# What are the mean and standard deviation of a probability density function given by Pr(X=k) = (24^ke^-24)/(k!)  for k in {0,1,2... oo}?

Nov 9, 2016

$E \left(X\right) = 24$

$s d = 2 \sqrt{6}$

#### Explanation:

P(X=k)=(24^ke^-24)/(k!)for $k \in \left\{0 , 1 , 2 , \ldots \infty\right\}$

is a Poisson distribution which is of teh form

P(X=k)=(lambda^ke^-lambda)/(k!) for $k \in \left\{0 , 1 , 2 , \ldots \infty\right\}$

This means that

$E \left(X\right) = \lambda$

and

$V a r \left(X\right) = \lambda$

for teh given distribution

$E \left(X\right) = 24$

$V a r \left(X\right) = 24$

$s d = \sqrt{V a r \left(X\right)} = \sqrt{24} = 2 \sqrt{6}$