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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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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प्रश्न

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
बेरीज

उत्तर १

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

X = xi pi =P [X = xi] xi · pi xi 2·pi = xi × xi·pi
-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, Σxi · pi = -0.05 and Σxi2 · pi = 2.25

∴E (X) = Σxi · pi= -0.05

and V (X) = Σxi 2 ·pi - ( Σxi · pi)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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उत्तर २

Expected value of X 5
= E(X) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]

= (–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) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]

= (–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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Probability Distributions - Exercise 7.1 [पृष्ठ २३२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 7 Probability Distributions
Exercise 7.1 | Q 9 | पृष्ठ २३२
बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 8 Probability Distributions
Exercise 8.1 | Q 9 | पृष्ठ १४१

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