IF the mean and S.D. of binomial distribution are 20 and 4 respectively. Than the number of trials , is

Options

A. 50
B. 25
C. 100
D. 80

Answer

To solve the problem, we need to find the number of trials \( n \) in a binomial distribution given the mean and standard deviation.

### Step-by-Step Solution:

1. **Understand the Mean and Standard Deviation of a Binomial Distribution**:
- The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
- The standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{n \cdot p \cdot q} \]
where \( q = 1 - p \) (the probability of failure).

2. **Given Values**:
- Mean \( \mu = 20 \)
- Standard Deviation \( \sigma = 4 \)

3. **Set Up the Equations**:
- From the mean:
\[ n \cdot p = 20 \quad \text{(1)} \]
- From the standard deviation:
\[ \sqrt{n \cdot p \cdot q} = 4 \quad \text{(2)} \]
- Squaring equation (2):
\[ n \cdot p \cdot q = 16 \quad \text{(3)} \]

4. **Express \( q \) in Terms of \( p \)**:
- Since \( q = 1 - p \), substitute \( q \) into equation (3):
\[ n \cdot p \cdot (1 - p) = 16 \quad \text{(4)} \]

5. **Substitute \( n \) from Equation (1) into Equation (4)**:
- From equation (1), we can express \( n \) as:
\[ n = \frac{20}{p} \]
- Substitute \( n \) into equation (4):
\[ \frac{20}{p} \cdot p \cdot (1 - p) = 16 \]
- Simplifying this gives:
\[ 20(1 - p) = 16 \]
- Expanding and rearranging:
\[ 20 - 20p = 16 \]
\[ 20p = 4 \]
\[ p = \frac{1}{5} \]

6. **Find \( n \)**:
- Substitute \( p \) back into equation (1) to find \( n \):
\[ n \cdot \frac{1}{5} = 20 \]
\[ n = 20 \cdot 5 = 100 \]

### Final Answer:
The number of trials \( n \) is \( 100 \).

---