S2 June 2007 Q8

EdexcelOld spec14 marksContinuous Random Variables

8. The continuous random variable \(X\) has probability density function given by

\[\mathrm{f}(x) = \begin{cases} \dfrac{1}{6}x & 0 \lt x \leqslant 3 \\[2mm] 2 - \dfrac{1}{2}x & 3 \lt x \lt 4 \\[2mm] 0 & \text{otherwise} \end{cases}\]

(a) Sketch the probability density function of \(X\). (3)

(b) Find the mode of \(X\). (1)

(d) Using your answer to part (c), find the median of \(X\). (3)

Scheme	Marks
	B1 B1 B1
	(3)

Scheme	Marks
Mode is \(x = 3\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\mathrm{F}(x) = \displaystyle\int_0^x \dfrac{1}{6}t\ \mathrm{d}t\) (for \(0 \leqslant x \leqslant 3\))	M1
\(= \dfrac{1}{12}x^2\)	A1
\(\mathrm{F}(x) = \displaystyle\int_3^x 2 - \dfrac{1}{2}t\ \mathrm{d}t; + \displaystyle\int_0^3 \dfrac{1}{6}t\ \mathrm{d}t\) (for \(3 \lt x \leqslant 4\))	M1; M1
\(= 2x - \dfrac{1}{4}x^2 - 3\)	A1
\(\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\[1mm] \dfrac{1}{12}x^2 & 0 \leqslant x \leqslant 3 \\[1mm] 2x - \dfrac{1}{4}x^2 - 3 & 3 \lt x \leqslant 4 \\[1mm] 1 & x \gt 4 \end{cases}\)	B1 ft B1
	(7)

Notes

1st M1 ignore limits for M

1st A1 must use limit of 0

2nd M1; 3rd M1 need limit of 3 and variable upper limit; need limit 0 and 3

B1 ft middle pair B1 ends

Mark scheme (d)

Scheme	Marks
\(\mathrm{F}(m) = 0.5\)	M1
\(\dfrac{1}{12}x^2 = 0.5\)	A1ft
\(x = \sqrt{6} = 2.45\)	A1
	(3)
	(14 marks)

Notes

M1 either eq

A1ft eq for their \(0 \leqslant x \leqslant 3\)

A1 \(\sqrt{6}\) or awrt 2.45