Questions on discrete random variables and expectation

Question 1

A discrete random variable X has P(X = x) = kx for x = 1, 2, 3, 4. Find the value of k.

Accepted Answer

The probabilities must total 1: k(1 + 2 + 3 + 4) = 10k = 1, so k = 1/10. M1 for summing the probabilities to 1, A1 for k = 1/10.

Question 2

X takes the values 1, 2, 3, 4 with probabilities 1/6, 1/3, 1/3, 1/6. Find E(X).

Accepted Answer

E(X) = 1(1/6) + 2(1/3) + 3(1/3) + 4(1/6) = 1/6 + 2/3 + 1 + 2/3 = 5/2. M1 for Σx P(X = x), A1 for the four products, A1 for 5/2. The distribution is symmetric about 2.5, so the mean sits exactly there, and no listing was strictly needed.

Question 3

A discrete random variable X has P(X = x) = c/x for x = 1, 2, 4, where c is a constant. Find c, and show that the resulting probabilities form a valid distribution.

Accepted Answer

The probabilities must total 1: c(1 + 1/2 + 1/4) = 7c/4 = 1, so c = 4/7. The probabilities are then 4/7, 2/7 and 1/7, each between 0 and 1 and totalling 1, so the distribution is valid. M1 for summing to 1, A1 for c = 4/7 with the check. Both conditions are needed: a total of 1 alone does not rule out a negative probability.

Question 4

X takes the values 1, 2 and 3 with probabilities 0.5, 0.3 and 0.2. Find E(X²) and E(2X + 3).

Accepted Answer

E(X²) = 1(0.5) + 4(0.3) + 9(0.2) = 0.5 + 1.2 + 1.8 = 3.5. E(X) = 0.5 + 0.6 + 0.6 = 1.7, so E(2X + 3) = 2(1.7) + 3 = 6.4. M1 for Σx² P(X = x), A1 for 3.5, B1 for 6.4. Squaring the mean instead gives 1.7² = 2.89, which is not E(X²): the function is applied to each value before the weighting.

Question 5

The random variable X takes the values 1, 2, 3 and 4 with probabilities 1/6, 1/3, 1/3 and 1/6. Find E(X²) and hence Var(X), as exact fractions.

Accepted Answer

E(X²) = 1(1/6) + 4(1/3) + 9(1/3) + 16(1/6) = 1/6 + 4/3 + 3 + 8/3 = 43/6.
E(X) = 5/2 by symmetry about 2.5, so Var(X) = E(X²) − [E(X)]² = 43/6 − 25/4 = 11/12, about 0.917. M1 for Σx² P(X = x), A1 for 43/6, M1 for E(X²) − [E(X)]², A1 for 11/12.
Subtract 25/4, not 5/2.

Question 6

The random variable X has E(X) = 5/2 and Var(X) = 11/12. Write down E(3X − 1) and Var(3X − 1).

Accepted Answer

E(3X − 1) = 3(5/2) − 1 = 13/2. Var(3X − 1) = 3² × 11/12 = 33/4. B1 for 13/2, M1 for multiplying the variance by 3², A1 for 33/4. The −1 shifts the centre and leaves the spread untouched, while the 3 enters the variance squared.

Question 7

A game pays out £X where X takes values 0, 1, 5 with probabilities 0.7, 0.2, 0.1. Find the expected payout, and state the largest fair stake.

Accepted Answer

E(X) = 0(0.7) + 1(0.2) + 5(0.1) = 0.7, so the expected payout is 70p. A stake above 70p loses money on average; a fair game charges exactly the expected payout. M1 for Σx P(X = x), A1 for an expected payout of 70p, B1 for a fair stake equalling the expected payout, B1 for 70p. Note that 70p is not a possible outcome, which is normal for a mean.

Question 8

A fair twenty-sided die shows each of the numbers 1 to 20 with equal probability. Using the standard results for Σr and Σr², find E(X) and Var(X) as exact fractions.

Accepted Answer

Pull the constant probability 1/20 outside each sum. E(X) = (1/20) × 20(21)/2 = 21/2. M1 for (1/n)Σr with the closed form, A1 for 21/2. E(X²) = (1/20) × 20(21)(41)/6 = 21 × 41/6 = 287/2, so Var(X) = 287/2 − (21/2)² = 574/4 − 441/4 = 133/4. M1 for E(X²) from Σr² and the subtraction of the squared mean, A1 for 133/4. The closed form (n² − 1)/12 gives 399/12 = 133/4, which confirms it; the mean 10.5 is the midpoint of the faces and, as with any die, is not a value the die can show.

Question 9

Y takes the values 1, 2, 3 and 4 with probabilities 0.2, 0.3, 0.4 and 0.1. Write down the running totals F(y) = P(Y is at most y). Explain why both 2 and 3 satisfy the definition of a median, and state the median under the convention that it is the smallest m for which F(m) is at least 0.5. State also the mode.

Accepted Answer

F(1) = 0.2, F(2) = 0.5, F(3) = 0.9, F(4) = 1.
A median m needs both P(Y is at most m) and P(Y is at least m) to be at least 0.5. For m = 2: F(2) = 0.5 and P(Y is at least 2) = 0.8. For m = 3: F(3) = 0.9 and P(Y is at least 3) = 0.5. Both pass, because the running total lands on exactly 0.5 at y = 2. Under the stated convention the median is 2. The mode is the most probable value, 3. M1 for the running totals, A1 for F(2) = 0.5 exactly, B1 for both 2 and 3 satisfying the definition, A1 for the median 2 under the convention with the mode 3. Name the convention you use; some texts would report 2.5 here.

Question 10

X takes the values 1, 2 and 4 with probabilities 4/7, 2/7 and 1/7. Find E(1/X) and E(X), and show that E(1/X) is not equal to 1/E(X).

Accepted Answer

E(1/X) = 1(4/7) + (1/2)(2/7) + (1/4)(1/7) = 4/7 + 1/7 + 1/28 = 16/28 + 4/28 + 1/28 = 21/28 = 3/4.
E(X) = 1(4/7) + 2(2/7) + 4(1/7) = 4/7 + 4/7 + 4/7 = 12/7, so 1/E(X) = 7/12, which is not 3/4.
M1 for weighting 1/x by the probabilities, A1 for 3/4, M1 for E(X), A1 for 12/7 with the comparison. E(g(X)) is g applied to each value and then averaged; it equals g(E(X)) only when g is linear.

Question 11

X takes the values 0, 1 and 2 with probabilities 0.25, 0.5 and 0.25, and Y = 4X − 3. Find E(Y) and Var(Y) directly from the probability distribution of Y, and verify them using E(X) and Var(X).

Accepted Answer

Y takes the values −3, 1 and 5 with probabilities 0.25, 0.5 and 0.25. E(Y) = −3(0.25) + 1(0.5) + 5(0.25) = −0.75 + 0.5 + 1.25 = 1. E(Y²) = 9(0.25) + 1(0.5) + 25(0.25) = 2.25 + 0.5 + 6.25 = 9, so Var(Y) = 9 − 1² = 8.
Check: E(X) = 0.5 + 0.5 = 1 and E(X²) = 0.5 + 1 = 1.5, so Var(X) = 0.5. Then E(4X − 3) = 4(1) − 3 = 1 and Var(4X − 3) = 16(0.5) = 8, agreeing with the direct values. M1 for the distribution of Y with the same probabilities, A1 for E(Y) = 1 and Var(Y) = 8, M1 for E(X) and Var(X), A1 for both checks. The probabilities move with the values: P(Y = −3) is P(X = 0), because Y = −3 exactly when X = 0.

Question 12

The discrete random variable X takes the values 1, 2, 3 and 4. P(X = 1) = P(X = 2) = a and P(X = 3) = P(X = 4) = b, and E(X) = 2.7. Find a and b, and hence find Var(5 − 2X).

Accepted Answer

Two unknowns need two equations, and the first is always that the probabilities total 1.
2a + 2b = 1.
For the second, E(X) = a(1) + a(2) + b(3) + b(4) = 3a + 7b = 2.7.
From the first equation a = 0.5 − b. Substituting: 3(0.5 − b) + 7b = 2.7, so 1.5 + 4b = 2.7 and b = 0.3, giving a = 0.2.
Check: the probabilities 0.2, 0.2, 0.3, 0.3 total 1, and E(X) = 0.2 + 0.4 + 0.9 + 1.2 = 2.7.
Now E(X²) = 0.2(1) + 0.2(4) + 0.3(9) + 0.3(16) = 0.2 + 0.8 + 2.7 + 4.8 = 8.5, so Var(X) = 8.5 − 2.7² = 8.5 − 7.29 = 1.21.
Finally Var(5 − 2X) = (−2)² × 1.21 = 4.84. B1 for 2a + 2b = 1, M1 for the expectation equation, A1 for a = 0.2 and b = 0.3, M1 for E(X²), A1 for Var(X) = 1.21, A1 for 4.84. The coefficient squares, so its sign disappears; a negative variance means the squaring was forgotten.

Question 13

Daily counts of a shop's sales have mean 3.1 and variance 1.2. Explain why a Poisson model is unsuitable for these counts, why a binomial model is not ruled out by the two numbers alone, and state the comparison a geometric model would fail.

Accepted Answer

Each family leaves a fingerprint in the pair of numbers. A Poisson distribution forces its variance to equal its mean, and 1.2 sits far below 3.1, so a Poisson model would overstate the day-to-day variability of these sales. B1 for the equality demanded and the direction of the mismatch. A binomial distribution has variance np(1 − p), always below its mean np, which is exactly the relationship the data show, so nothing in the pair rules a binomial out. B1 for the variance-below-mean fingerprint. A geometric distribution runs the other way, with variance well above its mean once p is small, so it fails the same comparison in the opposite direction. B1 for the geometric's fingerprint.

Question 14

X is equally likely to take each of the values 1, 2, …, n. Using the standard results Σr = n(n + 1)/2 and Σr² = n(n + 1)(2n + 1)/6, prove that E(X) = (n + 1)/2 and Var(X) = (n² − 1)/12. Check both results for a fair six-sided die.

Accepted Answer

Each value has probability 1/n, so E(X) = (1/n)Σr = (1/n) × n(n + 1)/2 = (n + 1)/2.
E(X²) = (1/n)Σr² = (1/n) × n(n + 1)(2n + 1)/6 = (n + 1)(2n + 1)/6.
Var(X) = E(X²) − [E(X)]² = (n + 1)(2n + 1)/6 − (n + 1)²/4. Take out (n + 1)/12: this is (n + 1)[2(2n + 1) − 3(n + 1)]/12 = (n + 1)(4n + 2 − 3n − 3)/12 = (n + 1)(n − 1)/12 = (n² − 1)/12.
For n = 6: E(X) = 7/2 = 3.5, and Var(X) = 35/12. Directly, E(X²) = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6, and 91/6 − 49/4 = 182/12 − 147/12 = 35/12, as the formula gives. M1 for (1/n)Σr with the closed form, A1 for (n + 1)/2, M1 for E(X²) from Σr², M1 for subtracting the squared mean and taking out the common factor, A1 for (n² − 1)/12 with the factorisation shown, B1 for both checks at n = 6.

Question 15

A model for the number of items X bought per customer proposes P(X = x) = k(5 − x) for x = 1, 2, 3, 4. Find k, and the mean and variance of X under the model. In a sample of 200 customers, 60 bought one item, 70 bought two, 40 bought three and 30 bought four. Decide, with reasons, whether the model is suitable.

Accepted Answer

The probabilities k(4 + 3 + 2 + 1) = 10k must total 1, so k = 1/10 and the probabilities are 0.4, 0.3, 0.2 and 0.1.
E(X) = 0.4 + 0.6 + 0.6 + 0.4 = 2. E(X²) = 0.4 + 1.2 + 1.8 + 1.6 = 5, so Var(X) = 5 − 4 = 1.
The model predicts frequencies 200 × (0.4, 0.3, 0.2, 0.1) = 80, 60, 40, 20 against the observed 60, 70, 40, 30, and the sample mean is (60 + 140 + 120 + 120)/200 = 2.2 against the model's 2. The model makes one item the modal purchase and has probabilities falling steadily, but the data peak at two items and have a heavier tail at four, 30 against 20. The model is not suitable: its shape disagrees with the data, not just its mean. M1 for k from the total, A1 for E(X) = 2 and Var(X) = 1, M1 for comparing predicted frequencies or the model mean with the data, A1 for the comparison of 80, 60, 40, 20 with 60, 70, 40, 30 or of 2 with 2.2, B1 for the decision with a reason about shape. A goodness-of-fit test would make the comparison formal; the direction of the disagreement is already visible in the frequencies.