If the standerd deviation of the binomial distribution `(q+p)^(16)` is 2, then mean is

Options

A. f16
B. 8
C. 4
D. 6

Answer

To solve the problem, we need to find the mean of a binomial distribution given that the standard deviation is 2 and the distribution is represented as \((q + p)^{16}\).

### Step-by-Step Solution:

1. **Identify the parameters of the binomial distribution:**
- The binomial distribution is represented as \( (q + p)^n \) where \( n \) is the number of trials. Here, \( n = 16 \).

2. **Recall the formula for the standard deviation of a binomial distribution:**
- The standard deviation (SD) is given by the formula:
\[ SD = \sqrt{n \cdot p \cdot (1 - p)} \]
- Here, \( p \) is the probability of success, and \( (1 - p) \) is the probability of failure.

3. **Set up the equation using the given standard deviation:**
- We know that \( SD = 2 \), so we can write:
\[ 2 = \sqrt{16 \cdot p \cdot (1 - p)} \]

4. **Square both sides to eliminate the square root:**
- Squaring both sides gives:
\[ 4 = 16 \cdot p \cdot (1 - p) \]

5. **Simplify the equation:**
- Dividing both sides by 16:
\[ \frac{4}{16} = p \cdot (1 - p) \]
- This simplifies to:
\[ \frac{1}{4} = p \cdot (1 - p) \]

6. **Rearranging the equation:**
- We can rearrange this to form a quadratic equation:
\[ p - p^2 = \frac{1}{4} \]
- Rearranging gives:
\[ p^2 - p + \frac{1}{4} = 0 \]

7. **Solve the quadratic equation using the quadratic formula:**
- The quadratic formula is given by:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1, b = -1, c = \frac{1}{4} \):
\[ p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \frac{1}{4}}}{2 \cdot 1} \]
- This simplifies to:
\[ p = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1 \pm 0}{2} = \frac{1}{2} \]

8. **Calculate the mean of the binomial distribution:**
- The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
- Substituting the values we have:
\[ \mu = 16 \cdot \frac{1}{2} = 8 \]

### Final Answer:
The mean of the binomial distribution is **8**.