Sum
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
P(x > 2)
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Solution
P( x > 2 )
= ` int_(2)^4 f (x) dx`
=` int_(2)^4 x/8 dx`
=`1/8[x^2/2]_2^4`
=`1/8 [16/2 - 4/2]`
= `1/8 xx 6`
=`3/4`
Is there an error in this question or solution?
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