In the following table, x is a discrete random variable and p(x) is the probability density function. The standard deviation of x is

x 1 2 3

p(x) 0.3 0.6 0.1

0.18

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0.36

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0.54

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0.6

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Solution

The correct option is D 0.6
For discrete distribution :

$Variance = E (x^{2}) - [E (x)]^{2}$

$E (x) = \sum x . p (x)$
= 1 × 0.3 + 2 × 0.6 + 3 × 0.1
= 1.8
$E (x^{2}) = \sum x^{2} p (x)$
$= 1^{2} \times 0.3 + 2^{2} \times 0.6 + 3^{2} \times 0.1$
= 3.6
$V a r (x) = 3.6 - (1.8)^{2}$
$V a r (x) = 0.36$
$σ = \sqrt{V a r (x)}$
$σ = 0.6$

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