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The mean and variance of a random variable X having a binomial distribution are 4 and 2 , then P\left ( x=1 \right )is 

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{32}

  • Option 3)

    \frac{1}{16}

  • Option 4)

    \frac{1}{8}

 

Answers (1)

 

Binomial Distribution(Statistical) -

Mean = np

Variance = npq

Standard \: deviation =\sqrt{npq}

-

 

 Mean np=4

npq=2

q=\frac{1}{2}

P=\frac{1}{2}

n=8

\therefore P(x=1)

\therefore 8_{C_{1}}.(\frac{1}{2})^{f} =\frac{8!}{7!}\times \frac{1}{128}

=\frac{8}{128}=\frac{1}{16}


Option 1)

\frac{1}{4}

Option is incorrect

Option 2)

\frac{1}{32}

Option is incorrect

Option 3)

\frac{1}{16}

Option is correct

Option 4)

\frac{1}{8}

Option is incorrect

Posted by

Vakul

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