S2 June 2012 Q7

EdexcelOld spec14 marksContinuous Random Variables

7. The continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by

\[\mathrm{f}(x) = \begin{cases} \dfrac{x^2}{45} & 0 \leqslant x \leqslant 3 \\[1ex] \dfrac{1}{5} & 3 \lt x \lt 4 \\[1ex] \dfrac{1}{3} - \dfrac{x}{30} & 4 \leqslant x \leqslant 10 \\[1ex] 0 & \text{otherwise} \end{cases}\]

(a) Sketch \(\mathrm{f}(x)\) for \(0 \leqslant x \leqslant 10\) (4)

(b) Find the cumulative distribution function \(\mathrm{F}(x)\) for all values of \(x\). (8)

Sketch: curve rising from (0, 0) to 0.2 at x = 3, horizontal line at 0.2 from x = 3 to 4, then a straight line falling to the x-axis at x = 10 — Mark scheme (a)

Scheme	Marks
	B1 B1 B1 B1dep 0.2,3,4,10
	(4)

Mark scheme (b)

Scheme	Marks
\(\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\ \dfrac{x^3}{135} & 0 \leqslant x \leqslant 3 \\[1ex] \dfrac{x}{5} - \dfrac{2}{5} & 3 \lt x \lt 4 \\[1ex] \dfrac{x}{3} - \dfrac{x^2}{60} - \dfrac{2}{3} & 4 \leqslant x \leqslant 10 \\ 1 & x \gt 10 \end{cases}\)	M1A1 M1A1 M1A1
1st M1 For \(0 \leqslant x \leqslant 3\), \(\mathrm{F}(x) = \displaystyle\int_0^x \frac{t^2}{45}\,\mathrm{d}t = \left[\frac{t^3}{135}\right]_0^x\)
2nd M1 For \(3 \lt x \lt 4\), \(\mathrm{F}(x) = \displaystyle\int_3^x \frac{1}{5}\,\mathrm{d}t + \frac{1}{5} = \left[\frac{t}{5}\right]_3^x + \frac{1}{5}\) or \(\mathrm{F}(x) = \displaystyle\int \frac{1}{5}\,\mathrm{d}x + \mathrm{C}\) and uses \(\mathrm{F}(3) = \dfrac{1}{5}\): \(\dfrac{1}{5} = \left[\dfrac{3}{5}\right] + C\)
3rd M1 For \(4 \leqslant x \leqslant 10\), \(\mathrm{F}(x) = \displaystyle\int_4^x \frac{1}{3} - \frac{x}{30}\,\mathrm{d}t + \frac{2}{5}\) or \(\mathrm{F}(x) = \displaystyle\int \frac{1}{3} - \frac{x}{30}\,\mathrm{d}x + \mathrm{C}\) and uses \(\mathrm{F}(4) = \dfrac{2}{5}\) or \(\mathrm{F}(10) = 1\) \(\mathrm{F}(x) = \left[\dfrac{t}{3} - \dfrac{t^2}{60}\right]_4^x + \dfrac{2}{5}\) \(\dfrac{2}{5} = \dfrac{4}{3} - \dfrac{4^2}{60} + \mathrm{C}\) or \(1 = \dfrac{10}{3} - \dfrac{10^2}{60} + \mathrm{C}\)
Top line of \(\mathrm{F}(x)\) ie 0 \(x \lt 0\)	B1
Bottom line of \(\mathrm{F}(x)\) ie 1 \(x \gt 10\)	B1
	(8)

Notes

For all the M marks, the attempt to integrate must have at least one \(x^n \to x^{n+1}\)

All A marks are for the correct expressions and ranges.

Do not penalise the use of \(\leqslant\) instead of \(\lt\) and \(\geqslant\) instead of \(\gt\).

1st M1 for attempt to integrate \(\displaystyle\int_0^x \frac{t^2}{45}\,\mathrm{d}t\) ignore limits

2nd M1 for attempt to integrate \(\displaystyle\int_3^x \frac{1}{5}\,\mathrm{d}t + \text{their F}(3)\) using correct limits. or for attempt to integrate \(\displaystyle\int \frac{1}{5}\,\mathrm{d}x + \mathrm{C}\) and substituting in 3 and putting = to their F(3) or substituting in 4 and putting = to their F(4) from their \(4 \leqslant x \leqslant 10\) line

3rd M1 for attempt to integrate \(\displaystyle\int_4^x \frac{1}{3} - \frac{x}{30}\,\mathrm{d}t + \text{their F}(4)\) using correct limits. or for attempt to integrate \(\displaystyle\int \frac{1}{3} - \frac{x}{30}\,\mathrm{d}t + \mathrm{C}\) and substituting in 4 and putting = to their F(4) or substituting in 10 and putting = 1

Mark scheme (c)

Scheme	Marks
\(\mathrm{F}(8) = \dfrac{8}{3} - \dfrac{8^2}{60} - \dfrac{2}{3}\)	M1
\(= \dfrac{14}{15} = 0.933\)	A1 cso
	(2)
	(14 marks)

Notes

M1 substituting 8 into the 4th line of their cdf or F(3) + F(4) – F(3) + F(8) – F(4) or \(1 - \displaystyle\int_8^{10} \frac{1}{3} - \frac{x}{30}\) (attempt to integrate needed) or use areas e.g \(1 - \dfrac{1}{2} \times 2 \times \dfrac{1}{15}\) or \(1 - \dfrac{1}{15}\)

A1 14/15 awrt 0.933 from correct working.

NB If using F(3) + F(4) – F(3) + F(8) – F(4) then \(\mathrm{F}(x)\) must be correct.