Question #346737

Let X be a random variable with probability distribution

X -1 0 1 2 3

F(X) 0.125 .50 0.20 0.05 0.125

a) Find E(X) and VAR(X).

b) Find the probability distribution of the random variable Y= 2X+1. Using the 

probability distribution of Y determine E(Y) and VAR(Y).


1
Expert's answer
2022-06-02T14:58:23-0400

a)


E(X)=0.125(1)+0.50(0)+0.20(1)E(X)=0.125(-1)+0.50(0)+0.20(1)

+0.05(2)+0.125(3)=0.55+0.05(2)+0.125(3)=0.55

E(X2)=0.125(1)2+0.50(0)2+0.20(1)2E(X^2)=0.125(-1)^2+0.50(0)^2+0.20(1)^2

+0.05(2)2+0.125(3)2=1.65+0.05(2)^2+0.125(3)^2=1.65

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

=1.650.552=1.3475=1.65-0.55^2=1.3475

b)


y11357F(Y)0.1250.500.200.050.125\def\arraystretch{1.5} \begin{array}{c:c} y & -1 & 1 & 3 & 5 & 7 \\ \hline F(Y) & 0.125 & 0.50 & 0.20 & 0.05 & 0.125 \\ \end{array}


E(Y)=0.125(1)+0.50(1)+0.20(3)E(Y)=0.125(-1)+0.50(1)+0.20(3)

+0.05(5)+0.125(7)=2.1+0.05(5)+0.125(7)=2.1

E(Y)=2E(X)+1E(Y)=2E(X)+1

E(Y2)=0.125(1)2+0.50(1)2+0.20(3)2E(Y^2)=0.125(-1)^2+0.50(1)^2+0.20(3)^2

+0.05(5)2+0.125(7)2=9.8+0.05(5)^2+0.125(7)^2=9.8

Var(Y)=E(Y2)(E(Y))2Var(Y)=E(Y^2)-(E(Y))^2

=9.82.12=5.39=9.8-2.1^2=5.39

Var(Y)=(2)2Var(X)Var(Y)=(2)^2Var(X)


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