Let x be a random variable with probability mass function

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k.|x|,}&{ifx = - 2,1,3}\\ {0,}&{otherwise} \end{array}} \right.\)

where, K is a constant. Then the variance of x is:

This question was previously asked in

SSC CGL Tier-II ( JSO ) 2016 Official Paper ( Held On : 2 Dec 2016 )

Option 2 : 5

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10 Questions
10 Marks
7 Mins

**Given**

f(x) = k.|x|, if x = - 2, 1, 3

0 otherwise

**Formula**

variance = E(x^{2}) - (E(x))^{2}

**Concept**

Total probability = 1

**Calculation**

**\(\mathop ∑ \limits_{x = \; - 2} f\left( x \right) + \;\mathop ∑ \limits_{x = 1} f\left( x \right) + \;\mathop ∑ \limits_{x = 3} f\left( x \right) = 1\)**

⇒ \(k\mathop ∑ \limits_{x = \; - 2} \left| X \right| + \;k\mathop ∑ \limits_{x = 1} \left| X \right| + \;k\mathop ∑ \limits_{x = 3} \left| X \right| = 1 \)

⇒ k|-2| + k|1| + k|3| = 1

⇒ 2k + k + 3k = 1

⇒ k = 1/6

We know that

Var x = E(x2) - (E(x))^{2}

E(x) = ∑xf(x)

⇒ E(x) = \(\mathop ∑ \limits_{x = \; - 2} xf\left( x \right) + \;\mathop ∑ \limits_{x = 1} xf\left( x \right) + \;\mathop ∑ \limits_{x = 3} xf\left( x \right) \)

⇒ E(x) = \(1/6\mathop ∑ \limits_{x = \; - 2} x\left| x \right| + \;1/6\mathop ∑ \limits_{x = 1} x\left| x \right| + \;1/6\mathop ∑ \limits_{x = 3} x\left| x \right| \)

⇒ E(x) = 1/6[ -2 × 2 + 1 × 1 + 3 × 3]

⇒ E(x) = 1/6[6]

⇒ E(x) = 1

Now E(x^{2}) = ∑x^{2}f(x)

⇒ E(x2) = \(\mathop \sum \limits_{x = \; - 2} \)x^{2}f(x) + \(\mathop \sum \limits_{x = \; 1}\)x^{2}f(x) + \(\mathop \sum \limits_{x = \; 3}\)x^{2}f(x)

⇒ E(x2) =1/6[ \(\mathop \sum \limits_{x = \; -2}\)x^{2}|x| + \(\mathop \sum \limits_{x = \; 1}\)x^{2}|x| + \(\mathop \sum \limits_{x = \; 3}\)x^{2}|x|

⇒ E(x2) = 1/6[(-2)^{2} × 2 + 1^{2} × 1 + 3^{2} × 3]

⇒ E(x2) = 1/6(8 + 1 + 27) = 6

⇒ Variance = 6 - (1)^{2}

**∴ Variance is 5**