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,x⩾Find 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.