S2 June 2014 Q2

EdexcelOld spec14 marksContinuous Random Variables

2. The length of time, in minutes, that a customer queues in a Post Office is a random variable, \(T\), with probability density function

\[\mathrm{f}(t) = \begin{cases} c(81 - t^2) & 0 \leqslant t \leqslant 9 \\ 0 & \text{otherwise} \end{cases}\]

where \(c\) is a constant.

(a) Show that the value of \(c\) is \(\dfrac{1}{486}\) (4)

(b) Show that the cumulative distribution function \(\mathrm{F}(t)\) is given by \[\mathrm{F}(t) = \begin{cases} 0 & t \lt 0 \\ \dfrac{t}{6} - \dfrac{t^3}{1458} & 0 \leqslant t \leqslant 9 \\ 1 & t \gt 9 \end{cases}\] (2)

A customer has been queueing for 3 minutes.

(d) Find the probability that this customer will be queueing for at least 7 minutes. (3)

Three customers are selected at random.

(e) Find the probability that exactly 2 of them had to queue for longer than 3 minutes. (3)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int_0^9 c\left(81 - t^2\right)\mathrm{d}t = 1\)	M1
\(c\left[81t - \dfrac{t^3}{3}\right]_0^9 = 1\)	A1
\(c\left[81 \times 9 - \dfrac{9^3}{3}\right] = 1\)	M1d
\(486c = 1\)
\(c = \dfrac{1}{486}\)	A1cso
	(4)

Notes

1st M1 Attempting to integrate, For attempt \(x^n \to x^{n+1}\) and \(c\) must remain as \(c\) or 1/486. Ignore limits

1st A1 Correct integration. Ignore limits.

2nd M1 dependent on previous M being awarded. Putting = 1 and substitution of 9 as a limit seen. Need at least one intermediate step before getting 486 or substitution of 1/486 and 9 seen and leading to an answer of 1

A1 \(c = \dfrac{1}{486}\) cso or if verifying, the statement \(c = \dfrac{1}{486}\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{F}(t) = \dfrac{1}{486}\displaystyle\int_0^t 81 - x^2\,\mathrm{d}x\)	M1
\(= \dfrac{1}{486}\left[81t - \dfrac{x^3}{3}\right]_0^t\)
\(= \dfrac{t}{6} - \dfrac{t^3}{1458}\)
\(\mathrm{F}(t) = \begin{cases} 0 & t \lt 0 \\ \dfrac{t}{6} - \dfrac{t^3}{1458} & 0 \leqslant t \leqslant 9 \\ 1 & t \gt 9 \end{cases}\)	A1cso
	(2)

Notes

M1 Attempting to integrate with correct limits or \(\displaystyle\int \mathrm{f}(t)\,\mathrm{d}t + \mathrm{C}\) and F(0) = 0 or F(9) = 1. Subst in \(c\) at some point

A1 \(\mathrm{F}(t)\) must be stated and cso. Condone use of \(\lt\) instead of \(\leqslant\) etc.

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(T \gt 3) = 1 - \left(\dfrac{3}{6} - \dfrac{3^3}{1458}\right)\)	M1
\(= \dfrac{14}{27}\) or awrt 0.519	A1
	(2)

Notes

M1 using or writing 1 – F(3) or \(\dfrac{1}{486}\displaystyle\int_3^9 81 - x^2\,\mathrm{d}x\) or \(1 - \mathrm{P}(X \leqslant 3)\)

A1 awrt 0.519

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(T \gt 7 \mid T \gt 3) = \dfrac{0.068587}{0.5185}\)	M1A1ft
\(= \dfrac{25}{189}\) or awrt 0.132	A1
	(3)

Notes

M1 \(\dfrac{\textit{a probability}}{\textit{their (c)}}\) where \(0 \lt\) a probability \(\lt\) their (c) \(\lt 1\). If a probability \(\geqslant\) their (c), give M0.

A1ft \(\dfrac{\;\frac{50}{729}\;}{\textit{their (c)}}\) or \(\dfrac{\text{awrt}\,0.0686}{\textit{their (c)}}\)

A1 \(\dfrac{25}{189}\) or awrt 0.132

Mark scheme (e)

Scheme	Marks
\({}^3C_2(0.5185)^2(1 - 0.5185) = \dfrac{2548}{6561}\) or awrt 0.388/ 0.387	M1A1ftA1
	(3)
	(14 marks)

Notes

M1 Allow (their ‘0.5185’)\(^2\)(1 – their ‘0.5185’)

A1ft Allow \({}^3\mathrm{C}_2\) (their ‘0.5185’)\(^2\)(1 – their ‘0.5185’)

A1 awrt 0.388 or 0.387