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Date May 2011 Marks available 2 Reference code 11M.2.hl.TZ1.7
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 7 Adapted from N/A

Question

A continuous random variable \(X\) has a probability density function given by the function \(f(x)\) , where

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k{{\left( {x + 2} \right)}^2},}&{ - 2 \leqslant x < 0} \\
  {k,}&{0 \leqslant x \leqslant \frac{4}{3}} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Find the value of \(k\) .

[2]
a.

Hence find

(i)     the mean of \(X\) ;

(ii)     the median of \(X\) .

[5]
b.

Markscheme

\(k\int_{ - 2}^0 {{{\left( {x + 2} \right)}^2}} {\text{d}}x + \int_0^{\frac{4}{3}} {k{\text{d}}x}  = 1\)     M1

\(\frac{{8k}}{3} + \frac{{4k}}{3} = 1\)

\(k = \frac{1}{4}\)     A1

Note: Only ft on positive values of \(k\).

[2 marks]

a.

(i)     \({\text{E}}(X) = \frac{1}{4}{\int_{ - 2}^0 {x\left( {x + 2} \right)} ^2}{\text{d}}x + \frac{1}{4}\int_0^{\frac{4}{3}} {x{\text{d}}x} \)     M1

\( = \frac{1}{4} \times \frac{{ - 4}}{3} + \frac{2}{9}\)

\( = - \frac{1}{9}\)   (\( - 0.111\))     A1

 

(ii)     median given by a such that \({\text{P}}(X < a) = 0.5\)

\(\frac{1}{4}{\int_{ - 2}^a {\left( {x + 2} \right)} ^2}{\text{d}}x = 0.5\)     M1

\(\left[ {\frac{{{{\left( {x + 2} \right)}^3}}}{3}} \right]_{ - 2}^a = 2\)     (A1)

\({\left( {a + 2} \right)^3} - 0 = 6\)

\(a = \sqrt[3]{6} - 2\)   (\(= - 0.183\))     A1

 

[5 marks]

b.

Examiners report

Many candidates recognised that integration was the appropriate technique to solve this question but the fact that the function was piecewise proved problematic for many. Good use of technology by some candidates was seen but few sketches of the function were made. A sketch would have been helpful to many candidates when attempting to solve (b (ii).

a.

Many candidates recognised that integration was the appropriate technique to solve this question but the fact that the function was piecewise proved problematic for many. Good use of technology by some candidates was seen but few sketches of the function were made. A sketch would have been helpful to many candidates when attempting to solve (b (ii).

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.1 » Concepts of population, sample, random sample and frequency distribution of discrete and continuous data.

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