Question

In a binomial distribution, the mean is 4 and variance is 3. Then, its mode is:

• A. 5
• B. 6
• C. 4
• D. None of these

Hint:

In binomial distributions, the mode is often the closest integer to \( (n+1)p \).

Answer 1

The Correct Option is C

The mean \( \mu \) of a binomial distribution is given by \( \mu = np \), and the variance is given by \( \sigma^2 = npq \). We are given that the mean is 4 and the variance is 3. Using these, we can solve for \(n\) and \(p\): \[ \mu = np = 4, \quad \sigma^2 = npq = 3 \] From this, we find \(p = \frac{1}{4}\), and \(n = 16\). Now, the mode \( M \) of a binomial distribution is given by: \[ M = \left( n + 1 \right) p \quad {if} \quad (n + 1)p { is an integer.} \] Substituting \(n = 16\) and \(p = \frac{1}{4}\): \[ M = (16 + 1)\left( \frac{1}{4} \right) = 4 \] Thus, the mode is 4.

Answer 2

Step 1: Understanding the Problem
In a binomial distribution, the mean and variance are related to the parameters \(n\) (number of trials) and \(p\) (probability of success) as follows:
\[ \text{Mean} = \mu = np \]
\[ \text{Variance} = \sigma^2 = np(1 - p) \]
We are given that the mean is 4 and the variance is 3. We need to find the mode of the distribution.

Step 2: Use the Given Mean and Variance
From the given data, we know:
\[ \mu = np = 4 \quad \text{and} \quad \sigma^2 = np(1 - p) = 3 \]
Using these two equations, we can solve for \(n\) and \(p\).

Step 3: Solve for \(n\) and \(p\)
From the first equation, we have:
\[ np = 4 \quad \Rightarrow \quad p = \frac{4}{n} \]
Substitute \(p = \frac{4}{n}\) into the variance equation:
\[ np(1 - p) = 3 \]
\[ n \times \frac{4}{n} \times \left(1 - \frac{4}{n}\right) = 3 \]
Simplifying:
\[ 4 \times \left(1 - \frac{4}{n}\right) = 3 \]
\[ 4 - \frac{16}{n} = 3 \]
\[ \frac{16}{n} = 1 \quad \Rightarrow \quad n = 16 \]
Now, substitute \(n = 16\) back into the equation \(np = 4\):
\[ 16p = 4 \quad \Rightarrow \quad p = \frac{4}{16} = \frac{1}{4} \]

Step 4: Finding the Mode
The mode of a binomial distribution is given by the integer part of \( (n + 1)p \), or \( \lfloor (n + 1)p \rfloor \). Substituting \(n = 16\) and \(p = \frac{1}{4}\):
\[ \text{Mode} = \lfloor (16 + 1) \times \frac{1}{4} \rfloor = \lfloor 17 \times \frac{1}{4} \rfloor = \lfloor 4.25 \rfloor = 4 \]

Step 5: Conclusion
Therefore, the mode of the binomial distribution is 4.