Unformatted text preview: = fX (u)du
−∞
1 A(1 − u2 )du =
−1 = A u− u3
3 1 =
−1 4A
,
3 so A = 3 .
4
The support of fX (the set on which it is not zero) is the interval [−1, 1]. To ﬁnd P {0.5 < X <
1.5}, we only need to integrate over the portion of the interval [0.5, 1.5] in the support of fX . That
is, we only need to integrate over [0.5, 1] :
1.5 P {0.5 < X < 1.5} = fX (u)du
0.5
1 =
0.5 = 3
4 3(1 − u2 )
du
4
u− u3
3 1 =
0 .5 3
4 2 11
−
3 24 = 5
.
32 Because the support of fX is the interval [−1, 1], we can immediately write down the partial
answer: 0 c ≤ −1
? −1 < c < 1
FX (c) = 1 c ≥ 1,
where the question mark represents what hasn’t yet been determined, namely, the value of FX (c)
for −1 < c < 1. So let −1 < c < 1. Then
FX (c) = P {X ≤ c}
c
3(1 − u2 )
=
du
4
−1
3
u3 c
u−
=
4
3
−1
= 2 + 3c − c3
.
4 This allows us to give the complete expression for FX : 0
c ≤ −1 2+3c−c3
FX (c) =
−1 < c < 1
4 1
c ≥ 1. 3.3. UNIFORM DISTRIBUTION 77
∞
−∞ ufX (u)du, The mean, µX , is zero, because µX =
over R is zero. (For example, ufX (u) = 9
32 and ufX (u) is an odd function so its integral and uf...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
 Spring '08
 Zahrn
 The Land

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