S2 January 2007 Q7

EdexcelOld spec14 marksContinuous Random Variables

7. The continuous random variable \(X\) has cumulative distribution function\[\mathrm{F}(x) = \begin{cases} 0, & x \lt 0, \\ 2x^2 - x^3, & 0 \leqslant x \leqslant 1, \\ 1, & x \gt 1. \end{cases}\]

(a) Find \(\mathrm{P}(X \gt 0.3)\). (2)

(b) Verify that the median value of \(X\) lies between \(x = 0.59\) and \(x = 0.60\). (3)

(d) Evaluate \(\mathrm{E}(X)\). (3)

(e) Find the mode of \(X\). (2)

(f) Comment on the skewness of \(X\). Justify your answer. (2)

Scheme	Marks
\(1 - \mathrm{F}(0.3) = 1 - (2 \times 0.3^2 - 0.3^3)\)	M1
\(= 0.847\)	A1
	(2)

Scheme	Marks
\(\mathrm{F}(0.60) = 0.5040\) \(\mathrm{F}(0.59) = 0.4908\)	M1A1
0.5 lies between therefore median value lies between 0.59 and 0.60.	B1
	(3)

Scheme	Marks
\(\mathrm{f}(x) = \begin{cases} -3x^2 + 4x, & 0 \leqslant x \leqslant 1, \\ 0, & \text{otherwise.} \end{cases}\)	M1A1
	(2)

Mark scheme (d)

Scheme	Marks
\(\displaystyle\int_0^1 x\mathrm{f}(x)\,\mathrm{d}x = \int_0^1 -3x^3 + 4x^2\,\mathrm{d}x\)	M1
\(= \left[\dfrac{-3x^4}{4} + \dfrac{4x^3}{3}\right]_0^1\)	M1
\(= \dfrac{7}{12}\) or \(0.58\dot{3}\) or 0.583 or equivalent fraction	A1
	(3)

Notes

1st M1 attempt to integrate \(x\mathrm{f}(x)\)

2nd M1 sub in limits

Scheme	Marks
\(\dfrac{\mathrm{df}(x)}{\mathrm{d}x} = -6x + 4 = 0\)	M1
\(x = \dfrac{2}{3}\) or \(0.\dot{6}\) or 0.667	A1
	(2)

Scheme	Marks
mean \(\lt\) median \(\lt\) mode, therefore negative skew.	B1,B1
	(2)
	(14 marks)