S2 January 2008 Q8

8. The continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by

\[\mathrm{f}(x) = \begin{cases} 2(x - 2) & 2 \leqslant x \leqslant 3 \\ 0 & \text{otherwise} \end{cases}\]

(a) Sketch \(\mathrm{f}(x)\) for all values of \(x\). (3)

(b) Write down the mode of \(X\). (1)

Find

(d) the median of \(X\). (4)

(e) Comment on the skewness of this distribution. Give a reason for your answer. (2)

Scheme	Marks
	B1 B1 B1
	(3)

Scheme	Marks
3	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\displaystyle\int_2^3 2x(x - 2)\,dx = \left[\dfrac{2x^3}{3} - 2x^2\right]_2^3\)	M1A1
\(= 2\tfrac{2}{3}\)	A1
	(3)

M1 attempt to find \(\displaystyle\int x\mathrm{f}(x)\mathrm{d}x\) for attempt we need to see \(x^n \to x^{n+1}\). ignore limits

A1 correct integration ignore limits

A1 accept \(2\tfrac{2}{3}\) or awrt 2.67 or \(2.\dot{6}\)

Mark scheme (d)

Scheme	Marks
\(\displaystyle\int_2^m 2(x - 2)\,dx = 0.5\) \(\left[x^2 - 4x\right]_2^m = 0.5\)	M1
\(m^2 - 4m + 4 = 0.5\) \(m^2 - 4m + 3.5 = 0\)	A1
\(m = \dfrac{4 \pm \sqrt{2}}{2}\)	M1
\(m = 2.71\)	A1
	(4)

M1 using \(\displaystyle\int \mathrm{f}(x)\mathrm{d}x = 0.5\)

A1 \(m^2 - 4m + 4 = 0.5\) oe

M1 attempting to solve quadratic.

A1 awrt 2.71 or \(\dfrac{4 + \sqrt{2}}{2}\) or \(2 + \dfrac{\sqrt{2}}{2}\) oe