 2 − 2 ln 1 = 2 ln 2 [1] (b) E(6X⁻¹) = ∫ (6/x)(2/x²) dx = 12 ∫ x⁻³ dx [1] = 12 [−1/(2x²)] from 1 to 2 [1] = 12(−1/8 + 1/2) = 9/2 [1] (c) E(X²) = ∫ x² · 2/x² dx = ∫ 2 dx = 2 [1]; Var(X) = 2 − (2 ln 2)² [1] = 0.0782 [1] [9] Examiner insight: In a “show that” part, display the integrand after simplifying and the integrated form; jumping to the printed answer leaves the reader nothing to credit.

3. E(4 − 2X) = 4 − 2(7.5) = −11 [1]; Var(4 − 2X) = (−2)² × 2.25 = 9 [1]; standard deviation = √9 = 3 [1] [3] Examiner insight: The constant 4 drops out of the variance and the −2 is squared, so a negative variance signals an error.

4. (a) E(X + Y) = 20 + 8 = 28 [1]; Var(X + Y) = 6.25 + 2.25 = 8.5 [1] (b) E(X − Y) = 20 − 8 = 12 [1]; Var(X − Y) = 6.25 + 2.25 = 8.5 [1] (c) The variances add in both cases, because (−1)² = 1 and subtracting Y does not remove its variability (the variables are independent). [1] [5] Examiner insight: Mention independence when you add variances; the formula depends on it and stating it shows you know why.

5. (a) E(S) = 5 × 2.4 = 12 [1]; Var(S) = 5 × 0.09 = 0.45 [1] (b) E(T) = 5 × 2.4 = 12 [1]; Var(T) = 25 × 0.09 = 2.25. The variances differ because T repeats one reading five times, which magnifies its error, while S combines five independent readings whose errors partly cancel. [1] [4] Examiner insight: Give a reason in words as well as the numbers when a question asks you to say why.

6. (a) Events at random at a constant rate give exponential gaps, so D ~ Exp(0.25) [1], with mean 1/0.25 = 4 km [1] (b) P(D > 6) = e^(−0.25 × 6) [1] = e^(−1.5) = 0.223 [1] (c) P(1 < D < 5) = e^(−0.25) − e^(−1.25) [2] = 0.7788 − 0.2865 = 0.492 [1] [7] Examiner insight: For an interval, subtract the two upper-tail values with the smaller x first; reversing them gives a negative probability.

7. λ = 1/12. (a) P(X < 8) = 1 − e^(−8/12) [2] = 1 − e^(−2/3) = 0.487 [1] (b) e^(−t/12) = 0.05 [1], t = 12 ln 20 = 35.9 [1] [5] Examiner insight: Convert the mean to λ = 1/mean at the start; using 12 as the rate is the most common source of wrong answers here.

8. (a) 1 − e^(−0.2x) = 0.5 so e^(−0.2x) = 0.5 [1]; x = ln 2 / 0.2 = 3.47 [1] (b) λ = 0.2, so E(X) = 5 and the standard deviation is 5 [1] (c) P(X > 7) = 1 − F(7) [1] = e^(−1.4) = 0.247 [1] [5] Examiner insight: P(X > x) is 1 − F(x), not F(x); say which tail the question asks for before you substitute.

9. E(X) = 1/0.25 = 4 and Var(X) = 1/0.25² = 16 [1]. E(2X + 3) = 2(4) + 3 = 11 [1]. E(X²) = Var(X) + (E(X))² = 16 + 16 = 32 [1]; E(5X²) = 5 × 32 = 160 [1]. Var(4X − 5) = 16 × 16 = 256 [1] [5] Examiner insight: Show how you got E(X²) from the variance and mean, as it is not equal to the square of the mean.

10. (a) E(X) = ∫ 3x²(8 − x)/256 dx from 0 to 8 [1] = (3/256) [8x³/3 − x⁴/4] [1] = (3/256)(4096/3 − 1024) = (3/256)(1024/3) = 4 [1] (b) E(X²) = (3/256) ∫ (8x³ − x⁴) dx [1] = (3/256) [2x⁴ − x⁵/5] from 0 to 8 [1] = (3/256)(8192 − 6553.6) = 96/5 [1]; Var(X) = 96/5 − 16 = 16/5 (3.2) [1] (c) E(3X² − 10) = 3(96/5) − 10 [1] = 238/5 (47.6) [1] (d) Var(5 − 2X) = (−2)² Var(X) [1] = 4 × 16/5 = 64/5 (12.8) [1] [11] Examiner insight: Part (b) is long, so keep exact fractions throughout; early rounding of E(X²) feeds into the variance and its multiples.

11. (a) λ = 1.5 and P(T > 1) = e^(−1.5) [1] = 0.223 [1] (b) Mean = 1/1.5 hours = 40 minutes [1]; the standard deviation is also 40 minutes [1] (c) N is Poisson with mean 1.5 × 2 = 3 [1]; P(N ≥ 2) = 1 − P(0) − P(1) = 1 − e^(−3)(1 + 3) [1] = 0.801 [1] (d) E(T₁ − T₂) = 2/3 − 2/3 = 0 [1]; Var(T₁ − T₂) = 2 × (2/3)² = 8/9 [1]; standard deviation = √(8/9) = 0.943 hours [1] (e) Var(2T₁) = 4 × 4/9 = 16/9 (1.78) and Var(T₁ + T₂) = 2 × 4/9 = 8/9 (0.889) [1]. Doubling one gap multiplies its variance by 4, whereas adding a second independent gap only doubles the variance, so Var(2T₁) is larger [1] [12] Examiner insight: Part (c) is a Poisson question in an exponential setting; say that you are using a Poisson count with mean λt so the method is clear.

Where marks are usually lost

• Forgetting to subtract μ² when finding a variance from E(X²).
• Using E(1/X) = 1/E(X). Integrate (1/x) f(x) instead.
• Computing Var(aX + b) as aVar(X) + b.
• Subtracting variances for a difference of independent variables.
• Treating T₁ + T₂ and 2T₁ as the same variable.
• Using the mean as λ in an exponential question instead of λ = 1/mean.
• Leaving the rate and the time in different units.
• Using F(x) when the question asks for an upper tail, or the reverse.
• Rounding before the final line, especially in long integrals.

Next steps

• Revise with the revision notes and the study guide.
• Practise finding probabilities from pdfs and cdfs with the probability density and cumulative distribution practice.
• Go to the OxfordAQA A-level Mathematics hub or print the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA: sections S2.2 (second part) and S2.3.