The sum of variance and mean of a binomial distribution for 10 trials is 7.5. Find the distribution.

Answer

To solve the problem step by step, we need to find the parameters of a binomial distribution given that the sum of its variance and mean is 7.5 and the number of trials \( n = 10 \).

### Step 1: Understand the properties of a binomial distribution
For a binomial distribution with \( n \) trials and probability of success \( p \):
- The mean \( \mu \) is given by:
\[ \mu = np \]
- The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = npq \]
where \( q = 1 - p \).

### Step 2: Set up the equation based on the given information
We know from the problem statement that:
\[ \text{Variance} + \text{Mean} = 7.5 \]
Substituting the expressions for mean and variance:
\[ npq + np = 7.5 \]
Given \( n = 10 \):
\[ 10pq + 10p = 7.5 \]

### Step 3: Simplify the equation
Dividing the entire equation by 10 to simplify:
\[ pq + p = 0.75 \]
Factoring out \( p \):
\[ p(q + 1) = 0.75 \]
Since \( q = 1 - p \), we can substitute \( q + 1 \):
\[ p(1 - p + 1) = 0.75 \]
This simplifies to:
\[ p(2 - p) = 0.75 \]

### Step 4: Rearranging the equation
Expanding the equation gives:
\[ 2p - p^2 = 0.75 \]
Rearranging this into standard quadratic form:
\[ p^2 - 2p + 0.75 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 0.75 \):
\[ p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 0.75}}{2 \cdot 1} \]
Calculating the discriminant:
\[ p = \frac{2 \pm \sqrt{4 - 3}}{2} \]
\[ p = \frac{2 \pm 1}{2} \]
This gives us two possible values for \( p \):
1. \( p = \frac{3}{2} \) (not valid since \( p \) cannot be greater than 1)
2. \( p = \frac{1}{2} \)

### Step 6: Find \( q \)
Since \( q = 1 - p \):
\[ q = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 7: Conclusion
The parameters of the binomial distribution are:
- \( n = 10 \)
- \( p = \frac{1}{2} \)
- \( q = \frac{1}{2} \)

Thus, the binomial distribution is:
\[ B(10, 0.5) \]