S2 June 2006 Q6

EdexcelOld spec16 marksContinuous Random Variables

6. The continuous random variable \(X\) has probability density function\[\mathrm{f}(x) = \begin{cases} \dfrac{1 + x}{k}, & 1 \leqslant x \leqslant 4, \\ 0, & \text{otherwise.} \end{cases}\]

(a) Show that \(k = \dfrac{21}{2}\). (3)

(b) Specify fully the cumulative distribution function of \(X\). (5)

(d) Find the value of the median. (3)

(e) Write down the mode. (1)

(f) Explain why the distribution is negatively skewed. (1)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int_1^4 \frac{1 + x}{k}\,\mathrm{d}x = 1\)	M1
\(\therefore \left[\dfrac{x}{k} + \dfrac{x^2}{2k}\right]_1^4 = 1\)	A1
\(k = \dfrac{21}{2}\ *\)	A1
	(3)

Notes

M1 \(\int \mathrm{f}(x) = 1\); Area = 1

1st A1 correct integral/correct expression

2nd A1 cso

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(X \leqslant x_0) = \displaystyle\int_1^{x_0} \frac{2}{21}(1 + x)\)	M1
\(= \left[\dfrac{2x}{21} + \dfrac{x^2}{21}\right]_1^{x_0}\)	A1
\(= \dfrac{2x_0 + x_0^2 - 3}{21}\) or \(\dfrac{(3 + x)(x - 1)}{21}\)	A1
\(\mathrm{F}(x) = \begin{cases} 0, & x \lt 1 \\ \dfrac{x^2 + 2x - 3}{21} & 1 \leqslant x \lt 4 \\ 1 & x \geqslant 4 \end{cases}\)	B1ft; B1
	(5)

Notes

M1 \(\int \mathrm{f}(x)\); variable limit or \(+C\)

1st A1 correct integral + limit of 1; may have \(k\) in

B1ft; B1 middle; ends

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(X) = \displaystyle\int_1^4 \frac{2x}{21}(1 + x)\,\mathrm{d}x\)	M1
\(= \left[\dfrac{x^2}{21} + \dfrac{2x^3}{63}\right]_1^4\)	A1
\(= \frac{171}{63} = 2\frac{5}{7} = \frac{19}{7} = 2.7142\ldots\)	A1
	(3)

Notes

M1 valid attempt \(\int x\mathrm{f}(x)\); \(x^2\) and \(x^3\)

1st A1 correct integration

2nd A1 awrt 2.71

Mark scheme (d)

Scheme	Marks
\(\mathrm{F}(m) = 0.5 \Rightarrow \dfrac{x^2 + 2x - 3}{21} = \dfrac{1}{2}\)	M1
\(\therefore 2x^2 + 4x - 27 = 0\) or equiv \(\therefore x = \dfrac{-4 \pm \sqrt{16 - 4 \cdot 2(-27)}}{4}\) \(\therefore x = -1 \pm 3.8078\ldots\)	M1
i.e. \(x = 2.8078\ldots\)	A1
	(3)

Notes

1st M1 putting their \(\mathrm{F}(x) = 0.5\)

2nd M1 attempt their 3 term quadratic

A1 awrt 2.81

Scheme	Marks
Mode \(= 4\)	B1
	(1)

Scheme	Marks
Mean < median < mode (\(\Rightarrow\) negative skew) Or Mean < median	B1

	(1)
	(16 marks)