If the sum of mean and variance of a binomial distribution is 4.8 for 5 trials. Find the distribution.

Answer

To solve the problem, we need to find the parameters of a binomial distribution given that the sum of the mean and variance is 4.8, and the number of trials (n) is 5.

### Step-by-Step Solution:

1. **Understand the parameters of the binomial distribution**:
- The binomial distribution is defined by two parameters:
- \( n \): number of trials
- \( p \): probability of success
- \( q \): probability of failure, where \( q = 1 - p \)

2. **Write the formulas for mean and variance**:
- 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:
\[ \sigma^2 = n \cdot p \cdot q \]

3. **Set up the equation based on the given information**:
- We know that the sum of the mean and variance is 4.8:
\[ \mu + \sigma^2 = 4.8 \]
- Substituting the formulas for mean and variance:
\[ n \cdot p + n \cdot p \cdot q = 4.8 \]
- Given \( n = 5 \):
\[ 5p + 5pq = 4.8 \]
- This simplifies to:
\[ 5p(1 + q) = 4.8 \]
- Since \( q = 1 - p \), we can substitute \( q \):
\[ 5p(1 + (1 - p)) = 4.8 \]
- This simplifies to:
\[ 5p(2 - p) = 4.8 \]

4. **Rearranging the equation**:
- Expanding the equation:
\[ 10p - 5p^2 = 4.8 \]
- Rearranging gives:
\[ 5p^2 - 10p + 4.8 = 0 \]

5. **Solving the quadratic equation**:
- We can use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = -10 \), and \( c = 4.8 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (-10)^2 - 4 \cdot 5 \cdot 4.8 = 100 - 96 = 4 \]
- Now substituting into the quadratic formula:
\[ p = \frac{10 \pm \sqrt{4}}{2 \cdot 5} = \frac{10 \pm 2}{10} \]
- This gives two possible values for \( p \):
\[ p = \frac{12}{10} = 1.2 \quad \text{(not valid since } p \leq 1\text{)} \]
\[ p = \frac{8}{10} = 0.8 \]

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

7. **Writing the binomial distribution**:
- The binomial distribution can be expressed as:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
- Substituting \( n = 5 \), \( p = 0.8 \), and \( q = 0.2 \):
\[ P(X = r) = \binom{5}{r} (0.8)^r (0.2)^{5-r} \]

### Conclusion:
The binomial distribution is given by:
\[ P(X = r) = \binom{5}{r} (0.8)^r (0.2)^{5-r} \]