Question

if P(1)= P(5) in poissons distribution .find the value of mean

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

꧁XxMissMasoomxX01 ꧂

Answer 2

Given:

P(1) = P(5)

To Find:

value of mean

Solution:

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

=> 2L = $L^{2}$

=> L = 0 or 2

For a 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

Therefore, the value of mean is 0.09.