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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

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