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
X |
-2 |
-1 |
0 |
1 |
2 |
3 |
P(x) |
0.2 |
0.12 |
0,1 |
0.24 |
0.1 |
0.24 |
$\therefore$ 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**