The probability density function, f, of a random variable X is given by f(x) = { k(6x - x²) for 0 ≤ x ≤ 6, 0 otherwise, where k is a constant.
- State the value of E(X) and show that Var(X) = 9/5.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] `E(X) = 3`. N.B. `E(X) = 108k` is B0 until correct k substituted in.
- [M1] Attempt to find k: `∫_0^6 k(6x - x²)dx = 1`. Attempt integration of f(x) and =1. Ignore limits at this stage.
- [A1] Correct value of `k = 1/36` (or `3/108`). Requires correct integration and substitution.
- [*M1] Attempt to find `E(X^2)`: `∫_0^6 x² * f(x) dx`. Attempt integration of `their k × x²f(x)`. Ignore limits at this stage. Accept in terms of k.
- [DM1] Calculation of Variance: `E(X^2) - (E(X))^2`. Their `E(X^2)` (from use of limits 0 and 6) minus their `(E(X))²`. Accept in terms of k: `388.8k – (108k)²`.
- [A1] `Var(X) = 9/5` or `1.8`. CWO. Must be convincingly obtained as AG.
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