a continuous random variable x is uniformly distributed over interval [2,5].What is the expected value of X
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Answer:
c.d.f. of the continuous variable is given by,
f(x)=∫
−1
x
3
y
2
dy
=[
9
y
3
]
−1
x
=
9
(x
3
+1)
,x∈R
Consider P(x<1)=f(1)=
9
(1)
3
+1
=
9
2
P(x≤−2)=0
P(x>0)=1−P(x≤0)
=1−f(0)
=1−(
9
1
)=
9
8
P(1<x<2)=f(2)−f(1)
=1−
9
2
=
9
7
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