In a binomial distribution, the mean is 4 and the variance is 3. What is the mode?

Options

A. 6
B. 5
C. 4
D. 3

Explanation

To find the mode of a binomial distribution given the mean and variance, we can follow these steps:

### Step 1: Understand the given information
We know that:
- Mean (μ) = 4
- Variance (σ²) = 3

In a binomial distribution, the mean (μ) and variance (σ²) are given by the formulas:
- Mean: \( \mu = n \cdot p \)
- Variance: \( \sigma^2 = n \cdot p \cdot q \) where \( q = 1 - p \)

### Step 2: Set up the equations
From the mean, we have:
\[ n \cdot p = 4 \tag{1} \]

From the variance, we have:
\[ n \cdot p \cdot q = 3 \tag{2} \]

### Step 3: Express \( q \) in terms of \( p \)
Since \( q = 1 - p \), we can substitute this into equation (2):
\[ n \cdot p \cdot (1 - p) = 3 \]

### Step 4: Substitute \( n \) from equation (1) into equation (2)
From equation (1), we can express \( n \) as:
\[ n = \frac{4}{p} \]

Now substitute this into equation (2):
\[ \frac{4}{p} \cdot p \cdot (1 - p) = 3 \]

This simplifies to:
\[ 4(1 - p) = 3 \]

### Step 5: Solve for \( p \)
Expanding the equation gives:
\[ 4 - 4p = 3 \]
\[ 4p = 1 \]
\[ p = \frac{1}{4} \]

### Step 6: Find \( q \)
Now that we have \( p \), we can find \( q \):
\[ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 7: Substitute \( p \) back to find \( n \)
Using equation (1):
\[ n \cdot \frac{1}{4} = 4 \]
\[ n = 4 \cdot 4 = 16 \]

### Step 8: Calculate the mode
The mode \( r \) of a binomial distribution can be found using the formula:
\[ r = \lfloor (n + 1)p \rfloor \text{ or } r = \lfloor (n + 1)p \rfloor - 1 \]

Calculating \( (n + 1)p \):
\[ (n + 1)p = 17 \cdot \frac{1}{4} = \frac{17}{4} = 4.25 \]

Thus, the mode can be:
\[ r = \lfloor 4.25 \rfloor = 4 \]
or
\[ r = \lfloor 4.25 \rfloor - 1 = 3 \]

Since the mode is the largest integer \( r \) such that the probability is maximized, we take:
\[ r = 4 \]

### Final Answer
Thus, the mode of the binomial distribution is **4**.