S2 June 2010 Q7

EdexcelOld spec15 marksContinuous Random Variables

7. The random variable \(Y\) has probability density function \(\mathrm{f}(y)\) given by

\[\mathrm{f}(y) = \begin{cases} ky(a - y) & 0 \leqslant y \leqslant 3 \\ 0 & \text{otherwise} \end{cases}\]

where \(k\) and \(a\) are positive constants.

(a)

(i) Explain why \(a \geqslant 3\)

(ii) Show that \(k = \dfrac{2}{9(a - 2)}\) (6)

Given that \(\mathrm{E}(Y) = 1.75\)

(b) show that \(a = 4\) and write down the value of \(k\). (6)

For these values of \(a\) and \(k\),

(d) write down the mode of \(Y\). (1)

Mark scheme (a)

Scheme	Marks
(i) \(\mathrm{f}(y) \geqslant 0\) or \(\mathrm{f}(3) \geqslant 0\) \(ky\left(a - y\right) \geqslant 0\) or \(3k(a - 3) \geqslant 0\) or \((a - y) \geqslant 0\) or \((a - 3) \geqslant 0\)	M1
\(a \geqslant 3\)	A1 cso
(ii) \(\displaystyle\int_0^3 k(ay - y^2)\,dy = 1\) integration	M1
\(\left[k\left(\dfrac{ay^2}{2} - \dfrac{y^3}{3}\right)\right]_0^3 = 1\) answer correct	A1
\(k\left(\dfrac{9a}{2} - 9\right) = 1\) answer = 1	M1
\(k\left[\dfrac{9a - 18}{2}\right] = 1\) \(k = \dfrac{2}{9\left(a - 2\right)}\) *	A1 cso
	(6)

Notes

(i) M1 for putting \(\mathrm{f}(y) \geqslant 0\) or \(\mathrm{f}(3) \geqslant 0\) or \(ky\left(a - y\right) \geqslant 0\) or \(3k(a - 3) \geqslant 0\) or \((a - y) \geqslant 0\) or \((a - 3) \geqslant 0\) or state in words the probability can not be negative o.e.
A1 need one of \(ky\left(a - y\right) \geqslant 0\) or \(3k(a - 3) \geqslant 0\) or \((a - y) \geqslant 0\) or \((a - 3) \geqslant 0\) and \(a \geqslant 3\)

(ii) M1 attempting to integrate (at least one \(y^n \to y^{n+1}\)) (ignore limits)
A1 Correct integration. Limits not needed. And equals 1 not needed.
M1 dependent on the previous M being awarded. Putting equal to 1 and have the correct limits. Limits do not need to be substituted.
A1 cso

Mark scheme (b)

Scheme	Marks
\(\displaystyle\int_0^3 k(ay^2 - y^3)\,\mathrm{d}y = 1.75\) Int \(\displaystyle\int xf(x)\)	M1
\(\left[k\left(\dfrac{ay^3}{3} - \dfrac{y^4}{4}\right)\right]_0^3 = 1.75\) Correct integration \(\displaystyle\int xf(x) = 1.75\) and limits 0,3	A1 M1dep
\(k\left(9a - \dfrac{81}{4}\right) = 1.75\) \(2\left(9a - \dfrac{81}{4}\right) = 15.75(a - 2)\) subst \(k\)	M1dep
\(2.25a = -31.5 + \dfrac{81}{2}\) a = 4 *	A1cso
\(k = \dfrac{1}{9}\)	B1
	(6)

Notes

M1 for attempting to find \(\displaystyle\int y\mathrm{f}(y)\,\mathrm{d}y\) (at least one \(y^n \to y^{n+1}\)) (ignore limits)
A1 correct Integration
M1 \(\displaystyle\int y\mathrm{f}(y) = 1.75\) and limits 0,3 dependent on previous M being awarded
M1 subst in for \(k\). dependent on previous M being awarded
A1 cso 4
B1 cao 1/9

Sketch: curve rising from the origin to a maximum and falling, ending at y = 3 above the axis — Mark scheme (c)

Sketch: full parabola from 0 to 4 — Mark scheme (c)

Scheme	Marks
mode = 2	B1
	(1)
	(15 marks)