Let X sim B(6, 1/2),then P[|X-4| leq 2] is | vedclass

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(C) Given $X \sim B(n, p)$ where $n=6$ and $p=1/2$. Thus $q = 1-p = 1/2$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom

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$\frac{57}{64}$

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(C) Given $X \sim B(n, p)$ where $n=6$ and $p=1/2$. Thus $q = 1-p = 1/2$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{6}{k} (1/2)^6 = \frac{\binom{6}{k}}{64}$. We need to find $P(|X-4| \leq 2)$. The inequality $|X-4| \leq 2$ is equivalent to $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$. Therefore,$P(2 \leq X \leq 6) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$. Alternatively,$P(2 \leq X \leq 6) = 1 - [P(X=0) + P(X=1)]$. Calculating $P(X=0) = \binom{6}{0} (1/2)^6 = 1/64$. Calculating $P(X=1) = \binom{6}{1} (1/2)^6 = 6/64$. Thus,$P(2 \leq X \leq 6) = 1 - (1/64 + 6/64) = 1 - 7/64 = 57/64$.

Let $X \sim B(6, 1/2)$,then $P[|X-4| \leq 2]$ is

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