| Date | May 2011 | Marks available | 2 | Reference code | 11M.1.sl.TZ1.4 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The probability distribution of a discrete random variable X is given by \[{\rm{P}}(X = x) = \frac{{{x^2}}}{{14}}{\text{, }}x \in \left\{ {1{\text{, }}2{\text{, }}k} \right\}{\text{, where}} k > 0\] .
Write down \({\rm{P}}(X = 2)\) .
Show that \(k = 3\) .
Find \({\rm{E}}(X)\) .
Markscheme
\({\rm{P}}(X = 2) = \frac{4}{{14}}\) \(\left( { = \frac{2}{7}} \right)\) A1 N1
[1 mark]
\({\rm{P}}(X = 1) = \frac{1}{{14}}\) (A1)
\({\rm{P}}(X = k) = \frac{{{k^2}}}{{14}}\) (A1)
setting the sum of probabilities \( = 1\) M1
e.g. \(\frac{1}{{14}} + \frac{4}{{14}} + \frac{{{k^2}}}{{14}} = 1\) , \(5 + {k^2} = 14\)
\({k^2} = 9\) (accept \(\frac{{{k^2}}}{{14}} = \frac{9}{{14}}\) ) A1
\(k = 3\) AG N0
[4 marks]
correct substitution into \({\rm{E}}(X) = \sum {x{\rm{P}}(X = x)} \) A1
e.g. \(1\left( {\frac{1}{{14}}} \right) + 2\left( {\frac{4}{{14}}} \right) + 3\left( {\frac{9}{{14}}} \right)\)
\({\rm{E}}(X) = \frac{{36}}{{14}}\) \(\left( { = \frac{{18}}{7}} \right)\) A1 N1
[2 marks]
Examiners report
Although many candidates were successful in working with the probability function, students had difficulty following the "show that" instruction of this question. Many substituted \(k = 3\) and worked backwards to show that the sum of probabilities is 1. Some would argue that \(k = 4\) does not work, but were unable to give a complete justification for \(k = 3\) . A good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided \({\rm{E}}(X)\) by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.
Although many candidates were successful in working with the probability function, students had difficulty following the "show that" instruction of this question. Many substituted \(k = 3\) and worked backwards to show that the sum of probabilities is 1. Some would argue that \(k = 4\) does not work, but were unable to give a complete justification for \(k = 3\) .
A good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided \({\rm{E}}(X)\) by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.
