Answer to Question #208746 in Statistics and Probability for Jojo

Question #208746

3.1 Consider the discrete random variable ( Y ) with the following probability distribution:


P( Y =y ) 0.05( y+1) = for y = 0,1,2,3,9


3.1.1 Express the probability distribution in tabular form


3.1.2 Find the variance of 2Y + 4



1
Expert's answer
2021-06-23T08:20:55-0400
"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}"

"E(Y)=\\mu=\\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"

"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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