AS June 2019 Q2

EdexcelAS paperCurrent spec9 marksContinuous Random Variables

2. Lloyd regularly takes a break from work to go to the local cafe. The amount of time Lloyd waits to be served, in minutes, is modelled by the continuous random variable \(T\), having probability density function

\[\mathrm{f}(t) = \begin{cases} \dfrac{t}{120} & 4 \leqslant t \leqslant 16 \\ 0 & \text{otherwise} \end{cases}\]

(a) Show that the cumulative distribution function is given by\[\mathrm{F}(t) = \begin{cases} 0 & t \lt 4 \\ \dfrac{t^2}{240} - c & 4 \leqslant t \leqslant 16 \\ 1 & t \gt 16 \end{cases}\]where the value of \(c\) is to be found. (2)

(b) Find the exact probability that the amount of time Lloyd waits to be served is between 5 and 10 minutes. (2)

(d) Find the value of \(k\) such that\[\mathrm{P}(T \lt k) = \frac{2}{3}\,\mathrm{P}(T \gt k)\]giving your answer to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\displaystyle\int \frac{t}{120}\,\mathrm{d}t = \frac{t^2}{240}\) and use of \(\mathrm{F}(4) = 0\) or \(\mathrm{F}(16) = 1\) or limits of \(t\) and 4 or attempt at area of trapezium allow 1 mistake. \(\dfrac{1}{2} \times (t - 4)\left(\dfrac{4}{120} + \dfrac{t}{120}\right)\)	M1	2.1
\(= \dfrac{t^2}{240} - \dfrac{1}{15}\)	A1	1.1b
	(2)

Notes

M1: for attempting to integrate and a correct method

A1: \(= \dfrac{t^2}{240} - \dfrac{1}{15}\) or \(= \dfrac{t^2}{240} - 0.0667\)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{F}(10) - \mathrm{F}(5) = \dfrac{100}{240} - \text{“}c\text{”} - \dfrac{25}{240} + \text{“}c\text{”}\)	M1	1.1b
\(= \dfrac{5}{16}\)	A1	1.1b
	(2)

Notes

M1: writing or using \(\mathrm{F}(10) - \mathrm{F}(5)\)

A1: awrt \(\dfrac{5}{16}\) or 0.3125 or exact equivalent

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{m^2}{240} - \dfrac{1}{15} = 0.5\)	M1	1.1b
\(m = 11.66\ldots.\) awrt 11.7	A1	1.1b
	(2)

Notes

M1: setting their \(\mathrm{F}(t) = 0.5\)

A1: awrt 11.7 or \(2\sqrt{34}\) or exact equivalent

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{F}(k) = \dfrac{2}{3}\big(1 - \mathrm{F}(k)\big)\) or \(\displaystyle\int_4^k \frac{t}{120}\,\mathrm{d}t = \frac{2}{3}\int_k^{16} \frac{t}{120}\,\mathrm{d}t\)	M1	3.1a
\(\dfrac{k^2}{240} - \text{“}\dfrac{1}{15}\text{”} = \dfrac{2}{3}\left(1 - \left(\dfrac{k^2}{240} - \text{“}\dfrac{1}{15}\text{”}\right)\right)\) or \(\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{3} \times \left(\dfrac{16}{15} - \dfrac{k^2}{240}\right)\)	dM1	1.1b
\(\dfrac{k^2}{144} = \dfrac{7}{9}\)
\(k = \sqrt{112}\) or awrt 10.6	A1	1.1b
	(3)
	(9 marks)

Notes

M1: Setting up a correct equation to solve the mathematical problem or setting up correct equation to find \(p\) and an attempt to solve

dM1: attempted to integrate and limits substituted or using “Their \(\mathrm{F}(k)\)” = “their \(p\)”

A1: \(\sqrt{112}\) or awrt 10.6

Alternative

Scheme	Marks
Let \(\mathrm{P}(T \lt k) = p\) then \(p = \dfrac{2}{3}(1 - p)\ \therefore\ p = \dfrac{2}{5}\)	(M1)
\(\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{5}\)	(dM1)
\(k = \sqrt{112}\) or awrt 10.6	(A1)