Find i) k ii) mean III) variance __

i) We must have

$\sum pi$ = 1

0.2 + k+ 0.1+ 2k +0.1 + 2k =1

0.4 +5k = 1

5k = 1 - 0.4

5k = 0.6

k= $\frac {0.6} {5}$

k= $\frac {3} {25}$ or k = 0.12

ii) Put k= $\frac {3} {25}$ or k = 0.12 in above table, we get

$\therefore$ Mean = E(X) = $\sum PiXi$

= (-2 $\times$ 0.2) +(-1 $\times$0.12)+(0$\times$0.1)+(1$\times$0.24)+(2$\times$0.1)+(3$\times$0.24)

= -0.4-0.12+0+0.24+0.2+0.72

= E(X) = Mean= 0.64

$\therefore E(X^2) = \sum PiXi^2$

= $\ 0.2(-2)^2 + 0.12 (-1)^2 + 0.1 (0) + 0.24 (1)^2 + 0.1(2)^2 + 0.24 (3)^2$

= 0.8 + 0.12 +0+0.24+0.4+2.16

= \$ E(X)^2 = 3.72$ Variance = $\sigma$ = E( $\ x^2) - [E(x)]^2$ = 3.72- $\ {0.64}^2$ = 3.72 -0.4096 **$\sigma$ = 3.3104**