Date | May 2012 | Marks available | 5 | Reference code | 12M.3sp.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
Command term | Determine | Question number | 4 | Adapted from | N/A |
Question
The continuous random variable X has probability density function f given by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{2x,}&{0 \leqslant x \leqslant 0.5,} \\
{\frac{4}{3} - \frac{2}{3}x,}&{0.5 \leqslant x \leqslant 2} \\
{0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]
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.
Markscheme
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]
Examiners report
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.