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Question
Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.
f (x) = k `(4 – x^2 )`, for –2 ≤ x ≤ 2 and = 0 otherwise.
P(x > 0)
Solution
Since, f is the p.d.f. of X,
` int_(-∞)^∞ f (x) dx` = 1
∴ ` int_(-∞)^-2 f (x) dx` +` int_(-2)^2 f (x) dx` + ` int_(2)^∞f (x) dx`= 1
∴ 0 + ` int_(-2)^2 k (4 -x^2) dx` = 1
∴ k ` int_(-2)^2 (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`
P(x > 0)
= ` int_(0)^∞ f (x) dx`
= ` int_(0)^2 f (x) dx`+ ` int_(2)^∞ f (x) dx`
= ` int_(0)^2 k (4-x^2) dx+ 0`
= k` int_(0)^2 (4-x^2) dx`
=`3/32[4x -x^3/3]_0^2` ..........[∵ k=`3/32`]
=`3/32 [8-8/3] = 3/32 xx16/3 = 1/2`
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