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Question
The following is the p.d.f. of r.v. X :
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise
P ( 1 < x < 2 )
Solution
P (1 < x < 2)
= ` int_(1)^2 f (x) dx`
= ` int_(1)^2 x/8 dx`
`1/8[x^2/2]_1^2`
=`1/8[4/2-1/2]`
=`3/16`
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