Week_11_Probability_Distributions
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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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d.
98
100
= 0
.
98
P
(8
.
8
−
c < h <
8
.
8 +
c
) = 0
.
98
P
(
h <
8
.
8 +
c
) = 0
.
98
P
(
z <
c
2
.
8
) = 0
.
98
c
=
2
.
33
c
= 2
.
33
×
2
.
8
c
= 6
.
524
Therefore,The value of c is 6.524
e.
z
=
10
−
8
.
8
2
.
8
z
=
1
.
2
2
.
8
z
= 0
.
43
P
(
x <
10) =
P
(
z <
0
.
43)
According to z
−
table
= 0
.
6664
P
(
Four trees <
10) =
P
(
x <
10)
.P
(
x <
10)
.P
(
x <
10)
.P
(
x <
10)
= [
P
(
x <
10)]
4
= (0
.
6664)
4
= 0
.
1972
P
(
atleast one >
10) = 1
−
P
(
Four trees <
10)
= 1
−
0
.
1972
= 0
.
8028
Therefore,The probability that at least one has a diameter exceeding 10 is 0.8028
49) Consider babies born in the “normal” range of 37–43 weeks ges-
tational age.
Extensive data supports the assumption that for such
babies born in the United States, birth weight is normally distributed
with mean 3432 g and standard deviation 482 g. (The article “Are Ba-
bies Normal?” (The American Statistician, 1999: 298–302) analyzed
4
data from a particular year; for a sensible choice of class intervals,
a histogram did not look at all normal, but after further investiga-
tions it was determined that this was due to some hospitals measuring
weight in grams and others measuring to the nearest ounce and then
converting to grams. A modified choice of class intervals that allowed
for this gave a histogram that was well described by a normal distri-
bution.)
a.
What is the probability that the birth weight of a randomly se-
lected baby of this type exceeds 4000 g?
Is between 3000 and 4000
g?
b.
What is the probability that the birth weight of a randomly se-
lected baby of this type is either less than 2000 g or greater than 5000
g?
c.
What is the probability that the birth weight of a randomly se-
lected baby of this type exceeds 7 lb?
d.
How would you characterize the most extreme .1percent of all
birth weights?
e.
If X is a random variable with a normal distribution and a is a
numerical constant (a is not equal to 0), then Y=aX also has a normal
distribution.
Use this to determine the distribution of birth weight
expressed in pounds (shape, mean, and standard deviation), and then
recalculate the probability from part (c). How does this compare to
your previous answer?
Given,
µ
= 3432
g
σ
= 482
g
z
=
x
−
µ
σ
a.
P
(
W <
4000) =
P
(
z >
4000
−
3432
482
)
=
P
(
z >
568
482
)
=
P
(
z >
1
.
18)
= 1
−
P
(
z <
1
.
18)
According to z
−
table
= 1
−
0
.
8810
= 0
.
1190
5
P
(3000
< W <
4000) =
P
(
W <
4000)
−
P
(
W <
3000)
=
P
(
z <
4000
−
3432
482
)
−
P
(
W <
3000
−
3432
482
)
=
P
(
z <
568
482
)
−
P
(
W <
−
432
482
)
=
P
(
z <
1
.
18)
−
P
(
z <
−
0
.
90)
According to z
−
table
= 0
.
8810
−
0
.
1841
= 0
.
6969
Therefore,The probability that the birth weight of a randomly selected baby of
this type exceeds 4000 g is 0.1190
Therefore,The probability that the birth weight of a randomly selected baby
of this type between 3000 and 4000 g is 0.6969
b.
=
P
(
W <
2000) +
P
(
W >
5000)
=
P
(
W <
2000) + 1
−
P
(
W <
5000)
= 1
−
P
(
z <
5000
−
3432
482
) +
P
(
z <
2000
−
3432
482
)
= 1
−
P
(
z <
1568
482
) +
P
(
z <
−
1432
482
)
= 1
−
P
(
z <
3
.
25)
−
P
(
z <
−
2
.
97)
According to z
−
table
= 1
−
0
.
9994 + (0
.
0015)
= 0
.
0006 + 0
.
0015
= 0
.
0021
Therefore,The probability that the birth weight of a randomly selected baby of
this type is either less than 2000 g or greater than 5000g is 0.0021
c.
1
lb
= 453
.
59
453
.
59
∗
7
7
lb
= 3175
.
15
g
P
(
W >
3175
.
15) = 1
−
P
(
W <
3175
.
15)
6
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= 1
−
P
(
z <
3175
.
15
−
3432
482
)
1
−
P
(
z <
−
256
.
85
482
)
= 1
−
P
(
z <
−
0
.
53)
According to z
−
table
= 1
−
0
.
2981
= 0
.
7019
Therefore,The probability that the birth weight of a randomly selected baby of
this type exceeds 7 lb is 0.7019
d.
A
=
0
.
1
100
A
= 0
.
001
The Lower Bound Probability will A/2
P
(
X
low
) =
0
.
001
2
P
(
X
low
) = 0
.
0005
Z
low
=
−
3
.
295
−
3
.
295 =
X
low
−
3432
482
−
3
.
295
×
482 =
X
low
−
3432
−
1588
.
19 =
X
low
−
3432
X
low
= 3432
−
1588
.
19
X
low
= 1843
.
81
1
−
X
low
=
X
upp
X
upp
= 1
−
0
.
0005
X
upp
= 0
.
9995
X
upp
= 3
.
30
3
.
30 =
X
upp
−
3432
482
X
upp
−
3432 = 3
.
30
×
482
X
upp
−
3432 = 1590
.
6
7
X
upp
= 1590
.
6 + 3432
X
upp
= 5022
.
6
Therefore,0.1 percent of babies weight less than 1843.81g or more than 5022.6g
e.
Given Y=ax
Normal Distribution
From Problem c; 1lb=453.59
Y
=
ax
Y
=
1
453
.
59
×
x
Y
= 0
.
002204
×
x
So a
= 0
.
002204
µ
= 3432
g σ
= 482
g
µ.Y
= 0
.
002204
×
3432
= 7
.
56
lb
σ.Y
= 0
.
002204
×
482
= 1
.
06
lb
From Problem C
P
(
z >
7) = 1
−
P
(
z <
7)
= 1
−
P
(
z <
7
−
7
.
56
1
.
06
)
= 1
−
P
(
z <
−
0
.
56
1
.
06
)
= 1
−
P
(
z <
−
0
.
56)
= 1
−
0
.
2981
= 0
.
7019
Therefore,we get similar probability of problem c
8