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Sum

Suppose error involved in making a certain measurement is a continuous r. v. X with p.d.f.

f(x) = `{("k"(4 - x^2), "for" -2 ≤ x ≤ 2),(0, "otherwise".):}`

compute P(X > 0)

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#### Solution

Given that f(x) represents a p.d.f. of r.v. X.

∴ `int_-2^2 f(x)*dx` = 1

∴ `int_-2^2 "k"(4 - x^2)*dx` = 1

∴ `"k"[4x - x^3/3]_-2^2` = 1

∴ `"k"[(8 - 8/3) - (-8 + 8/3)]` = 1

∴ `"k"(16/3 + 16/3)` = 1

∴ `"k"(32/3)` = 1

∴ k = `(3)/(32)`

F(x) = `int_-2^2 f(x)*dx`

= `int_-2^2"k"(4 - x^2)*dx`

= `(3)/(32)[4x - x^3/3]_-2^2`

= `(3)/(32)[4x - x^3/3 + 8 - 8/3]`

∴ F(x) = `(3)/(32)[4x - x^3/3 + 16/3]`

P(X > 0) = 1 – P(X ≤ 0)

= 1 – F(0)

= `1 - (3)/(32)(0 - 0 + 16/3)`

= `1 - (1)/(2)`

= `(1)/(2)`.

Concept: Probability Distribution of a Continuous Random Variable

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