AP EAMCET · Maths · Probability
For a binomial distribution \(B(n, p)\) mean \(=200\), standard deviation \(=10\), then \(n^2+\frac{1}{p^2}+\frac{1}{q^2}\) is equal to
- A \(160004\)
- B \(160006\)
- C \(160008\)
- D \(160002\)
Answer & Solution
Correct Answer
(C) \(160008\)
Step-by-step Solution
Detailed explanation
In binomial distribution mean \(=n p\) and standard deviation \(=\sqrt{n p q}\) \(\therefore \quad 200=n p\) and \(10=\sqrt{n p q}\) \(200=n p\) ...(i) \(100=n p q\) ...(ii) From Eqs. (i) and (ii), we get \(q=\frac{1}{2}, n=400\) and \(p=\frac{1}{2}\)…
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