If the p.m.f. of a random variable X is given | vedclass

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(B) To find the standard deviation,we first calculate the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - [E(X)]^2$. The probability distribut

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$\sqrt{2pq}$

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(B) To find the standard deviation,we first calculate the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - [E(X)]^2$. The probability distribution is:   $x_i$ $p_i$ $p_i x_i$ $p_i x_i^2$   $0$ $q^2$ $0$ $0$   $1$ $2pq$ $2pq$ $2pq$   $2$ $p^2$ $2p^2$ $4p^2$   Total $1$ $2pq + 2p^2$ $2pq + 4p^2$   Mean $\mu = E(X) = \sum p_i x_i = 2pq + 2p^2 = 2p(q+p)$. Since $p+q=1$,we have $\mu = 2p(1) = 2p$. Variance $\sigma^2 = E(X^2) - \mu^2 = (2pq + 4p^2) - (2p)^2 = 2pq + 4p^2 - 4p^2 = 2pq$. Standard deviation $\sigma = \sqrt{	ext{Variance}} = \sqrt{2pq}$.

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X=x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

If the p.m.f. of a random variable $X$ is given by the following table,then the standard deviation of $X$ is (given $p+q=1$):

$x$ $0$ $1$ $2$

$P(X=x)$ $q^2$ $2pq$ $p^2$

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If the p.m.f. of a random variable $X$ is given by the following table,then the standard deviation of $X$ is (given $p+q=1$): $x$ $0$ $1$ $2$ $P(X=x)$ $q^2$ $2pq$ $p^2$