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Engineering 0020: Probability and Statistics for Engineers 1
Spring, 2007; Solutions: Homework
#5
Chapter 4.4, 4.5
1.
Based on data for Question 81, p139.
a
Let X be the time between aircraft arrivals. X ~ exp(
λ
=8)
μ
x
= E(X) = 1/8 hr. = 7.5 minutes
b
P(X<5) = 1 – e
(8)(5/60)
= 0.4866
c
P(X>30) = e
(8)(0.5)
= 0.0183
d
P(X<5 or X>10) = P(X<5) + P(X>10) = 0.4866+ e
(8)(10/60)
=0.7502
e
P(X>30  last arrival occurred 30 min. ago) = P(X>30) = 0.0183
This follows from the "Memoryless" property of the Exponential Distribution.
86.
Compute the sample and Normal percentiles as shown below:
i
(
i0.5
)/
n
X
i
Z
i
1
0.025
47.1
1.96
2
0.075
68.1
1.44
3
0.125
68.1
1.15
4
0.175
90.8
0.93
5
0.225
103.6
0.76
6
0.275
106
0.60
7
0.325
115
0.45
8
0.375
126
0.32
9
0.425
146.6
0.19
10
0.475
229
0.06
11
0.525
240
0.06
12
0.575
240
0.19
13
0.625
278
0.32
14
0.675
278
0.45
15
0.725
289
0.60
16
0.775
289
0.76
17
0.825
367
0.93
18
0.875
385.9
1.15
19
0.925
392
1.44
20
0.975
505
1.96
A plot of X
i
vs. Z
i
is shown on the next page:
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View Full Document Normal Probability Plot
0
100
200
300
400
500
600
3.00
2.00
1.00
0.00
1.00
2.00
3.00
z percentile
observed data
The plot looks reasonably straight and an assumption of Normality seems to be a
plausible one.
The intercept (approx. 200+ ??) would be an estimate of the mean and the
slope would estimate the std. dev.
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This homework help was uploaded on 04/10/2008 for the course ENGR 0020 taught by Professor Rajgopal during the Spring '08 term at Pittsburgh.
 Spring '08
 Rajgopal

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