Suppose x ia binomial distribution with parameters n=8 and p=12 then Px-4≤2 is

P|X-4|≤2=P-2≤X-4≤2 =P(2≤X≤6) =1-P(X=0)+P(X=1)+P(X=7)+P(X=8) =1-c0 8 128+c1 8 128+c7 8 128+c8 8 128 =1-128 c0 8+c1 8+c7 8+c8 8 =1-128 1+8+8+1 =1-1828 =119128

Suppose x ia binomial distribution with parameters n=8 and p=12 then Px-4≤2 is

P|X-4|≤2=P-2≤X-4≤2 =P(2≤X≤6) =1-P(X=0)+P(X=1)+P(X=7)+P(X=8) =1-c0 8 128+c1 8 128+c7 8 128+c8 8 128 =1-128 c0 8+c1 8+c7 8+c8 8 =1-128 1+8+8+1 =1-1828 =119128

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Suppose x ia binomial distribution with parameters n=8 and $p = \frac{1}{2} then P (|x - 4| \leq 2)$ is

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$\frac{114}{123}$

$\frac{119}{128}$

$\frac{7}{33}$

$\frac{103}{124}$

answer is B.

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Detailed Solution

$P (| X - 4 | \leq 2) = P (- 2 \leq X - 4 \leq 2) = P (2 \leq X \leq 6) = 1 - [P (X = 0) + P (X = 1) + P (X = 7) + P (X = 8)] = 1 - [{}^{8}{c_{0}} {(\frac{1}{2})}^{8} + {}^{8}{c_{1}} {(\frac{1}{2})}^{8} + {}^{8}{c_{7}} {(\frac{1}{2})}^{8} + {}^{8}{c_{8}} {(\frac{1}{2})}^{8}] = 1 - {(\frac{1}{2})}^{8} ({}^{8}{c_{0}} + {}^{8}{c_{1}} + {}^{8}{c_{7}} + {}^{8}{c_{8}}) = 1 - {(\frac{1}{2})}^{8} (1 + 8 + 8 + 1) = 1 - \frac{18}{2^{8}} = \frac{119}{128}$