A-LevelMathematicsProbability and statistics9709/53 · May/June 2023 · Q37 Marks

The random variable X takes the values 1, 2, 3, 4. It is given that P(X = x) = kx(x + a), where k and a are constants.

(a)Given that P(X = 4) = 3P(X = 2), find the value of a and the value of k.[4]
(b)Draw up the probability distribution table for X, giving the probabilities as numerical fractions.[1]
(c)Given that E(X) = 3.2, find Var(X).[2]

Mark scheme

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

[M1] Using P(X= 4) = 3P(X= 2) to form an equation in a and k.
[A1] a = 2
[A1] If M0 scored, SC B1 for a = 2 www.
[M1] Using sum of probabilities = 1 to form an equation in k: k(1 + a) + 2k(2 + a) + 3k(3 + a)+ 4k(4 + a) = 1.
[A1] k = 1/50
[A1] If M0 scored, SC B1 for k = 1/50 www.
[B1 FT] X: 1, 2, 3, 4 and P(X): k(1+a), 2k(2+a), 3k(3+a), 4k(4+a)
[B1 FT] P(X) 3/50, 8/50, 15/50, 24/50. 0<p<1 for all outcomes, must be numerical.
[M1] Correct formula for variance method from their probability distribution table, 0 < their P(x) ≤ 1.
[M1] Accept (3+32+135+384)/50 - 256/25
[A1] [= 11.08 – 3.2² =] 0.84[0], 21/25
[A1] If M0 score SC B1 for 0.84 www.

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