A r.v. X assumes values 1, 2, 3, …, n with equal probabilities. If Var (X) = 4 E (X), then n =

Options

A. 49
B. 23
C. 25
D. 24

Answer

To solve the problem, we need to find the value of \( n \) given that the random variable \( X \) assumes values \( 1, 2, 3, \ldots, n \) with equal probabilities and that \( \text{Var}(X) = 4 \cdot E(X) \).

### Step-by-Step Solution

1. **Define the Random Variable and Its Probability**:
The random variable \( X \) takes values from \( 1 \) to \( n \). Since all values are equally likely, the probability \( P(X = k) \) for \( k = 1, 2, \ldots, n \) is:
\[ P(X = k) = \frac{1}{n} \]

2. **Calculate the Expectation \( E(X) \)**:
The expectation \( E(X) \) is calculated as:
\[ E(X) = \sum_{k=1}^{n} k \cdot P(X = k) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n} \sum_{k=1}^{n} k \]
The sum of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Thus, we have:
\[ E(X) = \frac{1}{n} \cdot \frac{n(n + 1)}{2} = \frac{n + 1}{2} \]

3. **Calculate the Variance \( \text{Var}(X) \)**:
The variance \( \text{Var}(X) \) is given by:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
First, we calculate \( E(X^2) \):
\[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X = k) = \sum_{k=1}^{n} k^2 \cdot \frac{1}{n} = \frac{1}{n} \sum_{k=1}^{n} k^2 \]
The sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, we have:
\[ E(X^2) = \frac{1}{n} \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{(n + 1)(2n + 1)}{6} \]
Now we can find the variance:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 = \frac{(n + 1)(2n + 1)}{6} - \left(\frac{n + 1}{2}\right)^2 \]
Simplifying \( (E(X))^2 \):
\[ (E(X))^2 = \left(\frac{n + 1}{2}\right)^2 = \frac{(n + 1)^2}{4} \]
Therefore:
\[ \text{Var}(X) = \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} \]

4. **Set Up the Equation**:
We know from the problem statement that:
\[ \text{Var}(X) = 4 \cdot E(X) \]
Substituting the expressions we found:
\[ \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} = 4 \cdot \frac{n + 1}{2} \]
Simplifying the right side:
\[ 4 \cdot \frac{n + 1}{2} = 2(n + 1) \]
Thus, we have:
\[ \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} = 2(n + 1) \]

5. **Multiply Through by 12 to Eliminate Fractions**:
\[ 2(n + 1)(2n + 1) - 3(n + 1)^2 = 24(n + 1) \]
Dividing through by \( n + 1 \) (assuming \( n + 1 \neq 0 \)):
\[ 2(2n + 1) - 3(n + 1) = 24 \]
Expanding and simplifying:
\[ 4n + 2 - 3n - 3 = 24 \]
\[ n - 1 = 24 \implies n = 25 \]

### Conclusion
The value of \( n \) is:
\[ \boxed{25} \]