A random variable X has the probability distr | vedclass

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(B) For a probability distribution, the sum of all probabilities must be $1$:$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$$	herefore K = \frac{1}{9

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$\frac{4}{3}$

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(B) For a probability distribution, the sum of all probabilities must be $1$:$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$$	herefore K = \frac{1}{9}$The mean $E(X) = \Sigma x_i P(x_i) = (1 	imes K) + (2 	imes 2K) + (3 	imes 3K) + (4 	imes 2K) + (5 	imes K)$$= K + 4K + 9K + 8K + 5K = 27K = 27 	imes \frac{1}{9} = 3$$E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 	imes K) + (2^2 	imes 2K) + (3^2 	imes 3K) + (4^2 	imes 2K) + (5^2 	imes K)$$= K + 8K + 27K + 32K + 25K = 93K = 93 	imes \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$Variance $Var(X) = E(X^2) - [E(X)]^2$$= \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31 - 27}{3} = \frac{4}{3}$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

Similar Questions

$A$ random variable $X$ has the following probability distribution:
$X = x$ $0$ $1$ $2$
$P(X = x)$ $4k - 10k^2$ $5k - 1$ $3k^3$

Then $P(X < 2)$ is:

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$ $0$ $2$ $4$ $6$ $8$

$P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$

The following is the probability distribution of $X$:

$X$ $0$ $1$ $2$ $3$

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

Let the mean and standard deviation of the probability distribution given by the table below be $\mu$ and $\sigma$ respectively. If $\sigma - \mu = 2$,then find the value of $\sigma$.
$X=x$ $-3$ $0$ $1$ $\alpha$
$P(X=x)$ $\frac{1}{4}$ $K$ $\frac{1}{4}$ $\frac{1}{3}$

$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ with probability $P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x$,where $k$ is a constant. Then $P(X=0)$ is

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$A$ random variable $X$ has the probability distribution given below. Its variance is:$X$$1$$2$$3$$4$$5$$P(X=x)$$K$$2K$$3K$$2K$$K$