Becky sometimes works in an office and sometimes works at home. The random variable X denotes the number of days that she works at home in any given week. It is given that P(X = x) = kx(x + 1), where k is a constant and x = 1, 2, 3 or 4 only.

Mark scheme

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

1. [M1] Using sum of probabilities = 1 to form an equation in k. Accept 1×2×k+2×3×k+ 3×4×k+4×5×k= 1.
2. [M1] X 1 2 3 4 P(X) 2k 6k 12k 20k Two correctly linked, accurate probabilities. May be in terms of k. May not be in a table.
3. [A1] Table with correct X values and correct probabilities.
4. [M1] [E(X) =] [E(X) = 1×2k+2×6k+3×12k+4×20k =130k] Accept unsimplified expression. May be calculated in variance. FT their table with 3 or more probabilities summing to 1 (0 < p < 1). If there are outcomes in the table without probabilities, condone and treat as p = 0.
5. [M1] [Var(X) = 1²×2k+22×6k+32 ×12k+42 ×20k-(130k)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 and the highlighted calculation (or less simplified) seen, Note: if table is correct, 454/40 or 227/20 or any calculation – (their E(X))² implies M1.
6. [A1] Answers for E(X) and Var(X) must be identified. E(X) may be identified by correct use in variance. Condone E, V, μ, σ etc. If A0 earned, SC B1 for identified correct final solutions.