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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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In a binomial distribution the mean is 15 and variance is 10, then parameter $n$ is

$\begin {array} {1 1} (A)\;28 & \quad (B)\;16 \\ (C)\;45 & \quad (D)\;25 \end {array}$

 

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1 Answer

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Given mean = $ np = 15$
Variance $ np(1-P)=10$
$ 1-P = \large\frac{10}{15}$
$ = \large\frac{2}{3}$
$ P = 1- \large\frac{2}{3}$
$ = \large\frac{1}{3}$
$ n = 15 \times 3 $
$ = 45$
Ans : (C)
answered Jan 16, 2014 by thanvigandhi_1
 

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