AS June 2025 Q4

EdexcelAS paperCurrent spec11 marksContinuous Random Variables

4. Subrat is modelling the time, \(t\) seconds, it takes for a computer to carry out a particular process.
He models the time using the continuous random variable \(T\) with cumulative distribution function

\[\mathrm{F}(t) = \begin{cases} 0 & t \lt 0 \\ at^4 - \dfrac{1}{8}t^3 + bt^2 & 0 \leqslant t \leqslant 4 \\ 1 & t \gt 4 \end{cases}\]

where \(a\) and \(b\) are constants.

(a) Find, in terms of \(a\) and \(b\), the probability that the computer takes less than 2 seconds to carry out the process. (2)

(b) Find, in terms of \(a\) and \(b\), the probability density function of \(T\) for all values of \(t\) (2)

The probability that the computer takes less than 2 seconds to carry out the process is \(\dfrac{11}{16}\)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{F}(2) \rightarrow \ a(2^4) - \dfrac{1}{8}(2^3) + b(2^2)\)	M1	3.4
\(16a + 4b - 1\)	A1	1.1b
	(2)

Notes

M1: Use of \(\mathrm{F}(2)\) but e.g. \(\mathrm{F}(4) - \mathrm{F}(2)\) is M0

A1: cao

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}(t) = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(at^4 - \dfrac{1}{8}t^3 + bt^2\right) = 4at^3 - \dfrac{3}{8}t^2 + 2bt\)	M1	1.1b
\(\mathrm{f}(t) = \begin{cases} 4at^3 - \dfrac{3}{8}t^2 + 2bt & 0 \leqslant t \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\)	A1	1.1b
	(2)

Notes

M1: attempt to differentiate \(\mathrm{F}(t)\) with at least one term correct

A1: both lines correct with correct limits

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{F}(4) = 1 \rightarrow 256a - 8 + 16b = 1\)	M1	3.1b
Solving simultaneously \(256a + 16b = 9 \qquad 16a + 4b = \dfrac{27}{16}\)	M1	1.1b
\(a = \dfrac{3}{256} \qquad b = \dfrac{3}{8}\)	A1	1.1b
\(\dfrac{\mathrm{d}}{\mathrm{d}t}\big(\mathrm{f}(t)\big) = 12at^2 - \dfrac{3}{4}t + 2b\)	M1	3.1b
\(\dfrac{9}{64}t^2 - \dfrac{3}{4}t + \dfrac{3}{4} = 0\) or e.g. \(3t^2 - 16t + 16 = 0\)	M1	2.1
\(t = \dfrac{4}{3}\) or \(t = 4\)	M1	1.1b
[\(t = 4\) is a minimum and \(\mathrm{f}(0) = 0\) so] mode of \(T\) is \(\dfrac{4}{3}\)	A1	3.2b
	(7)
	(11 marks)

Notes

M1: use of \(\mathrm{F}(4) = 1\)

M1: solving simultaneously. Method to eliminate one variable or implied by A1

A1: both values correct

M1: differentiating \(\mathrm{f}(t)\) ft their \(a\) and \(b\) values. Allow with letters \(a\) and \(b\) condone 1 slip

M1: (dep on 3^rd M1) setting equal to 0

M1: solving quadratic. Method must be seen or implied by the correct answer.

A1: selecting \(\dfrac{4}{3}\) as only solution