Show that $P = \frac{X}{n}$ is an unbiased estimator of $p$.

Show that ${\text{E}}\left( U \right) = \left( {n - 1} \right)p\left( {1 - p} \right)$.

Hence write down an unbiased estimator of Var(X).

${\text{E}}\left( P \right) = {\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}\left( X \right)$      M1

$= \frac{1}{n}\left( {np} \right) = p$      A1

so P is an unbiased estimator of $p$     AG

[2 marks]

${\text{E}}\left( {nP\left( {1 - P} \right)} \right) = {\text{E}}\left( {n\left( {\frac{X}{n}} \right)\left( {1 - \frac{X}{n}} \right)} \right)$

$= {\text{E}}\left( X \right) = \frac{1}{n}{\text{E}}\left( {{X^2}} \right)$      M1A1

use of ${\text{E}}\left( {{X^2}} \right) = {\text{Var}}\left( X \right) + {\left( {{\text{E}}\left( X \right)} \right)^2}$      M1

Note: Allow candidates to work with P rather than X for the above 3 marks.

$= np - \frac{1}{n}\left( {np\left( {1 - p} \right) + {{\left( {np} \right)}^2}} \right)$       A1

$= np - p\left( {1 - p} \right) - n{p^2}$

$= np\left( {1 - p} \right) - p\left( {1 - p} \right)$      A1

Note: Award A1 for the factor of $\left( {1 - p} \right)$.

$= \left( {n - 1} \right)p\left( {1 - p} \right)$     AG

[5 marks]

an unbiased estimator is $\frac{{{n^2}P\left( {1 - P} \right)}}{{n - 1}}\left( { = \frac{{nU}}{{n - 1}}} \right)$       A1

[1 mark]