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Standard XII

Mathematics

Standard Deviation about Mean for Continuous Frequency Distributions

The mean and ...

Question

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

\frac{37}{256}

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\frac{28}{256}

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\frac{128}{256}

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\frac{129}{256}

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Solution

The correct option is B $\frac{28}{256}$
np=4
npq=2
$\Rightarrow q = \frac{1}{2}, p = \frac{1}{2}, n = 8, h e r e r = 2$

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The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

The correct option is B 28256np=4 npq=2 ⇒q=12,p=12,n=8,herer=2

The correct option is B $\frac{28}{256}$
np=4
npq=2
$\Rightarrow q = \frac{1}{2}, p = \frac{1}{2}, n = 8, h e r e r = 2$