If the mean and variance of a binomial distribution are `15//4` and `15//16` then the number of trials is

Options

A. 5
B. 2
C. 4
D. 6

Answer

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

### Step-by-Step Solution:

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

2. **Set up the equations using the given values**:
- We are given:
\[ \mu = \frac{15}{4} \]
\[ \sigma^2 = \frac{15}{16} \]
- From the mean equation:
\[ n \cdot p = \frac{15}{4} \quad \text{(1)} \]
- From the variance equation:
\[ n \cdot p \cdot q = \frac{15}{16} \quad \text{(2)} \]

3. **Express \( q \) in terms of \( p \)**:
- Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = \frac{15}{16} \]

4. **Substitute \( n \cdot p \) from equation (1) into equation (2)**:
- From equation (1), we know \( n \cdot p = \frac{15}{4} \). Substitute this into equation (2):
\[ \frac{15}{4} \cdot (1 - p) = \frac{15}{16} \]

5. **Solve for \( 1 - p \)**:
- Multiply both sides by 16 to eliminate the fraction:
\[ 15 \cdot 16 \cdot (1 - p) = 15 \cdot 4 \]
- Simplifying gives:
\[ 16(1 - p) = 4 \]
- Dividing both sides by 16:
\[ 1 - p = \frac{4}{16} = \frac{1}{4} \]

6. **Find \( p \)**:
- Rearranging gives:
\[ p = 1 - \frac{1}{4} = \frac{3}{4} \]

7. **Substitute \( p \) back into equation (1) to find \( n \)**:
- Substitute \( p = \frac{3}{4} \) into equation (1):
\[ n \cdot \frac{3}{4} = \frac{15}{4} \]
- Multiply both sides by 4:
\[ 3n = 15 \]
- Dividing both sides by 3:
\[ n = 5 \]

### Final Answer:
The number of trials \( n \) is **5**.