AS June 2025 Q2

EdexcelAS paperCurrent spec9 marksContinuous Random Variables

2. The graph of the probability density function \(\mathrm{f}(x)\) of the continuous uniform random variable \(X\) is shown below.

Graph of f(x): f(x) = 0.2 for x from –3 to 2 and 0 elsewhere, x-axis from –4 to 4, vertical axis marked 0.1 and 0.3

(a) Write down the value of \(\mathrm{P}(X = 1)\) (1)

(b) Find \(\mathrm{E}(X)\) (1)

(d) Sketch the cumulative distribution function of \(X\) for \(-3 \leqslant x \leqslant 2\)
You should label any points where the sketch touches or crosses the coordinate axes. (3)

(e) Find \(\mathrm{P}(X^2 \gt 1.96)\) (3)

Scheme	Marks	AO
0	B1	1.1b
	(1)

Scheme	Marks	AO
\(\mathrm{E}(X)\left[= \dfrac{-3+2}{2}\right] = -\dfrac{1}{2}\)	B1	1.1b
	(1)

Scheme	Marks	AO
\(\mathrm{Var}(X)\left[= \dfrac{(2-(-3))^2}{12}\right] = \dfrac{25}{12}\)	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
Sketches straight line with positive gradient	M1	2.1
Labels (–3, 0) and ends graph at (2, 1)	A1	1.1b
Labels (0, 0.6)	A1	1.1b
	(3)

Notes

M1: Sketching straight line with positive gradient.

A1: Correct label and domain

A1: Correct label on vertical axis

Mark scheme (e)

Scheme	Marks	AO
\(X \gt 1.4\) or \(X \lt -1.4\)	M1	1.1b
\(\mathrm{P}(X \gt 1.4) + \mathrm{P}(X \lt -1.4) = \dfrac{2-1.4}{2-(-3)} + \dfrac{-1.4-(-3)}{2-(-3)}\) or \(\dfrac{0.6}{5} + \dfrac{1.6}{5}\) oe	M1	3.1a
\(\dfrac{11}{25}\)	A1	1.1b
	(3)
	(9 marks)

Notes

M1: Either correct region

M1: Attempt to find probability using both regions

A1: o.e. e.g. 0.44

NB Stating \(\mathrm{P}(-1.4 \lt X \lt 1.4) = 0.56\) can score M1A1A0