IE 300/GE 331 Homework 4 Sol
ASSIGNMENT:
48, 415, 436, 442, 462, 482, 4102, 4180 (from the textbook, 5
th
edition)
Problem Conversion Guide:
Problem number in 5
th
Edition
Corresponding Problem Number in 4
th
Edition
48
46
415
415
436
430
442
436
462
454
482
470
4102
488
4180
4150
48
The probability density function of the time to failure of an electronic component in a copier (in hours) is
g1858(g1876) =g1857
g2879
g3299
g3117g3116g3116g3116
/1000
for
g1876>0
. Determine the probability that
(a) A component lasts more than 3000 hours before failure.
(b) A component fails in the interval from 1000 to 2000 hours.
(c) A component fails before 1000 hours.
(d) Determine the number of hours at which 10% of all components have failed.
a)
05
.
0
1000
)
3000
(
3
3000
3000
1000
1000
=
=

=
=

∞
∞


∫
e
e
dx
e
X
P
x
x
b)
233
.
0
1000
)
2000
1000
(
2
1
2000
1000
2000
1000
1000
1000
=

=

=
=
<
<




∫
e
e
e
dx
e
X
P
x
x
c)
6321
.
0
1
1000
)
1000
(
1
1000
0
1000
0
1000
1000
=

=

=
=
<



∫
e
e
dx
e
X
P
x
x
d)
10
.
0
1
1000
)
(
1000
/
0
0
1000
1000
=

=

=
=
<



∫
x
x
x
e
e
dx
e
x
X
P
x
x
.
Then,
e
x

=
/
.
1000
09
, and x =

1000 ln 0.9 = 105.36.
415
Determine the cumulative function for the distribution in Exercise 41.
41 Suppose that
g1858(g1876)=g1857
g2879g3051
for
0<g1876
.
Since we have:
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IE 300/GE 331 Homework 4 Sol
g1858(g1876)={
0 g1876≤0
g1857
−g1876
g1876>0
We have
g1832(g1876)= g1842(g1850≤g1876)=g1516
g1858(g1873)g1856g1873=g1829−g1857
g2879g3051
g1858g1867g1870 g1876>0
g3051
g2879g2998
,
where C is a constant.
Since when x> +∞, F(x)=1, we have C=1, Hence,
g1832(g1876)={
0 g1876≤0
1−g1857
−g1876
g1876>0
436
The probability density function of the weight of packages delivered by a post office is
g1858(g1876)=
70/(69g1876
g2870
)
for
1<g1876<70
pounds.
(a) Determine the mean and the variance of weight.
(b) If the shipping cost is $2.50 per pound, what is the average shipping cost of a package?
(c) Determine the probability that the weight of a package exceeds 50 pounds.
(a)
3101
.
4
1
70

ln
69
70
69
70
)
(
)
(
70
1
70
1
=
=
=
=
∫
∫
x
dx
x
dx
x
xf
X
E
70
69
70
)
(
)
(
70
1
70
1
2
2
=
=
=
∫
∫
dx
dx
x
f
x
X
E
g1848g1853g1870(g1850)=
51.4230
18.5770
70
)
(
)
(
2
2
=

=

EX
X
E
(b) 2.5*4.3101=10.7753
(c)
0058
.
0
)
(
)
50
(
70
50
=
=
∫
dx
x
f
X
P
442
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 Spring '09
 Zafarani
 Normal Distribution, $2.50, 5mile, 2.41%, 5 miles

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