S2 June 2005 Q6

EdexcelOld spec18 marksContinuous Random Variables

6. A continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) where\[\mathrm{f}(x) = \begin{cases} k(4x - x^3), & 0 \leqslant x \leqslant 2, \\ 0, & \text{otherwise,} \end{cases}\]where \(k\) is a positive integer.

(a) Show that \(k = \dfrac{1}{4}\). (4)

Find

(b) \(\mathrm{E}(X)\), (3)

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

(e) Comment on the skewness of the distribution. (2)

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

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int_0^2 k(4x - x^3)\,\mathrm{d}x = 1\)	M1 A1
\(k\left[2x^2 - \dfrac{1}{4}x^4\right]_0^2 = 1\)	A1
\(k(8 - 4) = 1\) \(k = \dfrac{1}{4}\)	A1
	(4)

Notes

M1 A1 \(\int \mathrm{f}(x)\,\mathrm{d}x = 1\), all correct

2nd A1 [*]

3rd A1 cso

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}(X) = \displaystyle\int_0^2 x \cdot \frac{1}{4}(4x - x^3)\,\mathrm{d}x\)	M1
\(= \left[\dfrac{1}{3}x^3 - \dfrac{1}{20}x^5\right]_0^2\)	A1
\(= \dfrac{16}{15}\)	A1
	(3)

Notes

M1 \(\int x\mathrm{f}(x)\,\mathrm{d}x\)

1st A1 [*]

2nd A1 1.07 or \(1\frac{1}{15}\) or \(\frac{16}{15}\) or \(1.0\dot{6}\)

Mark scheme (c)

Scheme	Marks
At mode, \(\mathrm{f}'(x) = 0\)	M1
\(4 - 3x^2 = 0\)	M1
\(x = \dfrac{2}{\sqrt{3}}\)	A1
	(3)

Notes

1st M1 implied

2nd M1 attempt to differentiate

A1 \(\sqrt{\frac{4}{3}}\) or 1.15 or \(\frac{2}{\sqrt{3}}\) or \(\frac{2\sqrt{3}}{3}\)

Mark scheme (d)

Scheme	Marks
At median, \(\displaystyle\int_0^x \frac{1}{4}(4t - t^3)\,\mathrm{d}t = \frac{1}{2}\)	M1
\(\dfrac{1}{4}\left(2x^2 - \dfrac{1}{4}x^4\right) = \dfrac{1}{2}\)	M1
\(x^4 - 8x^2 + 8 = 0\) \(x^2 = 4 \pm 2\sqrt{2}\)	M1
\(x = 1.08\)	A1
	(4)

Notes

1st M1 \(\mathrm{F}(x) = \frac{1}{2}\) or \(\int \mathrm{f}(x)\,\mathrm{d}x = \frac{1}{2}\)

2nd M1 attempt to integrate

3rd M1 attempt to solve quadratic

A1 awrt 1.08

Scheme	Marks
	B1 B1
	(2)
	(18 marks)

Scheme	Marks
mean (1.07) \(\lt\) median (1.08) \(\lt\) mode (1.15)	M1
\(\Rightarrow\) negative skew	A1
	(2)