Question

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

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

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

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

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

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

$$=5.50$$

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

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

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

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