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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

  1. A \(160004\)
  2. B \(160006\)
  3. C \(160008\)
  4. D \(160002\)
Verified Solution

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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