The random variable \(X\) has a Poisson distribution with \(\mathrm { E } ( X ) = \lambda\).
Prove, from first principles, that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\).
Hence deduce that \(\operatorname { Var } ( X ) = \lambda\).
The independent Poisson random variables \(X _ { 1 }\) and \(X _ { 2 }\) are such that \(\mathrm { E } \left( X _ { 1 } \right) = 5\) and \(\mathrm { E } \left( X _ { 2 } \right) = 2\).
The random variables \(D\) and \(F\) are defined by
$$D = X _ { 1 } - X _ { 2 } \quad \text { and } \quad F = 2 X _ { 1 } + 10$$
Determine the mean and the variance of \(D\).
Determine the mean and the variance of \(F\).
For each of the variables \(D\) and \(F\), give a reason why the distribution is not Poisson.
The daily number of black printer cartridges sold by a shop may be modelled by a Poisson distribution with a mean of 5 .
Independently, the daily number of colour printer cartridges sold by the same shop may be modelled by a Poisson distribution with a mean of 2.
Use a distributional approximation to estimate the probability that the total number of black and colour printer cartridges sold by the shop during a 4 -week period ( 24 days) exceeds 175.