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) → \(\displaystyle\frac{62}{49}\)
The probabilities add up to 1:
\(k + 2k + k + 3k = 7k = 1\)
\( \displaystyle k = \frac{1}{7} \)
The mean:
\( \displaystyle E(X) = \frac{0(1) + 1(2) + 2(1) + 3(3)}{7} = \frac{13}{7} \)
The mean of the squares:
\( \displaystyle E(X^2) = \frac{0(1) + 1(2) + 4(1) + 9(3)}{7} = \frac{33}{7} \)
By the variance formula \(\mathrm{Var}(X) = E(X^2) - [E(X)]^2\):
\( \displaystyle \mathrm{Var}(X) = \frac{33}{7} - \frac{169}{49} \)
\( \displaystyle = \frac{231 - 169}{49} = \frac{62}{49} \)