Algebra

Probability

Medium

applied

A random variable X has the following probability distribution:

X 0 1 2 3
P(X) 0.2 0.1 0.3 0.4

The variance of X will be

X	0	1	2	3
P(X)	0.2	0.1	0.3	0.4

4.8

1.29

5.1

4.9

✅ Correct Option: 2

The variance formula is:

Variance $= E(X^2) - [E(X)]^2$

This requires finding $E(X)$ and $E(X^2)$ .

Calculate $E(X)$ :

$E(X) = \sum [x \times P(x)]$

X	P(X)	X × P(X)
0	0.2	0 × 0.2 = 0
1	0.1	1 × 0.1 = 0.1
2	0.3	2 × 0.3 = 0.6
3	0.4	3 × 0.4 = 1.2

$E(X) = 0 + 0.1 + 0.6 + 1.2$

$E(X) = 1.9$

Calculate $E(X^2)$ :

$E(X^2) = \sum [x^2 \times P(x)]$

X	X²	P(X)	X² × P(X)
0	0	0.2	0 × 0.2 = 0
1	1	0.1	1 × 0.1 = 0.1
2	4	0.3	4 × 0.3 = 1.2
3	9	0.4	9 × 0.4 = 3.6

$E(X^2) = 0 + 0.1 + 1.2 + 3.6$

$E(X^2) = 4.9$

Calculate variance:

Variance $= E(X^2) - [E(X)]^2$

Variance $= 4.9 - (1.9)^2$

Variance $= 4.9 - 3.61$

Variance $= 1.29$

Therefore, the variance of X is $1.29$ .

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

A random variable X has the following probability distribution:

X 0 1 2 3
P(X) 0.2 0.1 0.3 0.4

The variance of X will be