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.