The mean and standard deviation of a binomial | vedclass

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(B) For a binomial distribution, the mean is given by $\mu = np = 4$ and the variance is $\sigma^2 = npq = (\sqrt{3})^2 = 3$. Dividing the variance by

Vedclass · Accepted Answer

$1-\left(\frac{3}{4}\right)^{16}$

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(B) For a binomial distribution, the mean is given by $\mu = np = 4$ and the variance is $\sigma^2 = npq = (\sqrt{3})^2 = 3$. Dividing the variance by the mean, we get $\frac{npq}{np} = \frac{3}{4}$, which implies $q = \frac{3}{4}$. Since $p + q = 1$, we have $p = 1 - \frac{3}{4} = \frac{1}{4}$. Substituting $p = \frac{1}{4}$ into $np = 4$, we get $n(\frac{1}{4}) = 4$, so $n = 16$. We need to find $P(X \geq 1)$. Using the complement rule, $P(X \geq 1) = 1 - P(X = 0)$. For a binomial distribution, $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$. Thus, $P(X = 0) = {}^{16}C_{0} (\frac{1}{4})^{0} (\frac{3}{4})^{16} = 1 	imes 1 	imes (\frac{3}{4})^{16} = (\frac{3}{4})^{16}$. Therefore, $P(X \geq 1) = 1 - (\frac{3}{4})^{16}$.

The mean and standard deviation of a binomial variate $X$ are $4$ and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

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