Sketch the function f and show that the lower quartile is 0.5.

(i)     Determine E(X ).

(ii)     Determine ${\text{E}}({X^2})$.

Two independent observations are made from X and the values are added.

The resulting random variable is denoted Y .

(i)     Determine ${\text{E}}(Y - 2X)$ .

(ii)     Determine ${\text{Var}}\,(Y - 2X)$.

(i)     Find the cumulative distribution function for X .

(ii)     Hence, or otherwise, find the median of the distribution.

piecewise linear graph

correct shape     A1

with vertices (0, 0), (0.5, 1) and (2, 0)     A1

LQ: x = 0.5 , because the area of the triangle is 0.25     R1

[3 marks]

(i)     ${\text{E}}(X) = \int_0^{0.5} {x \times 2x{\text{d}}x + \int_{0.5}^2 {x \times \left( {\frac{4}{3} - \frac{2}{3}x} \right){\text{d}}x = \frac{5}{6}{\text{ }}( = 0.833...)} }$     (M1)A1

(ii)     ${\text{E}}({X^2}) = \int_0^{0.5} {{x^2} \times 2x{\text{d}}x + \int_{0.5}^2 {{x^2} \times \left( {\frac{4}{3} - \frac{2}{3}x} \right){\text{d}}x = \frac{7}{8}{\text{ }}( = 0.875)} }$     (M1)A1

[4 marks]

(i)     ${\text{E}}(Y - 2X) = 2{\text{E}}(X) - 2{\text{E}}(X) = 0$     A1

(ii)     ${\text{Var}}\,(X) = \left( {{\text{E}}({X^2}) - {\text{E}}{{(X)}^2}} \right) = \frac{{13}}{{72}}$     A1

$Y = {X_1} + {X_2} \Rightarrow {\text{Var}}\,(Y) = 2{\text{Var }}(X)$     (M1)

${\text{Var}}\,(Y - 2X) = 2{\text{Var}}\,(X) + 4{\text{Var}}\,(X) = \frac{{13}}{{12}}$     M1A1

[5 marks]

(i)     attempt to use $cf(x) = \int {f(u){\text{d}}u}$     M1

obtain $cf(x) = \left\{ {\begin{array}{*{20}{c}}   {{x^2},}&{0 \leqslant x \leqslant 0.5,} \\   {\frac{{4x}}{3} - \frac{1}{3}{x^2} - \frac{1}{3},}&{0.5 \leqslant x \leqslant 2,} \end{array}} \right.$     $\begin{array}{*{20}{c}}   {{\boldsymbol{A1}}} \\   {{\boldsymbol{A2}}} \end{array}$

(ii)     attempt to solve $cf(x) = 0.5$     M1

$\frac{{4x}}{3} - \frac{1}{3}{x^2} - \frac{1}{3} = 0.5$     (A1)

obtain 0.775     A1 

Note: Accept attempts in the form of an integral with upper limit the unknown median.

Note: Accept exact answer $2 - \sqrt {1.5}$ .

[7 marks]

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution $\frac{2}{4} = 0.5$. Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution $\frac{2}{4} = 0.5$. Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution $\frac{2}{4} = 0.5$. Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution $\frac{2}{4} = 0.5$. Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.