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

Options

A. 6
B. 8
C. 10
D. 12

Answer

To solve the problem, we need to find the mean of a binomial distribution given that the standard deviation is 2. Let's break this down step by step.

### Step 1: Understand the standard deviation of a binomial distribution
The standard deviation (SD) of a binomial distribution is given by the formula:
\[ SD = \sqrt{n \cdot p \cdot q} \]
where:
- \( n \) is the number of trials,
- \( p \) is the probability of success,
- \( q \) is the probability of failure, and \( q = 1 - p \).

### Step 2: Use the information given in the problem
From the problem, we know:
- The standard deviation is 2.
- The expression for the binomial distribution is \((p + q)^{16}\), which indicates \( n = 16 \).

### Step 3: Set up the equation for standard deviation
Since the standard deviation is given as 2, we can square it to eliminate the square root:
\[ 2^2 = n \cdot p \cdot q \]
This simplifies to:
\[ 4 = 16 \cdot p \cdot (1 - p) \]

### Step 4: Simplify the equation
Now, we simplify the equation:
\[ 4 = 16p(1 - p) \]
Dividing both sides by 16 gives:
\[ \frac{4}{16} = p(1 - p) \]
This simplifies to:
\[ \frac{1}{4} = p(1 - p) \]

### Step 5: Rearranging the equation
We can rearrange the equation:
\[ p - p^2 = \frac{1}{4} \]
Rearranging gives us a quadratic equation:
\[ p^2 - p + \frac{1}{4} = 0 \]

### Step 6: Solve the quadratic equation
To solve for \( p \), we can use the quadratic formula:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -1 \), and \( c = \frac{1}{4} \):
\[ p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \frac{1}{4}}}{2 \cdot 1} \]
This simplifies to:
\[ p = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1 \pm 0}{2} = \frac{1}{2} \]

### Step 7: Find the mean of the distribution
The mean of a binomial distribution is given by:
\[ \text{Mean} = n \cdot p \]
Substituting \( n = 16 \) and \( p = \frac{1}{2} \):
\[ \text{Mean} = 16 \cdot \frac{1}{2} = 8 \]

### Final Answer
Thus, the mean of the distribution is:
\[ \boxed{8} \]