The mean and variance of a binomial distribution are 8 and 4 respectively. What is P(X=1) equal to?

Options

A. `(1)/(2^(12))`
B. `(1)/(2^(8))`
C. `(1)/(2^(6))`
D. `(1)/(2^(4))`

Answer

To solve the problem, we need to find \( P(X = 1) \) for a binomial distribution where the mean is 8 and the variance is 4.

### Step-by-Step Solution:

1. **Identify the parameters of the binomial distribution**:
The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set up the equations**:
From the problem, we know:
\[ n \cdot p = 8 \quad \text{(1)} \]
\[ n \cdot p \cdot q = 4 \quad \text{(2)} \]

3. **Express \( q \) in terms of \( p \)**:
Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 4 \]

4. **Substitute \( n \cdot p \) from equation (1)**:
From equation (1), we have \( n \cdot p = 8 \). Substituting this into the modified equation (2):
\[ 8 \cdot (1 - p) = 4 \]

5. **Solve for \( p \)**:
\[ 8 - 8p = 4 \]
\[ 8p = 4 \]
\[ p = \frac{1}{2} \]

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

7. **Substitute \( p \) back to find \( n \)**:
Using equation (1):
\[ n \cdot \frac{1}{2} = 8 \]
\[ n = 16 \]

8. **Calculate \( P(X = 1) \)**:
The probability mass function for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n - r} \]
For \( r = 1 \):
\[ P(X = 1) = \binom{16}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{16 - 1} \]
\[ = 16 \cdot \frac{1}{2} \cdot \left(\frac{1}{2}\right)^{15} \]
\[ = 16 \cdot \frac{1}{2^{16}} \]
\[ = \frac{16}{2^{16}} = \frac{1}{2^{12}} \]

9. **Final answer**:
Thus, \( P(X = 1) = \frac{1}{2^{12}} \).

### Conclusion:
The answer is \( P(X = 1) = \frac{1}{2^{12}} \).