Question

If p(1)=p(2),then mean of poisson distribution

Answer 1

Answer:

If x is a Poisson variate, then its probability density function is p(x)=e−λλxx!,x=0,1,2,....∞p(x)=e−λλxx!,x=0,1,2,....∞ and λ>0λ>0

Its mean is λλ and variance is also λλ

Now, p(1)=p(2)⇒λ=λ22⇒λ(λ−2)=0⇒λ=2p(1)=p(2)⇒λ=λ22⇒λ(λ−2)=0⇒λ=2 as λ>0λ>0

Hence, mean and variance of the Poisson Distribution is 2.

Now, p(4)=e−λλ44!=e−2244!=1624e−2=23e−2p(4)=e−λλ44!=e−2244!=1624e−2=23e−2