The binomial distribution whose mean is 9 and whose standard deviation is \(\frac{3}{2}\) is equal to
The binomial distribution whose mean is 9 and whose standard deviation is \(\frac{3}{2}\) is equal to
\(\left(\frac{1}{4}+\frac{3}{4}\right)^{12}\)
\(\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)
\(\left(\frac{1}{2}+\frac{3}{2}\right)^{12}\)
\(\left(\frac{3}{2}+\frac{1}{2}\right)^{12}\)
Solution
We know that mean \(=n p=9\) (given)
and standard deviation \(=\sqrt{n p q}=\frac{3}{2}\)
\(\therefore \quad q=\frac{1}{4}\) and \(p=\frac{3}{4}\) and \(n=12\)
So, required binomial distribution \((p+q)^n\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)