AS June 2024 Q1

EdexcelAS paperCurrent spec8 marksContinuous Random Variables

1. A continuous random variable \(X\) has cumulative distribution function \(\mathrm{F}(x)\) given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt -1 \\ \dfrac{1}{5}(x+1)^2 & -1 \leqslant x \leqslant 0 \\ 1 - \dfrac{1}{20}(4-x)^2 & 0 \lt x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\]

(a) Find the probability density function, \(\mathrm{f}(x)\) (2)

(b)

(i) Sketch \(\mathrm{f}(x)\) (2)

(ii) Hence describe the skewness of the distribution. (1)

(c) Find, to 3 significant figures, the value of \(c\) such that\[\mathrm{P}(1 \lt X \lt c) = \mathrm{P}(c \lt X \lt 2)\] (3)

Mark scheme (a)

Scheme	Marks	AO
\(\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{5}(x+1)^2\right] = \frac{2}{5}(x+1)\) or \(\frac{\mathrm{d}}{\mathrm{d}x}\left[1 - \frac{1}{20}(4-x)^2\right] = \frac{1}{10}(4-x)\)	M1	2.1
\(\mathrm{f}(x) = \begin{cases} \frac{2}{5}(x+1) & -1 \leqslant x \leqslant 0 \\ \frac{1}{10}(4-x) & 0 \lt x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\)	A1	1.1b
	(2)

Notes

M1: Attempt to differentiate 2^nd line of cdf in the form \(k(x + 1)\) (oe)
or 3^rd line of cdf in the form \(m(4 - x)\) (oe)

A1: Correct probability density function.
(Condone missing “0 otherwise” and mis-use of \(\lt\) or \(\leqslant\) etc )

Mark scheme (b)

Scheme	Marks	AO
(i) Shape: Triangle with longer slant in 1^st quadrant	B1	1.1b
Labels: –1, 4 on horizontal axis and \(\frac{2}{5}\) on vertical axis	B1	1.1b
	(2)
(ii) (Probability density function has a longer tail to right,) so there is positive skew.	dB1	2.4
	(1)

Notes

(i) 1^st B1: Correct shape. Triangle must have both ends on axis

2^nd B1: Correct labels. All three needed.

(ii) dB1: Correct description of positive skew.
Dep on sketch that clearly shows positive skew.

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{P}(1 \lt X \lt 2) = \mathrm{F}(2) - \mathrm{F}(1)\) or \(2\mathrm{F}(c) = \mathrm{F}(2) + \mathrm{F}(1)\) \(= \left(1 - \dfrac{4}{20}\right) - \left(1 - \dfrac{9}{20}\right) = \dfrac{5}{20}\) or \(\dfrac{1}{4}\) or stating \(\mathrm{F}(c) = 0.675\)	M1	3.1a
So e.g. \(\mathrm{P}(c \lt X \lt 2) = \dfrac{1}{8} \ \Rightarrow\ \dfrac{1}{8} = \dfrac{(4-c)^2}{20} - \dfrac{4}{20}\)	M1	1.1b
\(c = \dfrac{8 - \sqrt{26}}{2}\) or \(1.45049\ldots\) awrt 1.45	A1	1.1b
	(3)
	(8 marks)

Notes

1^st M1: Suitable start to problem e.g. showing \(\mathrm{P}(1 \lt X \lt 2) = 0.25\) o.e.
e.g. \(\mathrm{F}(c) = 0.675\) or \(2\mathrm{F}(c) = 0.8 + 0.55\)

2^nd M1: Rearranging to form quadratic equation in \(c\) can ft their “0.25”
e.g. \(2c^2 - 16c + 19 = 0\)
Allow even if wrong part of \(\mathrm{F}(x)\) is used so M0M1A0 is possible.

A1: for awrt 1.45 only
NB Other root is 6.549… and if this is not rejected score A0