In a binomial distribution, n=5, the sum of the mean and variance is 1.8 . Then the distribution is

Answer

To solve the problem, we need to find the binomial distribution parameters given that \( n = 5 \) and the sum of the mean and variance is \( 1.8 \).

### Step-by-Step Solution:

1. **Understand the Mean and Variance of a Binomial Distribution**:
- For a binomial distribution with parameters \( n \) and \( p \):
- Mean \( \mu = np \)
- Variance \( \sigma^2 = npq \) where \( q = 1 - p \)

2. **Set Up the Equation**:
- We know from the problem that the sum of the mean and variance is \( 1.8 \):
\[ np + npq = 1.8 \]
- Substitute \( q \) with \( 1 - p \):
\[ np + np(1 - p) = 1.8 \]
- This simplifies to:
\[ np + np - np^2 = 1.8 \]
\[ 2np - np^2 = 1.8 \]

3. **Substitute \( n \)**:
- Since \( n = 5 \):
\[ 2(5)p - (5)p^2 = 1.8 \]
- This simplifies to:
\[ 10p - 5p^2 = 1.8 \]

4. **Rearrange the Equation**:
- Rearranging gives us:
\[ 5p^2 - 10p + 1.8 = 0 \]

5. **Multiply through by 10 to eliminate decimals**:
- To make calculations easier, multiply the entire equation by 10:
\[ 50p^2 - 100p + 18 = 0 \]

6. **Use the Quadratic Formula**:
- The quadratic formula is given by:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 50 \), \( b = -100 \), and \( c = 18 \):
\[ p = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 50 \cdot 18}}{2 \cdot 50} \]
\[ p = \frac{100 \pm \sqrt{10000 - 3600}}{100} \]
\[ p = \frac{100 \pm \sqrt{6400}}{100} \]
\[ p = \frac{100 \pm 80}{100} \]

7. **Calculate the Values of \( p \)**:
- This gives us two possible values for \( p \):
\[ p = \frac{180}{100} = 1.8 \quad (\text{not valid since } p \leq 1) \]
\[ p = \frac{20}{100} = 0.2 \]

8. **Calculate \( q \)**:
- Since \( q = 1 - p \):
\[ q = 1 - 0.2 = 0.8 \]

9. **Write the Binomial Distribution**:
- The binomial distribution can now be expressed as:
\[ P(X = x) = \binom{5}{x} (0.2)^x (0.8)^{5-x} \]
- Where \( x = 0, 1, 2, 3, 4, 5 \).

### Final Answer:
The binomial distribution is:
\[ P(X = x) = \binom{5}{x} (0.2)^x (0.8)^{5-x} \]