A2 June 2019 Q4

EdexcelCurrent spec8 marksContinuous Random Variables

4. The continuous random variable \(X\) has cumulative distribution function given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \leqslant 0 \\ k\left(x^3 - \dfrac{3}{8}x^4\right) & 0 \lt x \leqslant 2 \\ 1 & x \gt 2 \end{cases}\]

where \(k\) is a constant.

(a) Show that \(k = \dfrac{1}{2}\) (1)

(b) Showing your working clearly, use calculus to find

(i) \(\mathrm{E}(X)\)

(ii) the mode of \(X\)

(6)

Mark scheme (a)

Scheme	Marks	AO
\(k\left(2^3 - \dfrac{3}{8}2^4\right) = 1\) \(2k = 1\) \(k = \dfrac{1}{2}\) *	B1*	1.1b
Or \(\dfrac{1}{2}\left(2^3 - \dfrac{3}{8}2^4\right) = 1 \quad \therefore k = \dfrac{1}{2}\) *	(B1*)
	(1)

Notes

B1*: substituting \(x = 2\) into \(\mathrm{F}(x)\) and equating to 1 leading to \(k = \dfrac{1}{2}\) with no errors.
Minimum subst seen is \(k(8 - 6) = 1\) or \(0.5(8 - 6) = 1\)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}(x) = k\left[3x^2 - \dfrac{3}{2}x^3\right]\)	M1	2.1
(i) \(\displaystyle\int_0^2 x\mathrm{f}(x)\,\mathrm{d}x = k\int_0^2 \left(3x^3 - \frac{3}{2}x^4\right)\mathrm{d}x\)	M1d	1.1b
\(= \left[\dfrac{3x^4}{8} - \dfrac{3x^5}{20}\right]_0^2\)
\(= \dfrac{6}{5}\) or 1.2	A1	1.1b
(ii) \(3x - \dfrac{9x^2}{4} = 0\)	M1d	3.1a
\(x\left(3 - \dfrac{9x}{4}\right) = 0\)	M1d	1.1b
\(x = 0\) or \(\dfrac{4}{3}\) \(\therefore\) mode \(= \dfrac{4}{3}\)	A1	1.1b
	(6)

Notes

M1: Realising they need to find the pdf and attempting to differentiate \(k\left[x^3 - \dfrac{3}{8}x^4\right]\) at least 1 correct term

(i) M1d: dep on 1^st M1 Attempting to find \(\displaystyle\int_0^2 x(\text{their } \mathrm{f}(x))\,\mathrm{d}x\) At least one correct term ft their pdf

A1: \(\dfrac{6}{5}\) or 1.2 oe NB 1.2 with no working gains M0M0A0

(ii) M1d: dep on 1^st M1 for realising they need to differentiate their pdf. At least one correct term but ft their pdf

M1d: Dep on 3^rd M1. correct method for solving their differential of their pdf = 0 pdf must be of the form \(ax^2 + bx\)

A1: \(\therefore\) mode \(= \dfrac{4}{3}\) only. They must eliminate 0

Scheme	Marks	AO
Mode > mean implies it is negative skew	B1ft	2.4
	(1)
	(8 marks)