The graph of the probability density function f of a random variable X is symmetrical about the line x = 2. It is given that P(2 < X < 5) = 117/256.
- (a)Using only this information show that P(X >−1)= 245/256.[2]
- (b)It is now given that, for x in a suitable domain, f(x) = k(12+4x-x²), where k is a constant. Find the value of k.[3]
- (c)A different random variable X has probability density function g(x) = (9/7)(2+x-x²). The domain of X is all values of x for which g(x) ≥ 0. Find Var(X).[5]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.
- [M1] Use of symmetry about x = 2 to relate P(X > -1) to P(X < 5). (e.g. 1/2 + 117/256).
- [A1] Any correct numerical expression seen leading to 245/256 AG.
- [M1] Attempt to integrate f(x) = k(12+4x-x²) with any limits. (Integral: [k(12x + 2x² - x³/3)])
- [M1] Correct application of limits and equating to probability. (Use of limits 2 and 5 and equating their integration attempt to 117/256. Or limits -2 and 6 equated to 1. Or limits -1 and 6 equated to 245/256. Or limits -1 to 5 equated to 234/256. Oe. No mixed methods.)
- [A1] k = 3/256 or 0.0117.
- [B1] Domain of X found as -1 ≤ x ≤ 2 (x = -1 and x = 2 seen or implied).
- [B1] Mean = 0.5.
- [*M1] Attempt to integrate x²g(x) with correct limits. (Integral: (9/7) * [2x³/3 + x⁴/4 - x⁵/5] from -1 to 2)
- [DM1] Subtract their mean² from their ∫x²g(x)dx (both must be numerical).
- [A1] = 0.45 or 9/20.
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