AS June 2022 Q2

EdexcelAS paperCurrent spec5 marksContinuous Random Variables

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

Graph of f(x): zero up to x = 1, a straight line rising from (1, 0) to (7, 0.20), then a horizontal line at 0.10 from x = 7 to x = 11, and zero after 11

(a) Find \(\mathrm{P}(X \lt 4)\) (2)

(b) Specify the cumulative distribution function of \(X\) for \(7 \leqslant x \leqslant 11\) (3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{P}(X \lt 4) = \dfrac{(4-1) \times 0.1}{2}\) or \(\displaystyle\int_1^4 \tfrac{1}{30}(x-1)\,\mathrm{d}x\)	M1	2.1
\(= 0.15\)	A1	1.1b
	(2)

Notes

M1: Use of area of triangle or integration with limits to find required area
Condone \(\dfrac{4 \times 0.1}{2}\) or \(\displaystyle\int_1^4 \tfrac{1}{30}(x)\,\mathrm{d}x\) for M1

A1: 0.15 oe

Mark scheme (b)

Scheme

Marks

\(\dfrac{(7-1) \times 0.2}{2} + \ldots\)
\(\ldots\ 0.1(x - 7)\)

\(\displaystyle\int_1^7 \tfrac{1}{30}(x-1)\,\mathrm{d}x + \ldots\)
\(\displaystyle\ldots \int_7^x 0.1\,\mathrm{d}t\)

\(\displaystyle\int 0.1\,\mathrm{d}x\)
\(\mathrm{F}(7) = 0.1x + c\)
or
\(\mathrm{F}(11) = 0.1x + c\)

M1

M1

3.1a

1.1b

\(\mathrm{F}(x) = 0.1x - 0.1\) [for \(7 \leqslant x \leqslant 11\)]

1.1b

(3)

(5 marks)

Notes

M1: Complete method for finding the cdf including area from \(1 \leqslant x \lt 7\)
Condone one slip for the area from \(1 \leqslant x \lt 7\)

M1: Attempt at area from \(7 \leqslant x \leqslant 11\)

A1: \(0.1x - 0.1\)