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

Options

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

Answer

To solve the problem, we need to analyze the given information about the binomial distribution. The mean and variance of a binomial distribution can be expressed in terms of the number of trials (n) and the probability of success (p).

### Step-by-Step Solution:

1. **Understand the Mean and Variance of Binomial Distribution**:
- The mean (μ) of a binomial distribution is given by the formula:
\[ \mu = np \]
- The variance (σ²) of a binomial distribution is given by the formula:
\[ \sigma^2 = npq \]
where \( q = 1 - p \) is the probability of failure.

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

3. **Express q in terms of p**:
- Since \( q = 1 - p \), we can substitute this into equation (2):
\[ np(1 - p) = 4 \]

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

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

6. **Analyze the Result**:
- The value of p represents the probability of success in a binomial distribution, which must lie between 0 and 1 (inclusive). Since \( p = -\frac{1}{3} \) is not a valid probability, this indicates that the initial statement about the mean and variance cannot be true.

7. **Conclusion**:
- Therefore, the statement "The mean of a binomial distribution is 3 and variance is 4" is incorrect.

### Final Answer:
The correct conclusion is that the statement is never true.