A-LevelMathematicsProbability and statistics9709/61 · Oct/Nov 2018 · Q23 Marks
A random variable X has the probability distribution shown in the following table, where p is a constant. X -1 0 1 2 4 P(X = x) p p 2p 2p 0.1
- (i)Find the value of p.[1]
- (ii)Given that E(X) = 1.15, find Var(X).[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [B1] 6p +0.1 = 1; p = 0.15
- [M1] Var(X) =1 × p +1 × 2p + 4×2p + 16 × 0.1 – 1.152; Correct unsimplified formula, their p substituted (allow 1 error)
- [A1] 0.15 +0 +0.3 + 1.2 + 1.6 – 1.152 = 1.9275 = 1.93 (3sf)
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