Sum
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
Find P (x < 1·5)
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Solution
P (x < 1·5)
= ` int_(0)^1.5 f (x) dx`
= ` int_(0)^1.5 x/8 dx`
= `1/8[x^2/2]_0^1.5`
= `(1.5)^2/16 - 0`
=`((9/4))/16`
= `9/64`.
Is there an error in this question or solution?
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