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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 |
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.
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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