According to the Poisson distribution, the probability of a value to take the value k:
"P(X=k)=\\frac{\\lambda^k}{k!}e^{-\\lambda}"
We know
"E(X)=\\lambda=0.4"
Then:
"P(X>0)=1-P(0)=1-e^{-0.4}\\approx0.33"
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Dear Nana Kwame, please use the panel for submitting new questions.
A discrete random variable has probability mass function, P(X = n) =1 2n. Let Y = 1, for x even −1, for x odd Find the expected value of Y ; (E[y]).
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Dear Nana Kwame, please use the panel for submitting new questions.
A discrete random variable has probability mass function, P(X = n) =1 2n. Let Y = 1, for x even −1, for x odd Find the expected value of Y ; (E[y]).
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