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Consider the probability distribution of a random variable X: X 0 1 2 3 4 P(X) 0.1 0.25 0.3 0.2 0.15 Calculate VXV(X2) - Mathematics

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प्रश्न

Consider the probability distribution of a random variable X:

X 0 1 2 3 4
P(X) 0.1 0.25 0.3 0.2 0.15

Calculate `"V"("X"/2)`

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योग

उत्तर

Here, we have

X 0 1 2 3 4
P(X) 0.1 0.25 0.3 0.2 0.15

We know that: Var(X) = E(X2) – [E(X)]2

Where E(X) = `sum_("i" = 1)^"n" x_"i""p"_"i"` and E(X2) = `sum_("i" = 1)^"n" "p"_"i"x"i"^2`

∴ E(X) = 0 × 0.1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15

= 0 + 0.25 + 0.6 + 0.6 + 0.6

= 2.05

E(X2) = 0 × 0.1 + 1 × 0.25 + 4 × 0.3 + 9 × 0.2 + 16 × 0.15

= 0 + 0.25 + 1.2 + 1.8 + 2.40

= 5.65

 `"V"("X"/2) = 1/4"V"("X")`

= `1/4[5.65 - (2.05)^2]`

= `1/4[5.65 - 4.2025]`

= `1/4 xx 1.4475`

= 0.361875   .......`[because "V"("X"/2) = 1/2"V"("X")]`

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अध्याय 13: Probability - Exercise [पृष्ठ २७४]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 13 Probability
Exercise | Q 24. (i) | पृष्ठ २७४

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