If a random variable $x$ has the following probability distribution
$X = x_{i} : 0123$
$P (X = x_{i}) : 2 K^{2}, 3 K^{2}, 5 K^{2}, 6 K^{2}$
then the value of $K$ is

\frac{1}{4}

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\frac{- 1}{4}

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\pm \frac{1}{4}

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\frac{1}{2}

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Solution

The correct option is B $\pm \frac{1}{4}$
Using the fact , $\sum P (X) = 1$

$\Rightarrow 2 K^{2} + 3 K^{2} + 5 K^{2} + 6 K^{2} = 1$

$\Rightarrow K^{2} = \frac{1}{16} \Rightarrow K = \pm \frac{1}{4}$

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Similar questions

x

X = x_{i} : 0123

P (X = x_{i}) : 2 K^{2} 3 K^{2} 5 K^{2} 6 K^{2}

X = x_i :	0	1	2	3	4	5	6	7
P (X = x_i) :	0	2 p	2 p	3 p	p²	2 p²	7 p²	2 p

X = x_i :	0	1	2	3
P (X = x_i) :	2k⁴	3k² − 5k³	2k − 3k²	3k − 1

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

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