Question

If a random variable has a poisson distribution such that p(1)=p(2) then find the mean of distribution and p(4)

Answer 1

Answer:

Step-by-step explanation:

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Since it is Poisson distributed Random variable

P(X=k)=[λ^k.e^(-λ)]/k!

Given, P(X=1) we get

P(X=1)=[λ.e^(-λ)]/(1!)

P(X=1)=λe^(-λ)….(1)

P(X=2)=[λ².e^(-λ)]/2!

P(X=2)=[λ².e^(-λ)]/2….(2)

Given equation (1) and (2)

P(X=1)=P(X=2)

λ.e^(-λ) = [λ².(e^(-λ)]/2

2λe^(-λ)=λ²e^(-λ)

Since e^(±x) is always greater than zero we can cancel it on both the sides, we get

2λ=λ²

λ²-2λ=0

λ(λ-2)=0

λ=0 οr λ=2

But λ cannot be zero as it is a mean, hence λ=2

We require to find the probability at X=0 so,

P(X=k)=(λ^ke^(-λ))/k!

At X=0 the probability is derived by putting k=0

So, P(X=0)=(λ^(0).e^(-λ))/0!

P(X=0)=(1.e^(-λ))/1

P(X=0)=e^(-λ)

But λ=2, so,

P(X=0)=e-²

P(X=0)=1/e²

P(X=0)=0.135335283236

Answer 2

Answer:

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