Revision as of 00:53, 6 May 2023 by Admin (Created page with "'''Solution: C''' The four possible outcomes for which X + Y = 3 are given below, with their probabilities. <math display = "block"> \begin{align*} (0,3)&: e^{-1.7} \frac{...")
Exercise
May 06'23
Answer
Solution: C
The four possible outcomes for which X + Y = 3 are given below, with their probabilities.
[[math]]
\begin{align*}
(0,3)&: e^{-1.7} \frac{2.3^3e^{-2.3}}{3!} = 2.0278e^{-4} \\
(1,2)&: 1.7 \frac{e^{-1.7}}{1!} \frac{2.3^2e^{-2.3}}{2!} = 4.4965e^{-4} \\
(2,1) &: \frac{1.7^2 e^{-1.7}}{2!} \frac{2.3e^{-2.3}}{1!} = 3.3235e^{-4} \\
(3,0) &: \frac{1.7^3 e^{-1.7}}{3!}e^{-2.3} = 0.8188e^{-4}.
\end{align*}
[[/math]]
The conditional probabilities are found by dividing the above probabilities by their sum. They are, 0.1901, 0.4215, 0.3116, 0.0768, respectively. These apply to the X – Y values of –3, –1, 1,and 3. The mean is
–3(0.1901) –1(0.4215) + 1(0.3116) + 3(0.0768) = –0.4498.
The second moment is
9(0.1901) + 1(0.4215) + 1(0.3116) + 9(0.0768) = 3.1352.
The variance is 2.9329.