| Date | May 2012 | Marks available | 2 | Reference code | 12M.3sp.hl.TZ0.5 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
The random variable \(X \sim {\text{Po}}(m)\). Given that P(X = k −1) = P(X = k +1), where k is a positive integer,
show that \({m^2} = k(k + 1)\);
hence show that the mode of X is k .
Markscheme
\(\frac{{{m^{k - 1}}{e^{ - m}}}}{{(k - 1)!}} = \frac{{{m^{k + 1}}{e^{ - m}}}}{{(k + 1)!}}\) M1
\( \Rightarrow 1 = \frac{{{m^2}}}{{(k + 1)k}}\) A1
Note: Award A1 for any correct intermediate step.
\( \Rightarrow {m^2} = (k + 1)k\) AG
[2 marks]
\(\frac{{{\text{P}}(X = k)}}{{{\text{P}}(X = k - 1)}} = \frac{{{e^{ - m}} \times \frac{{{m^k}}}{{k!}}}}{{{e^{ - m}} \times \frac{{{m^{k - 1}}}}{{(k - 1)!}}}}\) M1
\( = \frac{m}{k}\) A1
\( = \frac{{\sqrt {k(k + 1)} }}{k}\) M1
\( = \sqrt {\frac{{k + 1}}{k}} > 1\) R1
so \({\text{P}}(X = k) > {\text{P}}(X = k - 1)\) R1
similarly \({\text{P}}(X = k) > {\text{P}}(X = k + 1)\) R1
hence k is the mode AG
[6 marks]
Examiners report
Most candidates were able to complete part (a). The remainder of the question involved some understanding of the shape of the distribution and some facility with algebraic manipulation.
Most candidates were able to complete part (a). The remainder of the question involved some understanding of the shape of the distribution and some facility with algebraic manipulation.
