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 find the mean.

\frac{33}{16}

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\frac{31}{16}

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\frac{35}{16}

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\frac{29}{16}

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Solution

The correct option is A $\frac{31}{16}$
Using $\sum P (X) = 1 \Rightarrow 16 K^{2} = 1 \Rightarrow K^{2} = \frac{1}{16}$

Thus mean $= \sum x_{i} P (X = x_{i}) = 0 + P (X = 1) + 2 P (X = 2) + 3 P (X = 3)$

$0 + 3 K^{2} + 10 K^{2} + 18 K^{2}$

$= 31 K^{2} = \frac{31}{16}$

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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}

K

X = x_i :	1	2	3	4
P (X = x_i) :	k	2k	3k	4k

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

p (X = x_{i}) : \frac{2}{6}, \frac{3}{6}, \frac{0}{6}, \frac{1}{6}

x

X = x_{i} : - 1, 0, + 1

p (X = x_{i}) : 0.4, 0.2, 0.4

X

X = x : 1234

p (X = x) : 0.4 0.3 0.2 0.1

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