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}#?

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Answer 1 · 2016-11-09T15:10:50

#E(X)=24#

#sd=2sqrt6#

#P(X=k)=(24^ke^-24)/(k!)#for #kin{0,1,2,...oo}#

is a Poisson distribution which is of teh form

#P(X=k)=(lambda^ke^-lambda)/(k!)# for #kin{0,1,2,...oo}#

This means that

# E(X)=lambda#

and

#Var(X)=lambda#

for teh given distribution

#E(X)=24#

#Var(X)=24#

#sd=sqrt(Var(X))=sqrt24=2sqrt6#