S2 January 2010 Q2

EdexcelOld spec10 marksContinuous Random Variables

2. A continuous random variable \(X\) has cumulative distribution function

\[\mathrm{F}(x) = \begin{cases} 0, & x \lt -2 \\ \dfrac{x + 2}{6}, & -2 \leqslant x \leqslant 4 \\ 1, & x \gt 4 \end{cases}\]

(a) Find \(\mathrm{P}(X \lt 0)\). (2)

(b) Find the probability density function \(\mathrm{f}(x)\) of \(X\). (3)

(d) Find the mean and the variance of \(X\). (3)

(e) Write down the value of \(\mathrm{P}(X = 1)\). (1)

Mark scheme (a)

Scheme	Marks
\(\mathrm{P}(X \lt 0) = \mathrm{F}(0)\)	M1
\(= \dfrac{2}{6} = \dfrac{1}{3}\)	A1
	(2)

Notes

M1 for attempting to find F(0) by a correct method eg subst 0 into F(\(x\)) or \(\displaystyle\int_{-2}^{0} \frac{1}{6}\,dx\)

Do NOT award M1 for \(\displaystyle\int_{-2}^{0} \frac{x + 2}{6}\,dx\) or \(\dfrac{1}{2} \times \dfrac{1}{3} \times 2\) both of which give the correct answer by using F(\(x\)) as the pdf

A1 1/3 o.e or awrt 0.333
Correct answer only with no incorrect working gets M1 A1

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}(x) = \dfrac{\mathrm{dF}(x)}{\mathrm{d}x}\)	M1
\(\mathrm{f}(x) = \begin{cases} \dfrac{1}{6} & -2 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\)	A1 B1
	(3)

Notes

M1 for attempting to differentiate F(\(x\)). (for attempt it must have no \(x\)s in)
A1 for the first line. Condone < signs
B1 for the second line. – They must have 0 \(x \lt -2\) and \(x \gt 4\) only.

Scheme	Marks
Continuous Uniform (Rectangular) distribution	B1
	(1)

Mark scheme (d)

Scheme	Marks
Mean = 1	B1
Variance is \(\dfrac{(4 - -2)^2}{12} = 3\)	M1 A1
	(3)

Notes

B1 for mean = 1

M1 for attempt to use \(\dfrac{[\pm(b - a)]^2}{12}\), they must subst in values and not just quote the formula, or using \(\displaystyle\int_{-2}^{4} x^2(\textit{their } f(x)) - (\textit{their mean})^2\), including limits. Must get \(x^3\) when they integrate.

A1 cao.