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Question

(D) The sum of all the probabilities in a probability distribution is always unity.$	herefore k + 3k + 3k + k = 1$$\Rightarrow 8k = 1$$\Rightarrow k

Vedclass · Accepted Answer

$\frac{1}{8}, \frac{3}{4}$

Vedclass · Answer

(D) The sum of all the probabilities in a probability distribution is always unity.$	herefore k + 3k + 3k + k = 1$$\Rightarrow 8k = 1$$\Rightarrow k = \frac{1}{8}$Now,calculate the mean $E(X)$:$E(X) = \sum x_i \cdot P(x_i)$$E(X) = 0\left(\frac{1}{8}ight) + 1\left(\frac{3}{8}ight) + 2\left(\frac{3}{8}ight) + 3\left(\frac{1}{8}ight)$$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$Next,calculate $E(X^2)$:$E(X^2) = \sum x_i^2 \cdot P(x_i)$$E(X^2) = 0^2\left(\frac{1}{8}ight) + 1^2\left(\frac{3}{8}ight) + 2^2\left(\frac{3}{8}ight) + 3^2\left(\frac{1}{8}ight)$$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$Finally,calculate the variance $\operatorname{Var}(X)$:$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$$\operatorname{Var}(X) = 3 - \left(\frac{3}{2}ight)^2$$\operatorname{Var}(X) = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$Thus,the values are $k = \frac{1}{8}$ and $\operatorname{Var}(X) = \frac{3}{4}$.

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = x_i)$	$k$	$3k$	$3k$	$k$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X = x_i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$P(X = x_i)$	$10k$	$9k$	$8k$	$8k$	$6k$	$5k$	$4k$	$3k$	$k$

If $X$ is a random variable with the distribution given below:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = x_i)$ $k$ $3k$ $3k$ $k$

Then the value of $k$ and its variance are respectively given by:

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If $X$ is a random variable with the distribution given below:$X = x_i$$0$$1$$2$$3$$P(X = x_i)$$k$$3k$$3k$$k$Then the value of $k$ and its variance are respectively given by: