Find the binomial distribution whose :
(i) mean is 4 and variance is 3
(ii) mean is 9 and variance is 6.

Answer

To find the binomial distribution for the given conditions, we will use the formulas for mean and variance of a binomial distribution. The mean (μ) and variance (σ²) of a binomial distribution are given by:

- Mean: \( \mu = np \)
- Variance: \( \sigma^2 = npq \)

Where:
- \( n \) = number of trials
- \( p \) = probability of success
- \( q \) = probability of failure (where \( q = 1 - p \))

### (i) Mean is 4 and Variance is 3

1. **Set up the equations:**
- From the mean: \( np = 4 \) (1)
- From the variance: \( npq = 3 \) (2)

2. **Substituting \( q \):**
- Since \( q = 1 - p \), we can rewrite equation (2):
\[ np(1 - p) = 3 \]

3. **Substituting \( np \) from equation (1) into equation (2):**
- Substitute \( np = 4 \) into \( np(1 - p) = 3 \):
\[ 4(1 - p) = 3 \]

4. **Solving for \( p \):**
\[ 4 - 4p = 3 \\ 4p = 1 \\ p = \frac{1}{4} \]

5. **Finding \( q \):**
\[ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \]

6. **Finding \( n \):**
- Substitute \( p \) back into equation (1):
\[ n \cdot \frac{1}{4} = 4 \\ n = 4 \cdot 4 = 16 \]

7. **Conclusion for (i):**
- The binomial distribution is:
\[ B(16, \frac{1}{4}) \]

### (ii) Mean is 9 and Variance is 6

1. **Set up the equations:**
- From the mean: \( np = 9 \) (1)
- From the variance: \( npq = 6 \) (2)

2. **Substituting \( q \):**
- Rewrite equation (2):
\[ np(1 - p) = 6 \]

3. **Substituting \( np \) from equation (1) into equation (2):**
- Substitute \( np = 9 \):
\[ 9(1 - p) = 6 \]

4. **Solving for \( p \):**
\[ 9 - 9p = 6 \\ 9p = 3 \\ p = \frac{1}{3} \]

5. **Finding \( q \):**
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

6. **Finding \( n \):**
- Substitute \( p \) back into equation (1):
\[ n \cdot \frac{1}{3} = 9 \\ n = 9 \cdot 3 = 27 \]

7. **Conclusion for (ii):**
- The binomial distribution is:
\[ B(27, \frac{1}{3}) \]

### Summary of Results:
- For (i): \( B(16, \frac{1}{4}) \)
- For (ii): \( B(27, \frac{1}{3}) \)