AP EAMCET · Maths · Binomial Theorem
The binomial distribution whose mean is 9 and whose standard deviation is \(\frac{3}{2}\) is equal to
- A \(\left(\frac{1}{4}+\frac{3}{4}\right)^{12}\)
- B \(\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)
- C \(\left(\frac{1}{2}+\frac{3}{2}\right)^{12}\)
- D \(\left(\frac{3}{2}+\frac{1}{2}\right)^{12}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)
Step-by-step Solution
Detailed explanation
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 \((p+q)^n\) \(=\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)
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