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Discrete random variables and expectation questions
A probability distribution is a set of weights. The mean is where they balance and the variance is how far they scatter. Both come out of one sum, and both survive being scaled and shifted in predictable ways.
6 original questions · 22 marks · the discrete random variables and expectation notes · Further Statistics 1
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A discrete random variable X has P(X = x) = kx for x = 1, 2, 3, 4. Find the value of k.
Worked 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. Checking validity before anything else is what makes every later answer trustworthy.X takes the values 1, 2, 3, 4 with probabilities 1/6, 1/3, 1/3, 1/6. Find E(X).
Worked 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.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.
Worked 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.
The square goes on the mean alone. Writing E(X² − X)² or subtracting 5/2 rather than its square is the slip that loses the accuracy mark here.The random variable X has E(X) = 5/2 and Var(X) = 11/12. Write down E(3X − 1) and Var(3X − 1).
Worked 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. Subtracting 1 from the variance as well is the standard error, and it is worth saying in your working that adding a constant cannot change a spread.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.
Worked 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.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).
Worked 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. Collecting the coefficients of a and of b before substituting keeps the algebra short.
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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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