The random variable X takes the values −2, 1, 2, 3. It is given that P(X = x) = kx², where k is a constant.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [B1] k = 1/18 (4k+ k +4k+9k = 18k = 1) SOI
2. [M1] Table with correct x values and at least one probability accurate using their k. Values need not be in order, lines may not be drawn, may be vertical, x and P(X=x) may be omitted. Condone any additional X values if probability stated as 0.
3. [A1] Remaining probabilities correct.
4. [M1] -8k+k+8k+27k. May be implied by use in Variance. Accept unsimplified expression. FT their table if probabilities sum to 1 or 0.999. SC B1 28k.
5. [M1] 16k+k+16k+81k – (their mean)². FT their table even if probabilities not summing to 1. Note: If table is correct, 114/18 – (their E(X))² M1. SC B1 114k – (their mean)².
6. [A1] Answers for E(X) and Var(X) must be identified. E(X) = 14/9 or 1.56, Var(X) = 317/81 or 3.91 (3.91 <= Var(X) <= 3.914)