AS June 2018 Q2

EdexcelAS paperCurrent spec8 marksContinuous Random Variables

2. The continuous random variable \(X\) has probability density function

\[\mathrm{f}(x) = \begin{cases} \dfrac{1}{8} & 1 \leqslant x \leqslant 9 \\ 0 & \text{otherwise} \end{cases}\]

(a) Write down the name given to this distribution. (1)

The continuous random variable \(Y = 5 - 2X\)

(b) Find \(\mathrm{P}(Y \gt 0)\) (2)

(d) Find \(\mathrm{P}(Y \lt 0 \mid X \lt 7.5)\) (3)

Scheme	Marks	AO
(Continuous) uniform or rectangular	B1	1.2
	(1)

Mark scheme (b)

Scheme	Marks	AO
\([\mathrm{P}(Y \gt 0) = \mathrm{P}(5 - 2X \gt 0) =]\ \mathrm{P}(X \lt 2.5)\) or \(\mathrm{f}(y) = \begin{cases} \dfrac{1}{16} & -13 \leqslant y \leqslant 3 \end{cases}\)	M1	1.1b
\(\dfrac{2.5 - 1}{8} = \underline{\dfrac{3}{16}}\) or \(\dfrac{3 - 0}{16} = \underline{\dfrac{3}{16}}\)	A1	1.1b
	(2)

Notes

M1 for using the distribution of \(X\) to obtain \(\mathrm{P}(X \lt 2.5)\) or for finding the distribution of \(Y\) in the range \(-13 \leqslant y \leqslant 3\)

A1 for \(\dfrac{3}{16}\) or 0.1875

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(Y) = 5 - 2\mathrm{E}(X)\ \left[= 5 - 2\left(\tfrac{1+9}{2}\right)\right]\) or \(\mathrm{E}(Y) = \dfrac{-13 + 3}{2}\)	M1	1.1b
\(= \underline{-5}\)	A1	1.1b
	(2)

Notes

M1 for use of \(\mathrm{E}(aX + b)\) or for use of \(\dfrac{a + b}{2}\) from the distribution of \(Y\)

A1 for –5

Mark scheme (d)

Scheme	Marks	AO
\([\mathrm{P}(Y \lt 0) \mid (X \lt 7.5)] = \dfrac{\mathrm{P}(2.5 \lt X \lt 7.5)}{\mathrm{P}(X \lt 7.5)}\)	M1	3.1a
\(= \dfrac{\;\dfrac{7.5 - 2.5}{8}\;}{\;\dfrac{7.5 - 1}{8}\;} \left[= \dfrac{0.625}{0.8125}\right]\)	M1	1.1b
\(= \underline{\dfrac{10}{13}}\)	A1	1.1b
	(3)
	(8 marks)

Notes

1^st M1 for a correct ratio expression

2^nd M1 for a correct numerical expression

A1 for \(\dfrac{10}{13}\)

SC: If M0M0A0 scored, then a correct numerator or correct denominator scores M0M1A0