8/19/2019 Hydrology Statistical Exam Grad 2014 Mid Term Sol

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100

2100

4100

6100

8100

10100

-3.1000 -2.1000 -1.1000 -0.1000 0.9000 1.9000 2.90000.999   0.5 0.4 0.3   0.20.60.950.99 0.050.10.70.9 0.8 0.0010.01

   F   l  o  w

   (  c   f  s   )

Excedance Probability

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8/19/2019 Hydrology Statistical Exam Grad 2014 Mid Term Sol

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100

1000

10000

100000

-3.1000 -2.1000 -1.1000 -0.1000 0.9000 1.9000 2.90000.999   0.5 0.4 0.3   0.20.60.950.99 0.050.10.70.9 0.8 0.0010.01

   F   l  o  w

   (  c   f  s   )

Excedance Probability

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8/19/2019 Hydrology Statistical Exam Grad 2014 Mid Term Sol

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c) The log-normal probability distribution (shown in 1b) seems most appropriate because it plots

rather cleanly along the indicated line. In contrast, the normal probability distribution shows areasonable amount of curvature (not a straight line).

d) Since the data appear to be log-normally distributed, I will use the sample statistics reported in 1b to

estimate the population curve as:

Return Period Probability K Q in log scale Q in ordinary scale

2 0.5 0 3.156338238 1433.303754

5 0.8 0.841621 3.399794083 2510.695723

10 0.9 1.281552 3.527052767 3365.524579

50 0.98 2.053749 3.750426357 5628.936598

100 0.99 2.326348 3.829281082 6749.647336

K

K

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8/19/2019 Hydrology Statistical Exam Grad 2014 Mid Term Sol

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Problem 2:

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