AS June 2024 Q3

EdexcelAS paperCurrent spec9 marksContinuous Random Variables

3. The continuous random variable \(Y\) has probability density function

\[\mathrm{f}(y) = \begin{cases} \dfrac{1}{24}(y+2)(4-y) & 0 \leqslant y \leqslant 3 \\ 0 & \text{otherwise} \end{cases}\]

(a) Show that the mode of \(Y\) is 1, justifying your reasoning. (2)

Given that \(\mathrm{P}(Y \lt 1) = \dfrac{13}{36}\)

(b) determine whether the median of \(Y\) is less than, equal to, or greater than 2
Give a reason for your answer. (2)

Given that \(\mathrm{E}(Y^2) = \dfrac{213}{80}\)

Computer sketch of the pdf of Y for 0 to 3: a curve rising slightly from about 0.33 to a maximum of about 0.375 at y = 1, then falling to about 0.21 at y = 3 — Mark scheme (a)

Scheme	Marks	AO
Sketch or differentiation to find mode of \(Y\) \(\mathrm{f}'(y) = \frac{1}{24}(2 - 2y) = 0\)	M1	3.1a
Mode occurs at \(Y = 1\) *	A1*	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
By symmetry \(\mathrm{P}(Y \lt 2) = 2 \times \dfrac{13}{36} = \dfrac{13}{18}\) ( or 0.72 or better)	M1	2.1
Median is less than 2 since \(\dfrac{13}{18} \gt \dfrac{1}{2}\)	A1	2.4
	(2)

Notes

M1: use of symmetry or other method e.g. calculator to determine \(\mathrm{P}(Y \lt 2)\)

A1: for median is less than 2 with correct reasoning

Alternative

M1 for attempt at \(\mathrm{F}(y) = \frac{1}{24}\left(-\dfrac{y^3}{3} + y^2 + 8y\right)\) and \(\mathrm{F}(y) = \frac{1}{2}\)
Condone missing \(\frac{1}{24}\) for M1

A1 for awrt 1.37 and comment with some evidence that \(\mathrm{F}(y) = \frac{1}{2}\) attempted

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(Y) = \displaystyle\int_0^3 \tfrac{1}{24}\, y(y+2)(4-y)\,\mathrm{d}y\) or \(\displaystyle\int_0^3 \tfrac{1}{24}\left(8y + 2y^2 - y^3\right)\mathrm{d}y\)	M1	1.1b
\(\mathrm{E}(Y) = \frac{1}{24}\left[4y^2 + \dfrac{2}{3}y^3 - \dfrac{y^4}{4}\right]_0^3 \rightarrow \frac{1}{24}\left[\left(36 + 18 - \frac{81}{4}\right) - (0)\right]\left[= \frac{45}{32}\right]\)	M1	1.1b
\(\mathrm{Var}(Y) = \mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2 = \dfrac{213}{80} - \left(\text{“}\dfrac{45}{32}\text{”}\right)^2 \left[= \frac{3507}{5120}\right]\)	M1	1.1b
\(\mathrm{Var}(2Y) = 4\mathrm{Var}(Y)\)	M1	3.1a
\(= \dfrac{3507}{1280} = 2.73984\ldots\) awrt 2.74	A1	1.1b
	(5)
	(9 marks)

Notes

1^st M1: Attempt to integrate \(y\mathrm{f}(y)\) – ignore limits here.
Need to see at least one \(y^n \rightarrow y^{n+1}\)

2^nd M1: Substitution of correct limits into integral of \(y\mathrm{f}(y)\)
Don’t need to see limits used if intention is clear and answer correct.

Must see algebraic integration for 1^st and 2^nd M1

3^rd M1: Attempt to find variance i.e. use of \(\mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2\) ft their \(\mathrm{E}(Y)\)

4^th M1: Use of \(\mathrm{Var}(2Y) = 2^2\,\mathrm{Var}(Y)\). Independent of 1^st ~ 3^rd M1

A1: (dep on all Ms) for \(\dfrac{3507}{1280}\) or awrt 2.74