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Question
Solve the following problem :
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 < – 0.5 or X > 0.5)
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 < – 0.5 or X > 0.5)
= 1 – P(– 0.5 ≤ X ≤ 0.5)
= 1 – [F(0.5) – F(– 0.5)]
= `1 - {3/32[4(0.5) - (0.5)^3/3 + 16/3] - 3/32[4(-0.5) - (0.5)^3/3 + 16/3]`
= `1 - 3/32(2 - 1/24 + 16/3 + 2 - 1/24 - 16/3)`
= `1 - 3/32(4 - 1/12)`
= `1 - (3)/(32) xx (47)/(12)`
= `1 - (47)/(128)`
= `(81)/(128)`.
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