The binomial distribution whose mean is 9 and | vedclass

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Question

(B) We know that the mean of a binomial distribution is given by $\mu = np = 9$. The standard deviation is given by $\sigma = \sqrt{npq} = \frac{3}{2}

Vedclass · Accepted Answer

$\left(\frac{3}{4}+\frac{1}{4}\right)^{12}$

Vedclass · Answer

(B) We know that the mean of a binomial distribution is given by $\mu = np = 9$. The standard deviation is given by $\sigma = \sqrt{npq} = \frac{3}{2}$. Squaring the standard deviation,we get $npq = \left(\frac{3}{2}ight)^2 = \frac{9}{4}$. Substituting $np = 9$ into the equation $npq = \frac{9}{4}$,we get $9q = \frac{9}{4}$,which implies $q = \frac{1}{4}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{4} = \frac{3}{4}$. Now,$np = 9 \implies n 	imes \frac{3}{4} = 9 \implies n = 9 	imes \frac{4}{3} = 12$. The binomial distribution is given by $(p + q)^n = \left(\frac{3}{4} + \frac{1}{4}ight)^{12}$.

The binomial distribution whose mean is $9$ and whose standard deviation is $\frac{3}{2}$ is equal to

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