#### Question

If *X* follows a binomial distribution with parameters *n* = 8 and *p* = 1/2, then *P* (|*X* − 4| ≤ 2) equals

##### Options

\[\frac{118}{128}\]

\[\frac{119}{128}\]

\[\frac{117}{128}\]

None Of these

#### Solution

\[\frac{119}{128}\]

\[n = 8, p = \frac{1}{2}\]

\[ \therefore q = 1 - \frac{1}{2} = \frac{1}{2}\]

\[\text{ Hence, the distribution is given by } \]

\[P(X = r) =^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} \]

\[P\left( \left| X - 4 \right| \right) \leq 2 \]

\[ = P( - 2 \leq X - 4 \leq 2) \]

\[ = P(2 \leq X \leq 6)\]

\[ = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)\]

\[ = \frac{^{8}{}{C}_2 + ^{8}{}{C}_3 + ^{8}{}{C}_4 + ^{8}{}{C}_5 + ^{8}{}{C}_6}{2^8}\]

\[ = \frac{28 + 56 + 70 + 56 + 28}{256}\]

\[ = \frac{238}{256}\]

\[ = \frac{119}{128}\]

Is there an error in this question or solution?

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If X Follows a Binomial Distribution with Parameters N = 8 and P = 1/2, Then P (|X − 4| ≤ 2) Equals(A)118 128(B) 119 128c) 117 128(D) None of These Concept: Bernoulli Trials and Binomial Distribution.

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