S2 January 2005 Q7

EdexcelOld spec17 marksContinuous Random Variables

7. The random variable \(X\) has probability density function\[\mathrm{f}(x) = \begin{cases} k(-x^2 + 5x - 4), & 1 \leqslant x \leqslant 4, \\ 0, & \text{otherwise.} \end{cases}\]

(a) Show that \(k = \frac{2}{9}\). (3)

Find

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

(d) the cumulative distribution function \(\mathrm{F}(x)\) for all \(x\). (5)

(e) Evaluate \(\mathrm{P}(X \leqslant 2.5)\), (2)

(f) Deduce the value of the median and comment on the shape of the distribution. (2)

Mark scheme (a)

Scheme	Marks
\(k\displaystyle\int_1^4 (-x^2 + 5x - 4)\,\mathrm{d}x = 1\)	M1
\(\therefore k\left[-\dfrac{x^3}{3} + \dfrac{5x^2}{2} - 4x\right]_1^4 = 1\)	A1
\(\Rightarrow \underline{k = \frac{2}{9}}\ *\)	A1 c.s.o.
	(3)

Notes

M1 use of \(\int \mathrm{f}(x)\,\mathrm{d}x = 1\)

1st A1 all correct integration with limits

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}(X) = \displaystyle\int_1^4 \tfrac{2}{9}(-x^3 + 5x^2 - 4x)\,\mathrm{d}x\)	M1
\(= \dfrac{2}{9}\left[-\dfrac{x^4}{4} + \dfrac{5x^3}{3} - \dfrac{4x^2}{2}\right]_1^4\)	A1
\(= \underline{\dfrac{5}{2}}\)	A1 cao
	(3)

Notes

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

1st A1 correct integration with limits

Mark scheme (c)

Scheme	Marks
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{f}(x) = \dfrac{2}{9}(-2x + 5) = 0;\ \Rightarrow \text{Mode} = \dfrac{5}{2}\)	M1; A1
	(2)

Notes

M1 differentiation of \(\mathrm{f}(x)\) & \(= 0\)

SC: \(\frac{5}{2}\) only; no working B1

Mark scheme (d)

Scheme	Marks
\(\mathrm{F}(x_0) = \displaystyle\int_1^{x_0} \tfrac{2}{9}(-x^2 + 5x - 4)\,\mathrm{d}x\)	M1
\(= \left[\dfrac{2}{9}\left(-\dfrac{x^3}{3} + \dfrac{5x^2}{2} - 4x\right)\right]_1^{x_0}\)	A1
\(= \dfrac{2}{9}\left\{-\dfrac{x_0^3}{3} + \dfrac{5x_0^2}{2} - 4x_0 + \dfrac{11}{6}\right\}\)	A1
\(\therefore \mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ \dfrac{2}{9}\left\{-\dfrac{x^3}{3} + \dfrac{5x^2}{2} - 4x + \dfrac{11}{6}\right\} & 1 \leqslant x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\)	B1 B1
	(5)

Notes

M1 use of \(\int \mathrm{f}(t)\,\mathrm{d}t\)

1st A1 integration with limits 1 & symbol

2nd A1 aef

1st B1 \(x \lt 1\); \(x \gt 4\)

2nd B1 \(1 \leqslant x \leqslant 4\)

Scheme	Marks
Median \(= 2.5\); Distribution is symmetrical	B1; B1
	(2)
	(17 marks)