Q80 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics

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
    $160006$
  • Correct answer
    C
    $160008$
  • D
    $160002$

Solution

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 $p=\frac{1}{2}$ $\therefore \quad n^2+\frac{1}{p^2}+\frac{1}{q^2}$ $(400)^2+(2)^2+(2)^2$ $=160000+4+4=160008$

Hence, the correct answer is option: 'C'.

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