Answer to Question #208741 in Statistics and Probability for Chalo

Question #208741

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


1
Expert's answer
2021-06-23T08:22:35-0400
"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}\n \\begin{array}{c:c}\n y & 0 & 1 & 2 & 3 & 9 \\\\ \\hline \n p(y) & 0.05 & 0.10 & 0.15 & 0.20 & 0.50\n\\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"


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