Question

In a binomial distribution, the mean is 8 and variance is 6. Then the mode is

• A. 5
• B. 8
• C. 4
• D. 7

Hint:

Mode of binomial distribution: If \((n+1)p\) is not an integer, mode = floor((n+1)p).
If it is an integer, there are two modes: \((n+1)p\) and \((n+1)p - 1\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a binomial distribution, mean \(\mu = np\) and variance \(\sigma^2 = npq\).
The mode is the most probable value.

Step 2: Key Formula or Approach:
For binomial distribution, mode is the integer \(m\) such that \((n+1)p - 1 \le m \le (n+1)p\).

Step 3: Detailed Explanation:
Given: \(\mu = np = 8\), \(\sigma^2 = npq = 6\).
From these, \(q = \frac{\sigma^2}{\mu} = \frac{6}{8} = 0.75\).
So \(p = 1 - q = 0.25\).
Then \(n = \frac{\mu}{p} = \frac{8}{0.25} = 32\).
Now, \((n+1)p = 33 \times 0.25 = 8.25\).
The mode is the integer part of 8.25, which is 8.
But if \((n+1)p\) is an integer, then both \(m\) and \(m-1\) are modes.
Here, 8.25 is not an integer, so mode = 8.
Thus, the mode is 8, which is option (B).