The mean and variance of a binomial distribut | vedclass

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(B) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$,where $p + q = 1$.Given $\mu = 4$ and $\

Vedclass · Accepted Answer

${}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$

Vedclass · Answer

(B) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$,where $p + q = 1$.Given $\mu = 4$ and $\sigma^2 = 3$,we have:$np = 4$ and $npq = 3$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{3}{4}$.Since $p = 1 - q$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$.Substituting $p = \frac{1}{4}$ into $np = 4$,we get $n \left( \frac{1}{4} \right) = 4$,which implies $n = 16$.The probability of getting exactly $x$ successes is given by $P(X = x) = {}^nC_x p^x q^{n-x}$.For $x = 6$,$P(X = 6) = {}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{16-6} = {}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$.Thus,the correct option is $B$.

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Then,the probability of getting exactly six successes in this distribution is:

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