MAE108 Homework 6 Solutions - Spring 2015
June 7, 2015
Problem 3.48
Define X as the capacity of a pile with µX = 100 tons and δX = 0.2, so σX = δXµX = 20 tons. Because Xfollows a lognormal distribution,
fX(x) =1√
2π(ζx)exp
[−1
2
(ln(x)− λ
ζ
)2],
ζ =√
ln(1 + δX
2)
=√
ln (1 + 0.22) = 0.198,
λ = ln(µX)− 1
2ζ2 = ln(100)− 1
20.1982 = 4.586.
a)
P (X > 100) = 1− P (X ≤ 100)
= 1− Φ
(ln 100− λ
ζ
)= 1− Φ
(ln 100− 4.586
0.198
)= 1− Φ (0.097) = 0.46.
b)
P (X > 100 |X > 75) =P (X > 100)
P (X > 75)
=0.46
1− Φ(
ln(75)−4.5860.198
)=
0.46
1− Φ(−1.356)
=0.46
Φ(1.356)= 0.504.
1
c)
P (X > 100 |X > 90) =P (X > 100)
P (X > 90)
=0.46
1− Φ(
ln(90)−4.5860.198
)=
0.46
1− Φ(−0.435)
=0.46
Φ(0.435)= 0.687.
Problem 3.57
Define X,Y as the head pressure at nodes A,B, respectively. We know µX = 10 units and δX = 0.2, soσX = δXµX = 2 units. Because X follows a lognormal distribution,
fX(x) =1√
2π(ζx)exp
[−1
2
(ln(x)− λ
ζ
)2],
ζ =√
ln(1 + δX
2)
=√
ln (1 + 0.22) = 0.198,
λ = ln(µX)− 1
2ζ2 = ln(10)− 1
20.1982 = 2.28.
a)
P (6 ≤ X ≤ 14) = Φ
(ln 14− λ
ζ
)− Φ
(ln 6− λ
ζ
)= Φ
(ln 14− 2.28
0.198
)− Φ
(ln 6− 2.28
0.198
)= Φ(1.81)− Φ(−2.47)
= Φ(1.81)− (1− Φ(2.47))
= 0.965− (1− 0.993) = 0.958.
b)
Define A as the event of 6 ≤ X ≤ 14 and B as the event of 6 ≤ Y ≤ 14. We know P (B) = 0.90,P (B|A) = 0.80, and P (A) = 0.958.
P (AB) = P (B|A)P (A)
= (1− P (B|A))(1− P (A))
= (1− 0.8)(1− 0.958) = 0.0084.
cI)
Now δX = 0.15 so our new values are
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ζ =√
(ln(1 + 0.152)) = 0.149
λ = ln(10)− 1
2· 0.1492 = 2.29
P (A) = Φ
(ln 14− 2.29
0.149
)− Φ
(ln 6− 2.29
0.149
)= Φ(2.34)− Φ(−3.34) = 0.99
P (AB) = (1− 0.8)(1− 0.99) = 0.002.
cII)
Now P (B) = 0.95 and P (B|A) = 0.90, so
P (AB) = (1− 0.90)(1− 0.958) = 0.0042.
Option I minimizes the probability of unsatisfactory water service.
Problem 4.7
Define R1, R2 as the maximum annual flood peak for rivers 1,2, respectively. Define C = R1 +R2, we knowR1 ∼ N(35, 10) and R2 ∼ N(25, 10).
a)
The mean and standard deviation of the sum of two distributions are given by
µC = µR1+ µR2
= 35 + 25 = 60 m3/sec.
σC =√σR1
2 + σR12 + 2Cov(R1, R2) =
√102 + 102 + 2 ∗ 0.5 ∗ 10 ∗ 10 = 17.32 m3/sec.
b)
P (C > 100) = 1− P (C ≤ 100)
= 1− Φ
(100− 60
17.32
)= 1− Φ(2.31) = 1− 0.9896 = 0.0104.
The return period is T = 10.0104 = 96 years.
c)
Define F as the event of flooding in a year. We model a 10 year period as a Bernouilli sequence, where
P (F10 ≥ 1) = 1− P (F10 = 0)
= 1−(
10
0
)0.01040(1− 0.0104)10
= 1− 1 · 1 · 0.901 = 0.10.
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d)
Now P (F10 ≥ 1) = 0.05, we need to find P (F ∗).
0.05 = 1−(
10
0
)P (F ∗)0(1− P (F ∗))10
0.05 = 1− (1− P (F ∗))10 =⇒ P (F ∗) = 1− 0.950.1 = 0.0051.
P (C > C) = 1− P (C ≤ C)
0.0051 = 1− Φ
(C − 60
17.32
)C − 60
17.32= Φ−1(0.995)
C = 104.69.
So the channel should be enlarged by 4.69 m3/sec.
Problem 4.9
Normal travel times are given by T1 ∼ N(30, 9), T2 ∼ N(20, 4), and T2 ∼ N(40, 12). In rush hour, time fromA to B is T̃2 ∼ N(30, 6).
a)
Define T = T1 + T2 + T3 as the round trip time under normal traffic conditions, we know the sum of normaldistributions gives a normal distribution.
µT = µT1+ µT2
+ µT3= 90
σT =√σT1
+ σT2+ σT3
= 15.52
P (T < 120) = Φ
(120− 90
15.52
)= Φ(1.93) = 0.973.
b)
Define X = T2 + T3 as the time from town A to the shopiing center under normal traffic conditions.
µX = µT2+ µT3
= 60
σX =√σT2
+ σT3= 12.65
P (X < 60) = Φ
(60− 60
12.65
)= Φ(0) = 0.50.
c)
Define F as the fraction of passengers who arrive at the shopping center in less than one hour during rushhour, and X̃ = T̃2 + T3 as the time from town A to the shopping center during rush hour.
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µX̃ = µT̃2+ µT3 = 70
σX̃ =√σT̃2
+ σT3 = 13.42
F =2
3P (T3 < 60) +
1
3P (X̃ < 60)
=2
3Φ
(60− 40
12
)+
1
3Φ
(60− 70
13.42
)=
2
3Φ(1.67) +
1
3Φ(−0.75)
=2
3Φ(1.67) +
1
3(1− Φ(0.75)) = 0.711.
d)
P (T2 < 60 |T3 > 45) =P (45 < T3 < 60)
P (T3 > 45)
=Φ(60−40
12
)− Φ
(45−40
12
)1− Φ
(45−40
12
)=
Φ(1.67)− Φ(0.42)
1− Φ(0.42)= 0.86.
Problem 6.3
We know x = 10, 000 cfs, s2 = 9 · 106cfs2, and n = 10.
a)
The two-sided 90% confidence interval is given by
〈µ〉1−α =
(x+ tα
2 ,n−1s√n, x+ t(1−α
2 ),n−1s√n
).
with α = 0.1.
〈µ〉0.9 =
(x+ t0.05,9
s√n, x+ t0.95,9
s√n
)=
(10000 + 1.83 · 3000√
10, 10000− 1.83 · 3000√
10
)= (8264, 11736).
b)
We now require
tα2 ,n−1
s√n≤ 1000
t0.95,n−13000√n≤ 1000
t0.95,n−1 ≤√n
3.
Using table A-3 we see that
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n = 15 =⇒ t0.95,14 = 1.76 � 1.24
n = 25 =⇒ t0.95,24 = 1.71 � 1.67
n = 27 =⇒ t0.95,26 = 1.71 ≤ 1.73.
We need 17 additional years of observation.
Problem 6.7
The PDF of ocean wave height is
fH(h) =h
α2· e 1
2 (h/α)2
, h ≥ 0.
Using the method of maximum likelihood,
L(h1, . . . , h10;α) =
10∏i=1
hiα2· e 1
2 (hi/α)2
ln(L) = ln(α−20) +
10∑i=1
ln(hi) +
10∑i=1
−1
2(hi/α)2
ln(L) = −20 ln(α) +10∑i=1
ln(hi)−1
2α2
10∑i=1
hi2.
To maximize this likelihood function,
∂ ln(L)
∂α=−20
α+
1
α3·
10∑i=1
hi2 = 0
−20α2 +
10∑i=1
hi2 = 0
−20α2 + 80.42 = 0
α = 2.0.
Problem 6.8
We know n = 10, x = 1n
∑10i=1 xi = 2.06, and s2 = 1
n−1
(∑10i=1 xi
2 − nx2)
= 0.216, so s = 0.465.
a)
We perform a one-sided hypothesis test with
H0 : µ = µ0
HA : µ > µ0
where µ0 = 2.0 mg/l. The test statistic is
t =x− µ0
s/√n
=2.06− 2.0
0.465/√
10
= 0.408.
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From table A-3 we find t0.95,9 = −1.833. Because t > t0.95,9, the test statistic is inside the region ofrejection, so the null hypothesis is rejected.
b)
If X ∼ N(µ, σ) then from our data, we estimate µ = x = 2.06 and σ = s = 0.465.
c)
〈µ〉1−α =
(x+ kα/2
σ√n, x+ k(1−α/2)
σ√n
)〈µ〉0.95 =
(x+ k0.025
σ√n, x+ k0.975
σ√n
)=
(2.06− Φ(0.975)
0.465√10
, x+ Φ(0.975)0.465√
10
)= (1.94, 2.18).
Problem 6.13
We know n = 10.
a)
Estimates of the sample mean and sample standard devation are given by
x =1
n
10∑i=1
xi
= 37.4
and
s2 =1
n− 1
(10∑i=1
xi2 − nx2
)
=1
10− 1
(14136− 10 · 37.42
)= 16.489 =⇒ s = 4.06.
b)
We perform a one-sided hypothesis test with
H0 : µ = µ0
HA : µ > µ0
where µ0 = 35 mpg. The test statistic is
t =x− µ0
s/√n
=37.4− 35
4.06/√
10
= 1.87.
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From table A-3 we find t0.98,9 = 2.4. Because t < t0.98,9, the test statistic is outside the region of rejection,so the null hypothesis is accepted.
c)
〈µ〉1−α =
(x+ tα
2 ,n−1s√n, x+ t(1− alpha
2 ),n−1
s√n
)〈µ〉0.95 =
(x+ t0.025,9
s√n, x+ t0.975,9
s√n
)=
(37.4− 2.26
4.06√10, x+ 2.26
4.06√10
)= (34.5, 40.3).
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