Math, asked by someo6413, 6 hours ago

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

Answers

Answered by Anonymous
5

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 ꧂

Answered by Raghav1330
1

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.

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