Week_11_Probability_Distributions
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School
Stevens Institute Of Technology *
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Course
556
Subject
Industrial Engineering
Date
Apr 3, 2024
Type
Pages
8
Uploaded by PrivateFlower11872
33) Mopeds (small motorcycles with an engine capacity below 50
cm3) are very popular in Europe because of their mobility, ease of
operation, and low cost. The article “Procedure to Verify the Max-
imum Speed of Automatic Transmission Mopeds in Periodic Motor
Vehicle Inspections” (J. of Automobile Engr., 2008: 1615–1623) de-
scribed a rolling bench test for determining maximum vehicle speed.
A normal distribution with mean value 46.8 km/h and standard devi-
ation 1.75 km/h is postulated. Consider randomly selecting a single
such moped.
a. What is the probability that maximum speed is at most 50 km/h?
b. What is the probability that maximum speed is at least 48 km/h?
c.
What is the probability that maximum speed differs from the
mean value by at most 1.5 standard deviations?
Answer
Given that,
µ
= 46
.
8
σ
= 1
.
75
z
=
x
−
µ
σ
a) maximum speed = 50
P
(
X
≤
50)
=
50
−
46
.
8
1
.
75
=
3
.
2
1
.
75
= 1
.
83
According to z
−
table
1
.
82 = 0
.
9664
Therefore,The probability that maximum speed at most 50 km/h is 0.9664
b) maximum speed = 48
P
(
X
≥
48)
=
−
48
−
46
.
8
1
.
75
1
=
−
1
.
2
1
.
75
=
−
0
.
69
According to z
−
table
0
.
68 = 0
.
2451
Therefore,The probability that maximum speed atleast 48 km/h is
c)
P
(
−
1
.
5
< z <
1
.
5)
=
P
(
z <
1
.
5)
−
P
(
z <
−
1
.
5)
= 0
.
9332
−
0
.
0668
= 0
.
8664
Therefore,The probability that maximum speed differs from the mean value by
at most 1.5 is 0.8664
35) Suppose the diameter at breast height (in.) of trees of a certain
type is normally distributed with m 5 8.8 and s 5 2.8, as suggested in
the article “Simulating a Harvester-Forwarder Softwood Thinning”
(Forest Products J., May 1997: 36–41).
a. What is the probability that the diameter of a randomly selected
tree will be at least 10 in.? Will exceed 10 in.?
b. What is the probability that the diameter of a randomly selected
tree will exceed 20 in.?
c. What is the probability that the diameter of a randomly selected
tree will be between 5 and 10 in.?
d.
What value c is such that the interval (8.8 2 c, 8.8 1 c) includes
98percent of all diameter values?
e.
If four trees are independently selected, what is the probabil-
ity that at least one has a diameter exceeding 10 in.?
Answer
Given,
µ
= 8
.
8
σ
= 2
.
8
z
=
x
−
µ
σ
a. x=10
z
=
10
−
8
.
8
2
.
8
z
=
1
.
2
2
.
8
2
z
= 0
.
428
P
(
h >
10) = 1
−
P
(
h <
10)
= 1
−
P
(
z <
0
.
43)
According to z
−
table
= 1
−
0
.
6664
= 0
.
3346
Therefore,The probability that the diameter of a randomly selected tree will be
at least 10 is 0.3346
b. x=20
z
=
20
−
8
.
8
2
.
8
z
=
11
.
2
2
.
8
z
= 4
P
(
h >
20) = 1
−
P
(
h <
20)
= 1
−
P
(
z <
4)
According to z
−
table
= 1
−
1
= 0
Therefore,The probability that the diameter of a randomly selected tree will
exceed 20 is 0
c.
P
(5
< x <
10) =
P
(
h <
10)
−
P
(
h <
5)
=
P
(
z <
10
−
8
.
8
2
.
8
)
−
P
(
z <
5
−
8
.
8
2
.
8
)
=
P
(
z <
1
.
2
2
.
8
−
P
(
z <
−
3
.
8
2
.
8
))
P
(
z <
0
.
43)
−
P
(
z <
−
1
.
36)
According to z
−
table
0
.
6664
−
0
.
0869
= 0
.
5795
Therefore,The probability that the diameter of a randomly selected tree will be
between 5 and 10 is 0.5795
3
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