In a binomial distribution the sum and difference of mean and variance are 1.8 and 0.2 respectively. Find the parameters.

Answer

To solve the problem, we need to find the parameters \( n \), \( p \), and \( q \) of a binomial distribution given the sum and difference of the mean and variance.

### Step-by-Step Solution:

1. **Understand the Mean and Variance of 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**:
- We are given that the sum of the mean and variance is \( 1.8 \):
\[ np + npq = 1.8 \quad \text{(Equation 1)} \]
- We are also given that the difference between the mean and variance is \( 0.2 \):
\[ np - npq = 0.2 \quad \text{(Equation 2)} \]

3. **Divide Equation 1 by Equation 2**:
- Dividing both equations gives:
\[ \frac{np + npq}{np - npq} = \frac{1.8}{0.2} \]
- Simplifying the right side:
\[ \frac{1.8}{0.2} = 9 \]
- Thus, we have:
\[ \frac{np(1 + q)}{np(1 - q)} = 9 \]
- Cancelling \( np \) from both sides (assuming \( np \neq 0 \)):
\[ \frac{1 + q}{1 - q} = 9 \]

4. **Solve for \( q \)**:
- Cross-multiplying gives:
\[ 1 + q = 9(1 - q) \]
- Expanding the right side:
\[ 1 + q = 9 - 9q \]
- Rearranging terms:
\[ 1 + q + 9q = 9 \]
\[ 10q = 8 \]
\[ q = \frac{8}{10} = 0.8 \]

5. **Find \( p \)**:
- Since \( p + q = 1 \):
\[ p = 1 - q = 1 - 0.8 = 0.2 \]

6. **Substitute \( p \) and \( q \) back into Equation 1**:
- Using Equation 1:
\[ np + npq = 1.8 \]
- Substitute \( p \) and \( q \):
\[ n(0.2) + n(0.2)(0.8) = 1.8 \]
- Simplifying:
\[ n(0.2) + n(0.16) = 1.8 \]
\[ n(0.36) = 1.8 \]
- Solving for \( n \):
\[ n = \frac{1.8}{0.36} = 5 \]

7. **Final Parameters**:
- Thus, the parameters are:
\[ n = 5, \quad p = 0.2, \quad q = 0.8 \]

### Summary of the Solution:
- The parameters of the binomial distribution are:
- \( n = 5 \)
- \( p = 0.2 \)
- \( q = 0.8 \)