Consider the following statement :
"The mean of a binomial distribution is 3 and variance is 4." Which of the following is correct regarding this statement ?

Options

A. It is always true
B. It is sometimes true
C. It is never true
D. No conclusion can be drawn

Explanation

To solve the problem, we need to analyze the given statement about the binomial distribution, specifically focusing on the mean and variance.

### Step-by-Step Solution:

1. **Understand the Mean and Variance of a Binomial Distribution:**
- The mean (μ) of a binomial distribution is given by the formula:
\[ \mu = N \cdot P \]
where \(N\) is the number of trials and \(P\) is the probability of success.
- The variance (σ²) of a binomial distribution is given by the formula:
\[ \sigma^2 = N \cdot P \cdot Q \]
where \(Q = 1 - P\) is the probability of failure.

2. **Set Up the Equations:**
- From the problem, we know:
\[ N \cdot P = 3 \quad \text{(1)} \]
\[ N \cdot P \cdot Q = 4 \quad \text{(2)} \]

3. **Express Q in Terms of P:**
- From equation (1), we can express \(Q\) as:
\[ Q = 1 - P \]
- Substitute \(Q\) into equation (2):
\[ N \cdot P \cdot (1 - P) = 4 \]

4. **Substituting from Equation (1):**
- From equation (1), we know \(N \cdot P = 3\). Substitute this into the modified equation (2):
\[ 3 \cdot (1 - P) = 4 \]

5. **Solve for P:**
- Rearranging gives:
\[ 3 - 3P = 4 \]
\[ -3P = 4 - 3 \]
\[ -3P = 1 \]
\[ P = -\frac{1}{3} \]

6. **Analyze the Result:**
- The probability \(P\) must be between 0 and 1. However, we found \(P = -\frac{1}{3}\), which is not a valid probability.

7. **Conclusion:**
- Since we derived an impossible value for \(P\), the original statement that the mean is 3 and variance is 4 cannot be true. Therefore, the correct option is that the statement is "never true."

### Final Answer:
The statement is **never true**.