Find expected value and variance of X for the following p.m.f. x -2 -1 0 1 2 P(X) 0.2 0.3 0.1 0.15 0.25 - Mathematics and Statistics

Question

Find expected value and variance of X for the following p.m.f.

x	-2	-1	0	1	2
P(X)	0.2	0.3	0.1	0.15	0.25

Sum

Solution 1

We construct the following table to calculate E (X) and V (X) :

X = x_i	_pi=P [X = x_i]	x_i · p_i	x_i ²·p_i = x_i × x_i·p_i
-2	0.2	-0.4	0.8
-1	0.3	-0.3	0.3
0	0.1	0	0
1	0.15	0.15	0.15
2	0.25	0.5	1
Total	1	-0.05	2.25

From the table, Σx_i · p_i = -0.05 and Σx_i² · p_i = 2.25

∴E (X) = Σx_i · p_i= -0.05

and V (X) = Σx_i²·p_i- ( Σx_i · p_i)²

= 2.25 - (-0.05)²

= 2.25 - 0.0025 = 2.2475

Hence, E (X)= -0.05 and V (X) = 2.2475.

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Solution 2

Expected value of X 5
= E(X) = $\sum_{i = 1}^{5} x_{i} . P_{i} (x_{i})$

= (–2) x (0.2) + (–1) x (0.3) + 0 x (0.1) + 1 x (0.15) + 2 x (0.25)
= – 0.4 – 0.3 + 0 + 0.15 + 0.5
= – 0.05

E(X²) = $\sum_{i = 1}^{5} x_{i} . P_{i} (x_{i})$

= (–2)² x (0.2) + (–1)2 x (0.3) + 02 x (0.1) + 12 x (0.15) + 22 x (0.25)
= 0.8 + 0.3 + 0 + 0.15 + 1
= 2.25
∴ Variance of X
= Var(X)
= E(X²) – [E(X)]²
= 2.25 – (– 0.05)²
= 2.2475.

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Probability Distribution of Discrete Random Variables

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Chapter 7: Probability Distributions - Exercise 7.1 [Page 232]

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Chapter 8 Probability Distributions
Exercise 8.1 | Q 9 | Page 141

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Find expected value and variance of X for the following p.m.f. x -2 -1 0 1 2 P(X) 0.2 0.3 0.1 0.15 0.25 - Mathematics and Statistics

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