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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