If the mean of a binomial distribution with 9 trials 6, then its variance is

Options

A. 2
B. 3
C. 4
D. `sqrt2`

Answer

To find the variance of a binomial distribution given the number of trials and the mean, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters of the binomial distribution**:
- The number of trials \( n = 9 \).
- The mean \( \mu = np = 6 \).

2. **Set up the equation for the mean**:
- From the mean formula, we have:
\[ np = 6 \]

3. **Substitute the value of \( n \)**:
- Substitute \( n = 9 \) into the equation:
\[ 9p = 6 \]

4. **Solve for \( p \)**:
- Rearranging gives:
\[ p = \frac{6}{9} = \frac{2}{3} \]

5. **Find \( q \)**:
- Since \( q = 1 - p \):
\[ q = 1 - \frac{2}{3} = \frac{1}{3} \]

6. **Calculate the variance**:
- The variance \( \sigma^2 \) of a binomial distribution is given by:
\[ \sigma^2 = npq \]
- Substitute the values of \( n \), \( p \), and \( q \):
\[ \sigma^2 = 9 \times \frac{2}{3} \times \frac{1}{3} \]

7. **Perform the calculations**:
- First, calculate \( 9 \times \frac{2}{3} = 6 \).
- Then, calculate \( 6 \times \frac{1}{3} = 2 \).
- Thus, the variance \( \sigma^2 = 2 \).

### Final Answer:
The variance of the binomial distribution is \( 2 \).

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