Chapter: Probability Distributions· Mathematics & Statistics (Commerce) Part 2
If \(X \sim P(m)\) with \(P(X = 1) = P(X = 2)\) then find the mean and \(P(X = 2)\). Given \(e^{-2} = 0.1353\)Solution : Since \(P(X = 1) = P(X = 2)\)\(\therefore \dfrac{e^{\square}m^1}{1!} = \dfrac{e^{-m}m^2}{\square}\)\(\therefore m = \square\)\(\t…
Q.6 (C)(ii)
If \(X \sim P(m)\) with \(P(X = 1) = P(X = 2)\) then find the mean and \(P(X = 2)\). Given \(e^{-2} = 0.1353\)
Solution : Since \(P(X = 1) = P(X = 2)\)
\(\therefore \dfrac{e^{\square}m^1}{1!} = \dfrac{e^{-m}m^2}{\square}\)
\(\therefore m = \square\)
\(\therefore P(X = 2) = \dfrac{e^{-2}\cdot 2^2}{2!} = \square\)