Answer in Statistics and Probability for Vishnu kishore #248363
October 24th, 2022
If the Poisson distribution is bimodal, the two modes are at the points x=λ-1 and x=λ, where λ is the parameter of the Poisson distribution.
“x=1 \nnx=2 \nnu03bb=2 \nnP(X=x) = frac{e^{-u03bb}u03bb^x}{x!} \nnP(X=1) = 2e^{-2} \nnP(X=2) = frac{e^{-2}2^2}{2!} = 2e^{-2}”
Required probability “= P(X=1) + P(X=2) = 2e^{-2} +2e^{-2} = 0.542”
