Question

If a random variable has a Poisson’s Distribution such that


P (4) = P (3) Find P (5)
e=2.1783

Answer 1

Answer:

PMF of the Poisson distribution, p(x) = Exp(-L)*(L^x)/x!

Given, p(1) = p(2)

So, Exp(-L)*(L^1)/1 = Exp(-L)*(L^2)/2

=>2L = L^2

=> L = 0 or 2

For a Poisson distribution, L > 0 and so L = 2

So, mean of the distribution = 2

P(4) = Exp(-2)*(2^4)/4! = Exp(-2)*16/24 = 0.09

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