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

\frac{1}{8}, \frac{22}{27}

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\frac{1}{8}, \frac{23}{27}

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\frac{1}{8}, \frac{8}{9}

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

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Solution

The correct option is D $\frac{1}{8}, \frac{3}{4}$
$\sum_{i = - 2}^{3} p_{i} = 1 k + 3 k + 3 k + k = 1 \Rightarrow k = \frac{1}{8} E [X] = \sum_{i = 0}^{3} x_{i} p_{i} = 0 \times P (X = 0) + 1 \times P (X = 1) + 2 \times P (X = 2) + 3 \times P (X = 3) 3 k + 6 k + 3 k = 12 k = \frac{12}{8} E [X^{2}] = \sum_{i = 0}^{3} x^{2} p_{i} = 3 k + 12 k + 9 k = 24 k = 3 V a r i a n c e = E [X^{2}] - {E [X]}^{2} = 3 - {(\frac{3}{2})}^{2} = 3 - \frac{9}{4} = \frac{3}{4}$

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Similar questions

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

x

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P (X = x)$	$k$	$3 k$	$5 k$	$7 k$	$9 k$	$11 k$	$13 k$

^{'} k^{'}

$X = x$	0	1	2	3	4	5	6	7
$P (X = x)$	$0$	$K$	$2 K$	$2 k$	$3 K$	$K^{2}$	$2 K^{2}$	$7 K^{2} + k$

P (0 < X < 5) =

X

K =

X = x_{1} : 1, 2, 3, 4

p (X = x_{1}) : 2 k, 4 k, 3 k, k

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