If X is a binomial variate with n=7 and P(X=3 | vedclass

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(D) For a binomial distribution $X \sim B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.Given $

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$\frac{21}{2^7}$

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(D) For a binomial distribution $X \sim B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.Given $n=7$ and $P(X=3) = P(X=4)$:$\binom{7}{3} p^3 q^4 = \binom{7}{4} p^4 q^3$Since $\binom{7}{3} = \binom{7}{4} = 35$,we have $35 p^3 q^4 = 35 p^4 q^3$.Dividing both sides by $35 p^3 q^3$ (assuming $p, q 
eq 0$),we get $q = p$.Since $p+q=1$,we have $p = q = \frac{1}{2}$.Now,we calculate $P(X=5)$:$P(X=5) = \binom{7}{5} (\frac{1}{2})^5 (\frac{1}{2})^{7-5} = \binom{7}{5} (\frac{1}{2})^7$.$\binom{7}{5} = \binom{7}{2} = \frac{7 	imes 6}{2} = 21$.Thus,$P(X=5) = 21 	imes \frac{1}{2^7} = \frac{21}{2^7}$.

If $X$ is a binomial variate with $n=7$ and $P(X=3)=P(X=4)$,then $P(X=5)$ is equal to:

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