9- A levee will be constructed to provide flood protection from a river. The annual peak streamflow in the river follows a log-Pearson Type III distribution with a log10 (Q[ft³/s]) mean, standard deviation and skewness coefficient of 2.2, 0.92 and 0.4 respectively based on 70 years of observed streamflow. If the 100-year return period flow is 40,380 ft³/s, what is the upper bound of this design flow in ft³/s with a 5% chance of exceedance?
9- A levee will be constructed to provide flood protection from a river. The annual peak streamflow in the river follows a log-Pearson Type III distribution with a log10 (Q[ft³/s]) mean, standard deviation and skewness coefficient of 2.2, 0.92 and 0.4 respectively based on 70 years of observed streamflow. If the 100-year return period flow is 40,380 ft³/s, what is the upper bound of this design flow in ft³/s with a 5% chance of exceedance?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
The Answer = 102,557 cfs
![9-
A levee will be constructed to provide flood protection from a river.
The annual peak streamflow in the river follows a log-Pearson Type III distribution
with a log10 (Q[ft³/s]) mean, standard deviation and skewness coefficient of 2.2,
0.92 and 0.4 respectively based on 70 years of observed streamflow. If the 100-year
return period flow is 40,380 ft³/s, what is the upper bound of this design flow in
ft³/s with a 5% chance of exceedance?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf5aa2e5-f0b0-4839-a4ef-6f98d6dc8551%2Fa08b54c8-7482-4bf6-aa26-0278268b2326%2Fj9szw2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:9-
A levee will be constructed to provide flood protection from a river.
The annual peak streamflow in the river follows a log-Pearson Type III distribution
with a log10 (Q[ft³/s]) mean, standard deviation and skewness coefficient of 2.2,
0.92 and 0.4 respectively based on 70 years of observed streamflow. If the 100-year
return period flow is 40,380 ft³/s, what is the upper bound of this design flow in
ft³/s with a 5% chance of exceedance?
Expert Solution

Step 1
The standard normal variable(z) for a 5% chance of exceedance can be calculated as the standard normal distribution value corresponding to the exceedance probability:
z = NormalInv(0.05) = 0.44
Mean
Standard deviation,
Step by step
Solved in 2 steps

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