of University California at Berkeley Department of Civil & Environmental Engineering
Structural Engineering Mechanics & Materials Instructor: Armen Der Kiureghian
Student name:__________________________
CE93 -- Engineering Data Analysis First Midterm Examination Wednesday, October 8, 2003 Work on all three problems. Write clearly and state any assumptions you make. The exam is closed books and notes. The problems have the following weights:
Problem 1 (35 points) ________________________ Problem 2 (30 points) ________________________ Problem 3 (35 points) ________________________
Exam grade (100 points) ______________________
Student name:__________________________________
Problem 1. (10+15+10= 35 points) Biochemical oxygen demand at 40 stations along a river measured in milligrams/liter are listed in the ascending order in the following table. 2.27 2.46 2.49 2.51 2.70 2.73 2.76 2.79 2.82 2.86 2.89 2.93 2.96 3.00 3.04 3.08 3.13 3.16 3.17 3.18 3.19 3.22 3.23 3.24 3.30 3.36 3.37 3.43 3.58 3.66 3.74 3.75 3.83 3.92 4.00 4.03 4.08 4.41 4.64 4.95
a) Determine the median and the first and third quartiles of the data and the inter-quartile range, IQR. b) Sketch a box plot of the data on the attached graph sheet. Clearly identify the key elements of the plot c) Are there any outliers? If so, identify the data point(s).
Solution: a) x 0. 50 = (3.18 + 3.19) / 2 = 3.185 x 0. 25 = ( 2.86 + 2.89) / 2 = 2.875 x 0. 75 = (3.66 + 3.74) / 2 = 3.700 IQR = 3.700 - 2.875 = 0.825 milligrams/liter median and second quartile milligrams/liter first quartile milligrams/liter third quartile milligrams/liter inter-quartile range
b)
x 0. 25 - 1.5 IQR = 2.875 - 1.5 0.825 = 1.638 milligrams/liter x 0. 75 + 1.5 IQR = 3.700 + 1.5 0.825 = 4.938 milligrams/liter Lower whisker ends at 2.27 (smallest data value within 1.5IQR below the 1st quartile). Upper whisker ends at 4.64 (largest data value within 1.5IQR above the 3rd quartile). See attached page for the box plot. There is one outlier, which is the data point 4.95.
a)
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Student name:__________________________________
4.95 outlier 4.64
3.700 3rd quartile
3.185 median 2.875 1st quartile
2.27
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Student name:__________________________________
Problem 2. (15+15 = 30 points) A contractor is estimating the probability of completing a construction job on time through winter months. She estimates the probability of on-time completion to be 95% if the weather remains good (G), 70% if it rains (R), and 50% if it snows (S). Long time forecast the suggests probabilities of these weather conditions to be 0.4, 0.4 and 0.2, respectively. a) compute the probability of on-time completion of the construction job. b) Suppose you are told that the job was not completed on time. Determine the probability that it snowed.
Solution:
Let C = one-time completion of construction job. We are given:
P(C | G ) = 0.95, P(C | R ) = 0.70, P(C | S ) = 0.50 P(G ) = 0.4, P ( R) = 0.4, P ( S ) = 0 .2
a)
P(C ) = P( C | G) P( G) + P( C | R) P( R ) + P(C | S ) P( S ) = ( 0.95)(0.4) + (0.70)( 0.4) + ( 0.50)( 0.20) = 0.76
b)
P( S | C ) =
P(C | S ) P( S ) P(C ) 1 - P (C | S ) = P( S ) 1 - P( C ) 1 - 0.50 = (0.20) 1 - 0.76 = 0. 417
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Student name:__________________________________ Problem 3. (15+5+10+5 = 35 points) A soil engineer estimates the depth X to the bedrock below the foundation of a building to be somewhere between 20 and 30 meters, but more likely towards the higher values. On this basis, he assigns the triangular probability density function shown below to express his uncertainty about the depth.
f X (x )
20
30
x (meters)
a) Determine the mean, the median, the mode, the standard deviation, and the coefficient of variation (c.o.v.) of X. b) What do you guess is the skewness coefficient of the distribution? (Do not make any calculations.) c) Compute and plot the cumulative distribution of X. d) What is the probability that the depth will be between 25 and 27 meters?
Solution: a) Determine the expression for PDF with unit area underneath: f X ( x ) = 0.02( x - 20) for 20 < x < 30 = 0 elsewhere
x3 ( 30) 3 ( 20) 3 X = 0.02x( x - 20)dx = 0.02 - 10 x 2 = 0.02 - 10(30) 2 - + 10(20) 2 3 3 3 20 20 = 26.667m mean
30 30
Find point that halves the area:
1 x - 20 2 ( x 0.50 - 20) = 0.5 ( x0.50 - 20) = 50 2 10 / 0.2 x 0. 50 = 20 + 50 = 27.071m median ~ = 30m mode x
x 4 20 3 (30) 4 20(30) 3 ( 20) 4 20(20) 3 E[ X ] = 0.02 x ( x - 20)dx = 0.02 - x = 0.02 - + 4 4 - 3 20 3 4 3 20
30 2 2 30
= 716.667m 2
mean - square
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Student name:__________________________________ X = 716.667 - 26.667 2 = 2.357 m standard deviation
X = 2.357 = 0.0884 26.667
c.o.v.
b) The distribution is strongly skewed to the left. My guess is that the skewness coefficient is around - 2.
c)
FX ( x) = 0.02( x - 20) dx = 0.01( x - 20) 2
20
x
20 < x < 30
= 1 30 x
FX (x)
zero slope 1.0
20
30
x (meters)
d) P( 25 < X < 27) = FX (27 ) - FX ( 25) = 0.01( 27 - 20) 2 - 0.01( 25 - 20) 2 = 0.24
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