If X sim Bleft(8, frac{1}{2}right),then P(|X- | vedclass

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(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8$,$p=\frac{1}{2}$,and $q=1-p=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequalit

Vedclass · Accepted Answer

$\frac{119}{128}$

Vedclass · Answer

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8$,$p=\frac{1}{2}$,and $q=1-p=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequality is equivalent to $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.Thus,$P(2 \leq X \leq 6) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.Alternatively,$P(2 \leq X \leq 6) = 1 - [P(X=0) + P(X=1) + P(X=7) + P(X=8)]$.Using the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} \left(\frac{1}{2}\right)^8$:$P(X=0) = \binom{8}{0} \left(\frac{1}{2}\right)^8 = 1 \cdot \frac{1}{256} = \frac{1}{256}$.$P(X=1) = \binom{8}{1} \left(\frac{1}{2}\right)^8 = 8 \cdot \frac{1}{256} = \frac{8}{256}$.$P(X=7) = \binom{8}{7} \left(\frac{1}{2}\right)^8 = 8 \cdot \frac{1}{256} = \frac{8}{256}$.$P(X=8) = \binom{8}{8} \left(\frac{1}{2}\right)^8 = 1 \cdot \frac{1}{256} = \frac{1}{256}$.Sum $= \frac{1+8+8+1}{256} = \frac{18}{256} = \frac{9}{128}$.Therefore,$P(2 \leq X \leq 6) = 1 - \frac{9}{128} = \frac{119}{128}$.

If $X \sim B\left(8, \frac{1}{2}\right)$,then $P(|X-4| \leq 2)=$

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