If the mean and variance of a binomial distribution are 9 and 6 respectively , then n = …..

Options

A. 36
B. 54
C. 18
D. 27

Answer

To solve the problem, we need to find the value of \( n \) given that the mean and variance of a binomial distribution are 9 and 6, respectively.

### Step-by-Step Solution:

1. **Understand the Mean and Variance of a Binomial Distribution**:
- The mean (\( \mu \)) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
- The variance (\( \sigma^2 \)) of a binomial distribution is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set Up the Equations**:
- From the problem, we know:
\[ n \cdot p = 9 \quad \text{(1)} \]
\[ n \cdot p \cdot q = 6 \quad \text{(2)} \]

3. **Express \( q \) in Terms of \( p \)**:
- Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 6 \]

4. **Substitute \( n \cdot p \) from Equation (1)**:
- From equation (1), we have \( n \cdot p = 9 \). Substitute this into the modified equation (2):
\[ 9 \cdot (1 - p) = 6 \]

5. **Solve for \( p \)**:
- Rearranging gives:
\[ 9 - 9p = 6 \]
\[ 9p = 3 \]
\[ p = \frac{1}{3} \]

6. **Find \( q \)**:
- Now, substitute \( p \) back to find \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

7. **Substitute \( p \) back to find \( n \)**:
- Now substitute \( p \) into equation (1):
\[ n \cdot \frac{1}{3} = 9 \]
\[ n = 9 \cdot 3 = 27 \]

### Final Answer:
Thus, the value of \( n \) is \( 27 \).

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