| Date | May 2013 | Marks available | 6 | Reference code | 13M.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find and Show that | Question number | 3 | Adapted from | N/A |
Question
A random variable \(X\) has probability density function \(f\) given by:\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{\lambda {e^{ - \lambda x}},}&{{\text{for }}x \geqslant 0{\text{ where }}\lambda > 0} \\
{0,}&{{\text{for }}x < 0.}
\end{array}} \right.\]
(i) Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .
A chicken crosses a road. It is known that cars pass the chicken’s crossing route, with intervals between cars measured in seconds, according to the random variable \(X\) , with \(\lambda = 0.03\) . The chicken, which takes \(10\) seconds to cross the road, starts to cross just as one car passes.
(ii) Find the probability that the chicken will reach the other side of the road before the next car arrives.
Later, the chicken crosses the road again just after a car has passed.
(iii) Show that the probability that the chicken completes both crossings is greater than \(0.5\).
A rifleman shoots at a circular target. The distance in centimetres from the centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if \(5 < X \le 10\) and no points if \(X > 10\) .
(i) Find the expected score when one bullet is fired at the target.
A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his value of \(\lambda \) .
(ii) Given that his expected score is \(6.5\), find his value of \(\lambda \) .
Markscheme
(i) \({\rm{P}}(X > a) = \int_a^\infty {\lambda {e^{ - \lambda x}}{\rm{d}}x} \) M1
\(\left[ { - {e^{ - \lambda x}}} \right]_a^\infty \) A1
\( = {e^{ - \lambda a}}\) A1
(ii) \({\rm{P}}(X > 10) = {e^{ - 0.3}}( = 0.74 \ldots )\) (M1)A1
(iii) probability of a safe double crossing \( = {e^{ - 0.6}}\) \(( = {0.74^2})\) \( = 0.55\) A1
which is greater than \(0.5\) AG
[6 marks]
(i) \({\rm{P}}(X \le 1) = 0.3296 \ldots \) (A1)
\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \) (A1)
\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \) (A1)
\({\rm{E(score)}} = 10 \times 0.3296 \ldots + 5 \times 0.5349 \ldots + 1 \times 0.1170 \ldots \) M1A1
\( = 6.09\) A1
Note: Accept probabilities in exponential form until the final decimal answer.
(ii) \({\rm{E(score)}}\) for X with unknown parameter can be expressed as \(10 \times (1 - {e^{ - \lambda }}) + 5 \times ({e^{ - \lambda }} - {e^{ - 5\lambda }}) + ({e^{ - 5\lambda }} - {e^{ - 10\lambda }})\) (M1)(A1)
attempt to solve \({\rm{E(score)}} = 6.5\) (M1)
obtain \(\lambda = 0.473\) A1
[10 marks]
Examiners report
This question was generally well done.
This question was generally well done.
