Answer:
a) E(y) = 0.8
b) The average subcharge is $165
Step-by-step explanation:
We are given the following distribution in the question:
y: Â Â Â Â Â 0 Â Â Â Â 1 Â Â Â Â 2 Â Â Â Â 3
P(y): Â 0.50 Â 0.25 Â 0.20 Â 0.05
a) E(y)
[tex]E(y) = \displaystyle\sum y_iP(y_i)\\\\E(y) = 0(0.50)+1(0.25)+2(0.20)+3(0.05)\\E(y) = 0.8[/tex]
b) Expected value of subcharge
Subcharge =
[tex]\$110y^2[/tex]
Expected value of subcharge =
[tex]=110E(y^2)\\\\=110\displaystyle\sum y_i^2P(y_i)\\\\=110(0^2(0.50)+1^2(0.25)+2^2(0.20)+3^2(0.05))\\=110(1.5)=165[/tex]
Thus, the average subcharge is $165