ME335Engineering StatisticsFall 2008
HW # 3
Solution
4.3 (a)

( )
1,
f x dx
x
x
x
x
4
0
1
x
ke
dx
x

x
x
,
4
1
k x
, k=1/4
(b)P{2<X<8}=
8
2
( )
f x dx
x
=
8
4
2
1
4
x
e
dx

=
1
2
2
e
e



=0.4712
(c) P{X>7}=
7
4
4
7
7
1
( )
0.1738
4
x
f x dx
e
dx
e
x


=
=
=
�
�
(d) P{X<1}=
1
1
1
4
4
0
1
( )
1
0.2212
4
x
f x dx
e
dx
e



=
=

=
�
�
(e) P{X>C}=
1
4
4
1
( )
0.5
4
C
C
C
f x dx
e
dx
e
x


=
=
�
�
�
(Obviously C>0)
4
ln(
)
ln(0.05)
4
C
C
e

=
= 
,
ln(0.05)
4ln(20)
11.98
C
= 
=
(f) P{X<d}=
d
4
4
0
1
( )
1
0.025
4
x
d
d
f x dx
e
dx
e



=
=

=
�
�
, d
0.1
W
4.9
From the additive property of the normal distribution, we know that, if you take
several normal random variables, Xi, with i=1,2,…,n, and form weighted sums
and/or differences among them, the result will always be another normal random
variable, i.e.
, Y is a normal random variable and its mean and variance
are:
,
.
For this question, here Y= A+B+C, A,B and C are independently normal random
variables, so Y is a normal random variable, and its mean and variance are as
follows:
,
Entering the formula Normal Dist(12.2, 12, 0.10488) to JMP to calculate the result.
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