If X is a random variable with probability distribution as given below.
|
$X$ |
0 |
1 |
2 |
3 |
|
$P(X)$ |
$k$ |
$2k$ |
$k$ |
$3k$ |
Then, the variance of the distribution is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{62}{49}$ **
Given probabilities:
$k,\;2k,\;k,\;3k$
Total probability:
$k + 2k + k + 3k = 7k = 1$
$k = \frac{1}{7}$
Now compute the mean:
$E(X) = 0\cdot\frac{1}{7} + 1\cdot\frac{2}{7} + 2\cdot\frac{1}{7} + 3\cdot\frac{3}{7}$
$= 0 + \frac{2}{7} + \frac{2}{7} + \frac{9}{7} = \frac{13}{7}$
Compute $E(X^2)$:
$E(X^2) = 0 + 1^2\cdot\frac{2}{7} + 2^2\cdot\frac{1}{7} + 3^2\cdot\frac{3}{7}$
$= \frac{2}{7} + \frac{4}{7} + \frac{27}{7} = \frac{33}{7}$
Variance:
$\text{Var}(X) = E(X^2) - [E(X)]^2$
$= \frac{33}{7} - \left(\frac{13}{7}\right)^2$
$= \frac{33}{7} - \frac{169}{49}$
$= \frac{231}{49} - \frac{169}{49} = \frac{62}{49}$
Variance = $\frac{62}{49}$