(B) Given,$p = 0.3, q = 1 - p = 0.7$. Mean $(\mu) = np = 0.3n$. Standard deviation $(\sigma) = \sqrt{npq} = \sqrt{n(0.3)(0.7)} = \sqrt{0.21n}$. Given

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(B) Given,$p = 0.3, q = 1 - p = 0.7$. Mean $(\mu) = np = 0.3n$. Standard deviation $(\sigma) = \sqrt{npq} = \sqrt{n(0.3)(0.7)} = \sqrt{0.21n}$. Given that $\mu = 3\sigma$. Substituting the values,we get $0.3n = 3\sqrt{0.21n}$. Dividing both sides by $3$,we get $0.1n = \sqrt{0.21n}$. Squaring both sides,we get $(0.1n)^2 = 0.21n$. $0.01n^2 = 0.21n$. $n^2 = 21n$. Since $n \neq 0$,we have $n = 21$.

Class 12 Mathematics Probability Binomial distribution

Medium

$A$ random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three times as large as the standard deviation of $X$,then $n=$

AP EAMCET 2022 All AP EAMCET

A
$9$
B
$21$
C
$27$
D
$3$

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