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Question

Let the random variable X have a binomial distribution with mean 8 and variance 4. If P(X2)=k216, then k is equal to:

A
1
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B
17
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C
121
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D
137
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Solution

The correct option is D 137
Given that binomial distribution with mean np=8 and variance npq=4
therefore, q=12n=16, p=12,
Hence P(X2)=P(X=0)+P(X=1)+P(X=2)
=16C0(12)16+16C1(12)1(12)15+16C2(12)2(12)14
=1+16+120216=137216=k216

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