Q.
The probability distribution of a random variable X is given below.
x = k
0
1
2
3
4
P(X = k)
0.1
0.4
0.3
0.2
0
The variance of X is
| x = k | 0 | 1 | 2 | 3 | 4 |
| P(X = k) | 0.1 | 0.4 | 0.3 | 0.2 | 0 |
TS EAMCET 2017
Solution:
We have,
$X$
$P(X)$
$P_{i} X_{i}$
$P_{i}X^{2}_{i}$
0
0.1
0
0
1
0.4
0.4
0.4
2
0.3
0.6
1.2
3
0.2
0.6
1.8
4
0
0
0
$\Sigma P_{i} X_{i}=1.6$
$\Sigma P_{i} X_{i}^{2}=3.4$
Variance $(\sigma^{2}) =\Sigma P_{i} X_{i}^{2}-\left(\Sigma P_{i} X_{i}\right)^{2} $
$=3.4-(1.6)^{2}=3.4-2.56=0.84 $
| $X$ | $P(X)$ | $P_{i} X_{i}$ | $P_{i}X^{2}_{i}$ |
|---|---|---|---|
| 0 | 0.1 | 0 | 0 |
| 1 | 0.4 | 0.4 | 0.4 |
| 2 | 0.3 | 0.6 | 1.2 |
| 3 | 0.2 | 0.6 | 1.8 |
| 4 | 0 | 0 | 0 |
| $\Sigma P_{i} X_{i}=1.6$ | $\Sigma P_{i} X_{i}^{2}=3.4$ |