Answer to Question #208741 in Statistics and Probability for Chalo

Question

Consider the discrete random variable (Y) with the following probability.P(Y=y)=0.05(y+1) for y=0,1,2,3,9

Answer 1

$p(Y=y)=0.05(1+y), y=0,1,2,3,9$

$p(Y=0)=0.05(1+0)=0.05$

$p(Y=1)=0.05(1+1)=0.10$

$p(Y=2)=0.05(1+2)=0.15$

$p(Y=3)=0.05(1+3)=0.20$

$p(Y=9)=0.05(1+9)=0.50$

Check

$0.05+0.10+0.15+0.20+0.50=1$

$\def\arraystretch{1.5} \begin{array}{c:c} y & 0 & 1 & 2 & 3 & 9 \\ \hline p(y) & 0.05 & 0.10 & 0.15 & 0.20 & 0.50 \end{array}$

$Var(Y)=E(Y^2)-\big(E(Y)\big)^2$

$E(Y)=\displaystyle\sum_{i=1}^5y_ip(y_i)=0(0.05)+1(0.10)$

$+2(0.15)+3(0.20)+9(0.50)=5.50$

$=5.50$

$E(Y^2)=\displaystyle\sum_{i=1}^5y_i^2p(y_i)=(0)^2(0.05)+(1)^2(0.10)$

$+(2)^2(0.15)+(3)^2(0.20)+(9)^2(0.50)=43.00$

$Var(Y)=43.00-\big(5.50\big)^2=12.75$

Or

$Var(Y)=\displaystyle\sum_{i=1}^5(y_i-\mu)^2p(y_i)=(0-5.50)^2(0.05)$

$+(1-5.50)^2(0.10)+(2-5.50)^2(0.15)$

$+(3-5.50)^2(0.20)+(9-5.50)^2(0.50)=12.75$

$Var(2Y+4)=2^2Var(Y)=2^2(12.75)=51.0$