A-LevelMathematicsProbability and statistics9709/62 · Feb/Mar 2023 · Q38 Marks

The diagram shows the graph of the probability density function, f, of a random variable X that takes values between x = 0 and x = 3 only. The graph is symmetrical about the line x = 1.5. [Figure 3.1]

(a)It is given that P(X < 0.6) = a and P(0.6 < X < 1.2) = b. Find P(0.6 < X < 1.8) in terms of a and b.[2]
(b)(i)It is now given that the equation of the probability density function of X is f(x) = { kx²(3-x)² 0 ≤ x ≤ 3, { 0 otherwise, where k is a constant. Show that k = 10/81.[3]
(b)(ii)Find Var(X).[3]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

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

[M1] 1-2(a + b) or 1-2a or 0.5-a-b or 1-(a+b) or a+a+b
[M1] OE. Seen or implied – may be on the diagram (or for correct un-simplified final expression).
[A1] P(0.6 ≤ X ≤ 1.8) = 1 – 2a - b
[A1] Accept 1 – (2a + b).
[M1] k * integral(9x^2 - 6x^3 + x^4)dx = 1
[M1] Attempt integrate f(x) ignore limits and ‘= 1'.
[A1] Correct integration seen, correct limits.
[A1] k × 81/10 = 1, k = 10/81
[A1] AG. Convincingly obtained. No errors seen. (Must see integration).
[M1] integral from 0 to 3 of x^2f(x) between 0 and 3 condone missing k.
[M1] Attempt integrate x^2f(x) between 0 and 3 condone missing k. Must see integration or correct answer of 18/7 seen or implied.
[M1] Their integral of x^2f(x) – 1.5^2 (or their mean^2).
[A1] = 9/28 or 0.321

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.