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 $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequality $|X-4| \

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$\frac{119}{128}$

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(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequality $|X-4| \leq 2$ implies $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.Thus,we need to calculate $P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.The probability mass function is $P(X=k) = {}^{8}C_{k} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{8-k} = {}^{8}C_{k} \left(\frac{1}{2}\right)^{8}$.Summing these probabilities:$P(2 \leq X \leq 6) = \left(\frac{1}{2}\right)^{8} \left[ {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} + {}^{8}C_{5} + {}^{8}C_{6} \right]$.Calculating the combinations:${}^{8}C_{2} = 28, {}^{8}C_{3} = 56, {}^{8}C_{4} = 70, {}^{8}C_{5} = 56, {}^{8}C_{6} = 28$.Sum $= 28 + 56 + 70 + 56 + 28 = 238$.Therefore,$P(2 \leq X \leq 6) = \frac{238}{256} = \frac{119}{128}$.

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

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