S2 June 2010 Q4

EdexcelOld spec10 marksContinuous Random Variables

4. The lifetime, \(X\), in tens of hours, of a battery has a cumulative distribution function \(\mathrm{F}(x)\) given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ \dfrac{4}{9}(x^2 + 2x - 3) & 1 \leqslant x \leqslant 1.5 \\ 1 & x \gt 1.5 \end{cases}\]

(a) Find the median of \(X\), giving your answer to 3 significant figures. (3)

(b) Find, in full, the probability density function of the random variable \(X\). (3)

A camping lantern runs on 4 batteries, all of which must be working. Four new batteries are put into the lantern.

(d) Find the probability that the lantern will still be working after 12 hours. (2)

Mark scheme (a)

Scheme	Marks
\(\dfrac{4}{9}\left(m^2 + 2m - 3\right) = 0.5\)	M1
\(m^2 + 2m - 4.125 = 0\) \(m = \dfrac{-2 \pm \sqrt{4 + 16.5}}{2}\)	M1
\(m = 1.26,\ -3.264\) (median =) 1.26	A1
	(3)

Notes

M1 putting \(\mathrm{F}(x) = 0.5\)
M1 using correct quadratic formula. If use calc need to get 1.26 (384... )
A1 cao 1.26 must reject the other root.
If they use Trial and improvement they have to get the correct answer to gain the second M mark.

Mark scheme (b)

Scheme	Marks
Differentiating \(\dfrac{\mathrm{d}\left(\frac{4}{9}\left(x^2 + 2x - 3\right)\right)}{\mathrm{d}x} = \dfrac{4}{9}\left(2x + 2\right)\)	M1 A1
\(\mathrm{f}(x) = \begin{cases} \dfrac{8}{9}\left(x + 1\right) & 1 \leqslant x \leqslant 1.5 \\ 0 & \text{otherwise} \end{cases}\)	B1ft
	(3)

Notes

M1 attempt to differentiate. At least one \(x^n \to x^{n-1}\)
A1 correct differentiation
B1 must have both parts- follow through their \(\mathrm{F}^{\prime}(x)\) Condone <

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(X \geqslant 1.2) = 1 - \mathrm{F}(1.2)\)	M1
\(= 1 - 0.3733\) \(= \dfrac{47}{75},\ 0.6267\) awrt 0.627	A1
	(2)

Notes

M1 finding/writing \(1 - \mathrm{F}(1.2)\) may use/write \(\displaystyle\int_{1.2}^{1.5} \frac{8}{9}(x + 1)\,\mathrm{d}x\) or \(1 - \displaystyle\int_{1}^{1.2} \frac{8}{9}(x + 1)\,\mathrm{d}x\) or \(\displaystyle\int_{1.2}^{1.5} \text{"their f}(x)\text{"}\,\mathrm{d}x\). Condone missing d\(x\)
A1 awrt 0.627

Mark scheme (d)

Scheme	Marks
\((0.6267)^4 = 0.154\) awrt 0.154 or 0.155	M1 A1
	(2)
	(10 marks)

Notes

M1 \((\text{c})^4\) If expressions are not given you need to check the calculation is correct to 2sf.
A1 awrt 0.154 or 0.155