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Date November 2009 Marks available 8 Reference code 09N.2.hl.TZ0.5
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 5 Adapted from N/A

Question

The annual weather-related loss of an insurance company is modelled by a random variable X with probability density functionf(x)={2.5(200)2.5x3.5,xFind the median.

Markscheme

\int_{200}^M {\frac{{2.5{{\left( {200} \right)}^{2.5}}}}{{{x^{3.5}}}}{\text{d}}x}  = 0.5     M1A1A1

Note: Award M1 for the integral equal to 0.5

   A1A1 for the correct limits.

 

\frac{{ - {{200}^{2.5}}}}{{{M^{2.5}}}}\left( {\frac{{ - {{200}^{2.5}}}}{{{{200}^{2.5}}}}} \right) = 0.5     M1A1A1

Note: Award M1 for correct integration

   A1A1 for correct substitutions.

 

\frac{{ - {{200}^{2.5}}}}{{{M^{2.5}}}} + 1 = 0.5 \Rightarrow {M^{2.5}} = 2{\left( {200} \right)^{2.5}}     (A1)

M = 264     A1

 

[8 marks]

Examiners report

Many students used incorrect limits to the integral, although many did correctly let the integral equal to 0.5.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Definition and use of probability density functions.
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