Revision as of 20:28, 3 May 2023 by Admin (Created page with "'''Solution: D''' Let <math>N</math> be the number of claims filed. We are given <math display = "block"> P[N = 2] = \frac{e^{-\lambda}\lambda^2}{2!} = 3 \frac{e^{-\lambda...")
Exercise
ABy Admin
May 03'23
Answer
Solution: D
Let [math]N[/math] be the number of claims filed. We are given
[[math]]
P[N = 2] = \frac{e^{-\lambda}\lambda^2}{2!} = 3 \frac{e^{-\lambda}\lambda^4}{4!} = 3 \cdot P[N=4] 24 \lambda^2 = 6 \lambda^4
[[/math]]
which implies that [math]\lambda = 2 [/math]. Therefore, [math]\operatorname{Var}[N] = \lambda = 2 [/math].