S2 June 2013 Q4

EdexcelOld spec9 marksContinuous Random Variables

4. A continuous random variable \(X\) is uniformly distributed over the interval \([b, 4b]\) where \(b\) is a constant.

(a) Write down \(\mathrm{E}(X)\). (1)

(b) Use integration to show that \(\mathrm{Var}(X) = \dfrac{3b^2}{4}\). (3)

Given that \(b = 1\) find

(d) the cumulative distribution function of \(X\), \(\mathrm{F}(x)\), for all values of \(x\), (2)

(e) the median of \(X\). (1)

Scheme	Marks
\(\mathrm{E}(X) = \frac{5b}{2}\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\mathrm{Var}(X) = \mathrm{E}(X^2) - (\mathrm{E}(X))^2\)
\(= \displaystyle\int_b^{4b} \frac{x^2}{3b}\,\mathrm{d}x - \left(\frac{5b}{2}\right)^2\)	M1
\(= \left[\dfrac{x^3}{9b}\right]_b^{4b} - \dfrac{25b^2}{4}\)	M1d
\(= \dfrac{63b^3}{9b} - \dfrac{25b^2}{4}\)
\(= \dfrac{3b^2}{4}\)	A1cso
	(3)

Notes

Alt 4(b)

Scheme	Marks
\(\mathrm{Var}(X) = \displaystyle\int_a^b \frac{(x - \bar{x})^2}{b - a}\,\mathrm{d}x\)
\(= \displaystyle\int_b^{4b} \frac{4x^2 - 20bx + 25b^2}{12b}\,\mathrm{d}x\)	M1
\(= \left[\dfrac{\frac{4x^3}{3} - 10bx^2 + 25b^2x}{12b}\right]_b^{4b}\)	M1
\(= \dfrac{9b^3}{12b}\)
\(= \dfrac{3b^2}{4}\)	A1cso
	(3)

NB remember the answer is given (AG) so they must show their working

1st M1 for using \(\displaystyle\int \frac{x^2}{3b}\,\mathrm{d}x\) - (their (a))\(^2\) limits not needed and condone missing d\(x\). NB need not use the letter \(x\) but if they use \(b\) instead do not award if they cancel down to \(\dfrac{b}{3}\)

NB Check they have subtracted (their(a))\(^2\)

2nd M1 dependent on previous M being awarded. For some correct integration \(x^n \to x^{n+1}\) and correct limits substituted at some point. condone \(4b^3\) instead of \((4b)^3\)

A1 for correct solution with no incorrect working seen.

Scheme	Marks
\(\mathrm{Var}(3 - 2X) = 4\mathrm{Var}(X)\)	M1
\(= 3b^2\)	A1
	(2)

Mark scheme (d)

Scheme	Marks
\(\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ \dfrac{x - 1}{3} & 1 \leqslant x \leqslant 4 \\ 1 & x \gt 4 \end{cases}\)	B1B1
	(2)

Notes

1st B1 top and bottom line. Allow use of \(\leqslant\) instead of \(\lt\) and \(\geqslant\) instead of \(\gt\)

2nd B1 middle row. Allow use of \(\lt\) instead of \(\leqslant\)

Scheme	Marks
\(\frac{x - 1}{3} = 0.5\) so \(x = 2.5\)	B1
	(1)
	(9 marks)