| Date | May 2013 | Marks available | 3 | Reference code | 13M.2.hl.TZ2.7 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
The length, X metres, of a species of fish has the probability density function
\[f(x) = \left\{ {\begin{array}{*{20}{r}}
{a{x^2},}&{{\text{for }}0 \leqslant x \leqslant 0.5} \\
{0.5a(1 - x),}&{{\text{for }}0.5 \leqslant x \leqslant 1} \\
{0,}&{{\text{otherwise }}{\text{.}}}
\end{array}} \right.\]
Show that a = 9.6.
Sketch the graph of the distribution.
Find \({\text{P}}(X < 0.6)\).
Markscheme
\(\int_0^{0.5} {a{x^2}{\text{d}}x + \int_{0.5}^1 {0.5a(1 - x){\text{d}}x = 1} } \) M1A1
\(\frac{{5a}}{{48}}\) (or equivalent) or \(a \times 0.104 \ldots = 1\) A1
Note: Award M1 for considering two definite integrals.
Award A1 for equating to 1.
Award A1 for a correct equation.
The A1A1 can be awarded in any order.
a = 9.6 AG
[3 marks]
correct shape for \(0 \leqslant x \leqslant 0.5\) and \(f(0.5) \approx 2.4\) A1
correct shape for \(0.5 \leqslant x \leqslant 1\) and \(f(1) = 0\) A1
[2 marks]
attempting to find \({\text{P}}(X < 0.6)\) (M1)
direct GDC use or eg \({\text{P}}(0 \leqslant X \leqslant 0.5) + {\text{P}}(0.5 \leqslant X \leqslant 0.6)\) or \(1 - {\text{P}}(0.6 \leqslant X \leqslant 1)\)
\({\text{P}}(X < 0.6) = 0.616{\text{ }}\left( { = \frac{{77}}{{125}}} \right)\) A1
[2 marks]
Examiners report
Part (a) was generally well done. Common errors usually involved not recognizing that the sum of the two integrals was equal to one, premature rounding or not showing full working to conclusively show that a = 9.6.
Part (b) was not well done with many graphs poorly labelled and offering no reference to domain and range.
Part (c) was reasonably well done. The most common error involved calculating an incorrect probability from an incorrect definite integral.
