A2 October 2021 Q3

3. The continuous random variable \(X\) has cumulative distribution function given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 2 \\ 1.25 - \dfrac{2.5}{x} & 2 \leqslant x \leqslant 10 \\ 1 & x \gt 10 \end{cases}\]

(a) Find \(\mathrm{P}(\{X \lt 5\} \cup \{X \gt 8\})\) (2)

(b) Find the median of \(X\). (2)

(d)

(i) Sketch the probability density function of \(X\).

(ii) Describe the skewness of the distribution of \(X\).

(3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{F}(5) + (1 - \mathrm{F}(8)) \qquad \left[\dfrac{3}{4} + \left(1 - \dfrac{15}{16}\right)\right]\)	M1	2.1
\(= \dfrac{13}{16}\)	A1	1.1b
	(2)

M1: Equivalent correct probability statement, e.g. \([1 - (\mathrm{F}(8) - \mathrm{F}(5))]\)

A1: \(\dfrac{13}{16}\) oe

Scheme	Marks	AO
\(\mathrm{F}(m) = 0.5 \qquad \left[1.25 - \dfrac{2.5}{m} = 0.5\right]\)	M1	1.1b
\(m = \dfrac{10}{3}\)	A1	1.1b
	(2)

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{f}(x)\left[= \tfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{F}(x))\right] = 2.5x^{-2}\)	M1	2.1
\(\mathrm{E}(X^2)\left[= \displaystyle\int_2^{10} x^2\mathrm{f}(x)\,\mathrm{d}x\right] = \displaystyle\int_2^{10} 2.5\,\mathrm{d}x\)	M1	1.1b
\(= 20\)	A1	1.1b
	(3)

M1: Realising that \(\mathrm{f}(x)\) is required and attempting to differentiate \(\mathrm{F}(x)\)

M1: Use of \(\displaystyle\int_2^{10} x^2\mathrm{f}(x)\,\mathrm{d}x\)

A1: 20 cao

Scheme	Marks	AO
(i)	M1	1.1b
2 and 10 correctly labelled on horizontal axis	A1	2.1
(ii) Therefore positive skew.	A1	2.2b
	(3)
	(10 marks)