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

The probability distribution has:

• $X = 0$ with probability $k$
• $X = 1$ with probability $2k$
• $X = 2$ with probability $k$
• $X = 3$ with probability $3k$

Since the sum of all probabilities equals 1:

$k + 2k + k + 3k = 1$

$7k = 1$

$k = \dfrac{1}{7}$

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The expected value is calculated as:

$E(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot k + 3 \cdot 3k$

$E(X) = 0 + 2k + 2k + 9k$

$E(X) = 13k$

$E(X) = 13 \cdot \dfrac{1}{7}$

$E(X) = \dfrac{13}{7}$

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The expected value of $X^2$ is:

$E(X^2) = 0^2 \cdot k + 1^2 \cdot 2k + 2^2 \cdot k + 3^2 \cdot 3k$

$E(X^2) = 0 + 2k + 4k + 27k$

$E(X^2) = 33k$

$E(X^2) = 33 \cdot \dfrac{1}{7}$

$E(X^2) = \dfrac{33}{7}$

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The variance is:

$\text{Var}(X) = E(X^2) - [E(X)]^2$

$\text{Var}(X) = \dfrac{33}{7} - \left(\dfrac{13}{7}\right)^2$

$\text{Var}(X) = \dfrac{33}{7} - \dfrac{169}{49}$

$\text{Var}(X) = \dfrac{231}{49} - \dfrac{169}{49}$

$\text{Var}(X) = \dfrac{62}{49}$

Therefore, the variance of the distribution is $\dfrac{62}{49}$.