In a binomial distribution,the mean is 15 and | vedclass

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(C) For a binomial distribution,the mean is given by $\mu = np = 15$. The variance is given by $\sigma^2 = np(1-p) = 10$. Substituting the value of $n

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) For a binomial distribution,the mean is given by $\mu = np = 15$. The variance is given by $\sigma^2 = np(1-p) = 10$. Substituting the value of $np$ into the variance equation: $15(1-p) = 10$. $1-p = \frac{10}{15} = \frac{2}{3}$. Therefore,$p = 1 - \frac{2}{3} = \frac{1}{3}$. Now,substitute $p$ back into the mean equation: $n 	imes \frac{1}{3} = 15$. $n = 15 	imes 3 = 45$. Thus,the parameter $n$ is $45$.

In a binomial distribution,the mean is $15$ and the variance is $10$. Then the parameter $n$ is:

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