The random variable takes the values 1,2,3,…….m. If `P(X=n)=(1)/(m)` to each n, then the variance of X is

Options

A. `((m+1)(2m+1))/(6)`
B. `(m^(2)-1)/(12)`
C. `(m+1)/(2)`
D. `(m^(2)+1)/(12)`

Answer

To solve the problem, we need to find the variance of a discrete random variable \( X \) that takes values from 1 to \( m \) with equal probability \( P(X = n) = \frac{1}{m} \) for each \( n \).

### Step 1: Define the random variable and its probability
The random variable \( X \) takes values \( 1, 2, 3, \ldots, m \) and the probability for each value is given as:
\[ P(X = n) = \frac{1}{m} \quad \text{for } n = 1, 2, \ldots, m \]

### Step 2: Calculate the expected value \( E(X) \)
The expected value \( E(X) \) is calculated as:
\[ E(X) = \sum_{n=1}^{m} n \cdot P(X = n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} \]
This simplifies to:
\[ E(X) = \frac{1}{m} \sum_{n=1}^{m} n = \frac{1}{m} \cdot \frac{m(m + 1)}{2} = \frac{m + 1}{2} \]

### Step 3: Calculate \( E(X^2) \)
Next, we calculate \( E(X^2) \):
\[ E(X^2) = \sum_{n=1}^{m} n^2 \cdot P(X = n) = \sum_{n=1}^{m} n^2 \cdot \frac{1}{m} \]
This simplifies to:
\[ E(X^2) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m + 1)(2m + 1)}{6} = \frac{(m + 1)(2m + 1)}{6} \]

### Step 4: Calculate the variance \( Var(X) \)
The variance \( Var(X) \) is given by the formula:
\[ Var(X) = E(X^2) - (E(X))^2 \]
Substituting the values we calculated:
\[ Var(X) = \frac{(m + 1)(2m + 1)}{6} - \left(\frac{m + 1}{2}\right)^2 \]
Calculating \( \left(\frac{m + 1}{2}\right)^2 \):
\[ \left(\frac{m + 1}{2}\right)^2 = \frac{(m + 1)^2}{4} \]
Now substituting this back into the variance formula:
\[ Var(X) = \frac{(m + 1)(2m + 1)}{6} - \frac{(m + 1)^2}{4} \]

### Step 5: Simplify the variance expression
To combine the two fractions, we need a common denominator, which is 12:
\[ Var(X) = \frac{2(m + 1)(2m + 1)}{12} - \frac{3(m + 1)^2}{12} \]
This simplifies to:
\[ Var(X) = \frac{(m + 1)(2m + 1 - 3(m + 1))}{12} \]
Calculating \( 2m + 1 - 3(m + 1) \):
\[ 2m + 1 - 3m - 3 = -m - 2 \]
Thus:
\[ Var(X) = \frac{(m + 1)(-m - 2)}{12} = \frac{-(m^2 + 1)}{12} \]
This can be rewritten as:
\[ Var(X) = \frac{m^2 - 1}{12} \]

### Final Answer
The variance of \( X \) is:
\[ \boxed{\frac{m^2 - 1}{12}} \]