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:

For most symmetric or near-symmetric binomial distributions, the mode is either equal to the mean or lies very close to it.
Since the mean here is 8, 8 is the most likely value for the mode.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A binomial distribution is characterized by two parameters: the number of trials \(n\), and the probability of success in a single trial \(p\).
The mode of a binomial distribution is the value of \(x\) that has the highest probability.

Step 2: Key Formula or Approach:
For a binomial distribution:
- \(\text{Mean} = n p\)
- \(\text{Variance} = n p q\), where \(q = 1 - p\)
The mode of the distribution is determined by the value of the term \((n + 1) p\):
- If \((n + 1) p\) is a non-integer, the unique mode is the greatest integer value less than or equal to \((n + 1) p\), denoted as \(\lfloor (n + 1) p \rfloor\).
- If \((n + 1) p\) is an integer (say \(k\)), the distribution is bimodal with modes at \(k\) and \(k - 1\).

Step 3: Detailed Explanation:
We are given:
\[ \text{Mean} = n p = 8 \]
\[ \text{Variance} = n p q = 6 \]
We find \(q\) by dividing the variance by the mean:
\[ q = \frac{n p q}{n p} = \frac{6}{8} = 0.75 \]
Now, calculate \(p\):
\[ p = 1 - q = 1 - 0.75 = 0.25 \]
Next, find \(n\) using our mean value:
\[ n p = 8 \implies n \times 0.25 = 8 \implies n = 32 \]
Now, evaluate the term \((n + 1) p\) to determine the mode:
\[ (n + 1) p = (32 + 1) \times 0.25 \]
\[ (n + 1) p = 33 \times 0.25 = 8.25 \]
Since 8.25 is not an integer, the distribution has a single unique mode.
This mode is the greatest integer value less than or equal to 8.25:
\[ \text{Mode} = \lfloor 8.25 \rfloor = 8 \]
The mode of the binomial distribution is 8.
This matches the second option.

Step 4: Final Answer:
Therefore, the correct option is (B).