The random variable X takes the values –2, 2 and 3. It is given that P(X = x) = k(x² – 1), where k is a constant.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] [3k +3k+8k = 1,so]k = 1/14
2. [B1 FT] Table with correct values of x, and at least one correct probability linked with outcome. FT their k. Condone any additional X values if probability stated as 0.
3. [B1 FT] The outcomes in the table must be -2, 2 and 3. 2 further correct probabilities in table or 3 correct probabilities not in table linked to outcomes, or 3 correct FT probabilities in table using their k, or 3 incorrect probabilities summing to 1 in table if k not stated.
4. [M1] [E(X)=-2x(3/14)+2x(3/14)+3x(8/14)] Accept unsimplified expression. May be calculated in variance. FT their table with 3 probabilities summing to 0.999 ≤ total < 1 (0 < p < 1) or in terms of k.
5. [M1] [Var(X)=(-2)²x(3/14)+2²x(3/14)+3²x(8/14)-(their E(X))²] Appropriate variance formula using their (E(X))² value. FT their table with 3 or more probabilities (0 <p<1) which need not sum to 1, or in terms of k with an expression no more evaluated than shown.
6. [A1] E(X) = 5/7, 1.71, 1 5/7 AND Var(X) = 192/49, 3.92, 3 45/49. Answers for E(X) and Var(X) must be identified. E(X) may be identified by correct use in Variance (condone E, V, μ, σ², etc.). If A0 earned, SC B1 for identified correct final answers.