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

For a binomial distribution, the mean \( \mu \) is \( \mu = np \) and the variance is \( \sigma^2 = npq \).Given the mean is 4 and the variance is 3, we can determine \(n\) and \(p\):\[\mu = np = 4, \quad \sigma^2 = npq = 3\]Solving these equations yields \(p = \frac{1}{4}\) and \(n = 16\).The mode \( M \) of a binomial distribution is calculated as:\[M = \left( n + 1 \right) p \quad {if} \quad (n + 1)p { is an integer.}\]Substituting the values \(n = 16\) and \(p = \frac{1}{4}\):\[M = (16 + 1)\left( \frac{1}{4} \right) = 4\]Therefore, the mode is 4.