AS June 2022 Q4

EdexcelAS paperCurrent spec9 marksContinuous Random Variables

4. A random variable \(X\) has probability density function given by

\[\mathrm{f}(x) = \begin{cases} 0.8 - 6.4x^{-3} & 2 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]

The median of \(X\) is \(m\)

(a) Show that \(m^3 - 3.625m^2 + 4 = 0\) (3)

(b)

(i) Find \(\mathrm{f}'(x)\)

(ii) Explain why the mode of \(X\) is 4 (2)

Given that \(\mathrm{E}(X^2) = 10.5\) to 3 significant figures,

Mark scheme (a)

Scheme	Marks	AO
\(\displaystyle\int_2^m (0.8 - 6.4x^{-3})\,\mathrm{d}x = 0.5\)	M1	2.1
\(\left[0.8x + 3.2x^{-2}\right]_2^m = 0.5\)	M1	1.1b
\(0.8m + \dfrac{3.2}{m^2} - \left(0.8(2) + \dfrac{3.2}{2^2}\right) = 0.5 \rightarrow 0.8m + \dfrac{3.2}{m^2} - 2.9 = 0 \rightarrow\) \(m^3 - 3.625m^2 + 4 = 0\)*	A1*cso	1.1b
	(3)

Notes

M1: Integral = 0.5 (ignore limits)

M1: Integration with limits

A1*cso: Given answer with at least one line of intermediate working.

Mark scheme (b)

Scheme	Marks	AO
(i) \(\mathrm{f}'(x) = 19.2x^{-4}\)	B1	1.1b
(ii) Since \(\mathrm{f}'(x) \gt 0\), \(\mathrm{f}(x)\) is increasing (the pdf has its maximum value at the upper end of the interval), the mode is 4	B1	2.4
	(2)

Notes

(i) B1: \(19.2x^{-4}\)

(ii) B1: Correct reasoning and conclusion (allow equivalent correct reasoning e.g. no turning points with a sketch of \(\mathrm{f}(x)\)).
Do not allow unsupported comments on their own to score e.g. ‘\(x = 4\) is the highest point on \(\mathrm{f}(x)\)’

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(X) = \displaystyle\int_2^4 x(0.8 - 6.4x^{-3})\,\mathrm{d}x\)	M1	1.1b
\(\mathrm{E}(X) = \left[0.4x^2 + 6.4x^{-1}\right]_2^4\) \(= 0.4(4^2) + 6.4(4^{-1}) - \left(0.4(2^2) + 6.4(2^{-1})\right)\ [= 3.2]\)	M1	1.1b
\(\mathrm{Var}(X) = 10.5 - \text{‘}3.2\text{’}^2\)	M1	1.1b
\(\mathrm{Var}(X) = 0.26\)	A1	1.1b
	(4)
	(9 marks)

Notes

M1: Multiplying out \(x\mathrm{f}(x)\) and attempt to integrate

M1: correct use of limits (implied by 3.2oe)

M1: Use of \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\)

A1: 0.26