OxfordAQA International AS and A-level Mathematics: Discrete random variables: expectation and variance (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 S1.2 (Discrete random variables) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. This content is part of International AS, Unit PSM1 (section 3.2, S1). Learn the method first in the study guide or the revision notes. See also the course hub and the checklist.

Questions

1. The discrete random variable X has the distribution below, where k is a constant. Find k and P(X ≥ 2).

x	0	1	2	3
P(X = x)	0.1	k	3k	0.3

[3]

2. The random variable Y has the distribution below. Find E(Y), Var(Y) and the standard deviation of Y, to 3 significant figures where not exact.

y	−1	0	2	5
P(Y = y)	0.2	0.3	0.4	0.1

[5]

3. E(X) = 8 and Var(X) = 5. Find E(2X + 3), Var(2X + 3) and Var(−3X). [4]

4. The random variable R has the distribution below.

r	1	2	5
P(R = r)	0.4	0.4	0.2

(a) Find E(5R²). [2] (b) Find E(10/R). [2] (c) Find 10/E(R) and state whether it equals your answer to part (b). [1]

5. For R in question 4, find Var(6/R). [4]

6. The random variable N has P(N = n) = k/n for n = 1, 2, 4. Find the exact value of k, E(N) and Var(N). [6]

7. X and Y are independent random variables with E(X) = 5, Var(X) = 2, E(Y) = 3 and Var(Y) = 4. (a) Find E(2X − 3Y). [1] (b) Find Var(2X − 3Y). [2] (c) Find the standard deviation of 2X − 3Y to 3 significant figures. [1] (d) State, with a reason, whether Var(2X + 3Y) equals Var(2X − 3Y). [1]

8. A spinner gives a score X of 1 or 3, each with probability 1/2. (a) Find E(X) and Var(X). [2] (b) Let D = 2X. Find E(D) and Var(D). [2] (c) Two independent spins give scores X₁ and X₂. Let T = X₁ + X₂. Find E(T) and Var(T). [2] (d) Explain why D and T have equal means but different variances. [1]

9. A shop sells X laptops per day, with the distribution below. E(X) = 2.1.

x	0	1	2	3	4
P(X = x)	0.1	a	0.3	b	0.1

(a) Find a and b. [4] (b) Find Var(X). [3] (c) Each laptop earns £40 and there is a fixed daily cost of £55, so the daily profit is P = 40X − 55. Find E(P) and the standard deviation of P, to 3 significant figures where not exact. [2]

10. Independent variables X and Y have the distributions below.

x	1	2	3
P(X = x)	0.5	0.3	0.2

y	0	2
P(Y = y)	0.6	0.4

(a) Find E(X) and Var(X). [3] (b) Find E(Y) and Var(Y). [2] (c) Find E(X + Y), Var(X + Y), E(X − Y) and Var(X − Y). [3] (d) Find P(X + Y = 3). [2]

11. A game gives a score Z with P(Z = −2) = 0.3, P(Z = 0) = 0.5 and P(Z = 3) = 0.2. (a) Find E(Z) and Var(Z). [3] (b) A player takes 12 independent turns and T is the total score. Find E(T) and the standard deviation of T. [2] (c) Hence find E(T²). [1]

Answers

1. 0.1 + k + 3k + 0.3 = 1 [1], so 4k = 0.6 and k = 0.15 [1]. P(X ≥ 2) = 3(0.15) + 0.3 = 0.45 + 0.3 = 0.75 [1] [3] Examiner insight: Show the sum-to-1 equation. Writing only the value of k leaves nothing to credit if it is wrong.

2. E(Y) = −0.2 + 0 + 0.8 + 0.5 = 1.1 [1]. E(Y²) = 0.2 + 0 + 1.6 + 2.5 = 4.3 [1]. Var(Y) = 4.3 − 1.1² [1] = 3.09 [1]. Standard deviation = √3.09 = 1.76 [1] [5] Examiner insight: A negative y value squares to a positive number, which is a common slip. Showing E(Y²) on its own line makes it easier to credit the method.

3. E(2X + 3) = 2(8) + 3 = 19 [1]. Var(2X + 3) = 2² × 5 [1] = 20 [1]. Var(−3X) = (−3)² × 5 = 45 [1] [4] Examiner insight: Answers that add 3 to the variance, or give a negative value for Var(−3X), show the same misunderstanding; write the squared coefficient explicitly.

4. (a) E(R²) = 1(0.4) + 4(0.4) + 25(0.2) = 0.4 + 1.6 + 5 = 7 [1], so E(5R²) = 35 [1] (b) 10/R takes 10, 5, 2 [1]: E = 10(0.4) + 5(0.4) + 2(0.2) = 4 + 2 + 0.4 = 6.4 [1] (c) E(R) = 0.4 + 0.8 + 1 = 2.2, so 10/E(R) = 4.55 (3 s.f.), which does not equal 6.4 [1] [5] Examiner insight: Apply the function to each value before weighting. Putting E(R) into the function gives a different, wrong value even if the arithmetic is correct.

5. E(6/R) = 6(0.4) + 3(0.4) + 1.2(0.2) = 2.4 + 1.2 + 0.24 = 3.84 [1]. E(36/R²) = 36(0.4) + 9(0.4) + 1.44(0.2) [1] = 14.4 + 3.6 + 0.288 = 18.288 [1]. Var = 18.288 − 3.84² = 18.288 − 14.7456 = 3.5424 [1] [4] Examiner insight: Non-linear functions need the full E(g²) − (E(g))² route. Using a² Var(R) with a = 6 would give the wrong answer here.

6. k(1 + 1/2 + 1/4) = 1 [1], so (7/4)k = 1 and k = 4/7 [1]. E(N) = k(1 + 1 + 1) = 3k = 12/7 [1]. E(N²) = k(1 + 2 + 4) = 7k = 4 [1]. Var(N) = 4 − (12/7)² [1] = 4 − 144/49 = 52/49 [1] [6] Examiner insight: Write the table of probabilities (4/7, 2/7, 1/7) before using it. Leaving the answer as an exact fraction avoids rounding errors.

7. (a) 2(5) − 3(3) = 1 [1] (b) Var = 2² × 2 + 3² × 4 [1] = 8 + 36 = 44 [1] (c) √44 = 6.63 [1] (d) Yes: the variances add in both cases, because the coefficients are squared [1] [5] Examiner insight: Mention independence when you add variances. Subtracting on the variance line is a slip to watch for in this type of question.

8. (a) E(X) = 2 [1]. E(X²) = 0.5 + 4.5 = 5, so Var(X) = 5 − 4 = 1 [1] (b) E(D) = 4 [1]; Var(D) = 4 × 1 = 4 [1] (c) E(T) = 2 + 2 = 4 [1]; Var(T) = 1 + 1 = 2 [1] (d) D doubles one spin, so deviations double and variance is multiplied by 4. T adds two independent spins, so deviations partly cancel [1] [7] Examiner insight: A comparison needs a reason, not only the two numbers. Refer to one observation scaled and two independent observations added.

9. (a) a + b + 0.5 = 1 gives a + b = 0.5 [1]. E(X) = a + 0.6 + 3b + 0.4 = 2.1, so a + 3b = 1.1 [1]. Subtracting gives 2b = 0.6 [1], so b = 0.3 and a = 0.2 [1] (b) E(X²) = 0.2 + 1.2 + 2.7 + 1.6 = 5.7 [1]. Var(X) = 5.7 − 2.1² [1] = 5.7 − 4.41 = 1.29 [1] (c) E(P) = 40(2.1) − 55 = £29 [1]. Var(P) = 1600 × 1.29 = 2064, so standard deviation = £45.4 [1] [9] Examiner insight: Set up both equations in a and b clearly. The fixed cost of 55 affects the mean but not the variance, so check it has been left out of the variance step.

10. (a) E(X) = 0.5 + 0.6 + 0.6 = 1.7 [1]. E(X²) = 0.5 + 1.2 + 1.8 = 3.5 [1]. Var(X) = 3.5 − 2.89 = 0.61 [1] (b) E(Y) = 2(0.4) = 0.8 [1]. E(Y²) = 4(0.4) = 1.6, so Var(Y) = 1.6 − 0.64 = 0.96 [1] (c) E(X + Y) = 2.5 and E(X − Y) = 0.9 [1]. Var(X + Y) = 0.61 + 0.96 = 1.57 [1]. Var(X − Y) = 1.57 also, since variances add [1] (d) (X, Y) = (1, 2) or (3, 0) [1]. P = 0.5(0.4) + 0.2(0.6) = 0.2 + 0.12 = 0.32 [1] [10] Examiner insight: List the combinations that give the required total before multiplying probabilities. It is easy to miss the second combination in part (d).

11. (a) E(Z) = −0.6 + 0 + 0.6 = 0 [1]. E(Z²) = 4(0.3) + 9(0.2) = 1.2 + 1.8 = 3 [1]. Var(Z) = 3 − 0 = 3 [1] (b) E(T) = 12 × 0 = 0 [1]. Var(T) = 12 × 3 = 36, so standard deviation = 6 [1] (c) Since Var(T) = E(T²) − (E(T))², E(T²) = 36 + 0 = 36 [1] [6] Examiner insight: The turns are separate observations, so use n Var(Z), not n² Var(Z). Rearranging the variance formula for part (c) is quicker than listing the distribution of T.

Where marks are usually lost

• Using (E(X))² in place of E(X²), or forgetting to subtract μ² in the variance.
• Failing to check that probabilities sum to 1 before using the table.
• Writing Var(aX + b) = a Var(X) + b, or forgetting to square a, including when a is negative.
• Subtracting variances for a difference of independent variables.
• Replacing E(g(X)) by g(E(X)) for reciprocals and squares.
• Treating a function of a random variable as linear when it is not, such as Var(6/R).
• Confusing nX with a sum of n independent copies of X.
• Rounding intermediate means and then carrying the rounded value into the variance.
• Giving variance when the standard deviation was asked for.

Next steps

• Review the revision notes and the study guide.
• Return to the course hub and tick off topics on 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, section S1.2 Discrete random variables. Published by OxfordAQA; International AS exams from May/June 2018 and International A-level exams from May/June 2019.