Which of the following pairs can be the mean and the variance, respectively, of a binomial distribution?
(A) Mean 4, variance 3
(B) Mean 3, variance 4
(C) Mean 5, variance 5
(D) Mean 6, variance 2
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
For a binomial distribution, mean = np and variance = npq, so
\( \displaystyle q = \frac{\text{variance}}{\text{mean}} \)
Since \(0 \lt p \lt 1\), q must also lie strictly between 0 and 1, and n = mean / p must be a positive whole number.
So a pair is possible only when variance ÷ mean is less than 1 and n comes out a whole number.
Checking the statements
(A) Mean 4, variance 3 — Correct. \(q = \frac{3}{4}\), \(p = \frac{1}{4}\), \(n = 4 \div \frac{1}{4} = 16\).
(B) Mean 3, variance 4 — Incorrect. \(q = \frac{4}{3} \gt 1\), which is impossible.
(C) Mean 5, variance 5 — Incorrect. \(q = 1\) gives p = 0, so there could be no success at all.
(D) Mean 6, variance 2 — Correct. \(q = \frac{1}{3}\), \(p = \frac{2}{3}\), \(n = 6 \div \frac{2}{3} = 9\).