Question

If the mean and variance of a binomial variate are 12 & 4, then the distribution is _____

• A. \( \left(\frac{1}{3}+\frac{2}{3}\right)^{15} \)
• B. \( \left(\frac{1}{3}+\frac{2}{3}\right)^{16} \)
• C. \( \left(\frac{1}{3}+\frac{2}{3}\right)^{17} \)
• D. \( \left(\frac{1}{3}+\frac{2}{3}\right)^{18} \)

Hint:

Binomial Distribution Properties. Mean = np, Variance = npq. Use these to find parameters: \(q = Variance / Mean\), \(p = 1 - q\), \(n = Mean / p\). The distribution can be written as \((q+p)^n\).

Answer 1

The Correct Option is D

For a binomial distribution B(n, p): - Mean (\(\mu\)) = np - Variance (\(\sigma^2\)) = npq, where q = 1-p.
We are given: Mean = 12 \(\implies np = 12\) Variance = 4 \(\implies npq = 4\) Now we can find q and p: $$ \frac{\text{Variance}}{\text{Mean}} = \frac{npq}{np} = q $$ $$ q = \frac{4}{12} = \frac{1}{3} $$ Since \(p + q = 1\): $$ p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3} $$ Now find n using the mean equation: $$ np = 12 $$ $$ n \left(\frac{2}{3}\right) = 12 $$ $$ n = 12 \times \frac{3}{2} = 6 \times 3 = 18 $$ So, the parameters of the binomial distribution are \(n=18\) and \(p=2/3\).
The distribution is often represented symbolically by the binomial expansion \((q+p)^n\).
In this case, \(q=1/3\) and \(p=2/3\), so the distribution is: $$ \left(\frac{1}{3}+\frac{2}{3}\right)^{18} $$ This matches option (4).