Q.
A random variable $X$ has the probability distribution given below.
X
1
2
3
4
5
P(X = x)
K
2K
3K
2K
K
Its variance is
| X | 1 | 2 | 3 | 4 | 5 |
| P(X = x) | K | 2K | 3K | 2K | K |
EAMCETEAMCET 2014
Solution:
Given distribution is
X
1
2
3
4
5
P(X = x)
K
2K
3K
2K
K
$\therefore $ Variance $=\Sigma x_{i}^{2} \,p-\left(\sum x_{i}\, p\right)^{2}$
$=(1 k+8 k+27 k+32 k+25 k)-(k+4 k+9 k+8 k+5 k)^{2}$
$=(93 k)-(27 k)^{2}=\left(93 \times \frac{1}{9}\right)-\left(27 \times \frac{1}
{9}\right)^{2}\,\,\,\left(\because \Sigma p=1, \therefore k=\frac{1}{9}\right)$
$=\frac{93}{9}-9=\frac{93-81}{9}=\frac{12}{9}=\frac{4}{3}$
| X | 1 | 2 | 3 | 4 | 5 |
| P(X = x) | K | 2K | 3K | 2K | K |