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Statistics & Flood Frequency Chapter 3 Dr. Philip B. Bedient Rice University 2006
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Predicting FLOODS
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Flood Frequency Analysis Statistical Methods to evaluate probability exceeding a particular outcome - P (X >20,000 cfs) = 10% Statistical Methods to evaluate probability exceeding a particular outcome - P (X >20,000 cfs) = 10% Used to determine return periods of rainfall or flows Used to determine return periods of rainfall or flows Used to determine specific frequency flows for floodplain mapping purposes (10, 25, 50, 100 yr) Used to determine specific frequency flows for floodplain mapping purposes (10, 25, 50, 100 yr) Used for datasets that have no obvious trends Used for datasets that have no obvious trends Used to statistically extend data sets Used to statistically extend data sets
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Random Variables Parameter that cannot be predicted with certainty Parameter that cannot be predicted with certainty Outcome of a random or uncertain process - flipping a coin or picking out a card from deck Outcome of a random or uncertain process - flipping a coin or picking out a card from deck Can be discrete or continuous Can be discrete or continuous Data are usually discrete or quantized Data are usually discrete or quantized Usually easier to apply continuous distribution to discrete data that has been organized into bins Usually easier to apply continuous distribution to discrete data that has been organized into bins
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Typical CDF Discrete Continuous F(x 1 ) = P(x < x 1 ) F(x 1 ) - F(x 2 )
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Frequency Histogram Probability that Q is 10,000 to 15, 000 = 17.3% Prob that Q < 20,000 = 1.3 + 17.3 + 36 = 54.6% 17.3 36 1.3 27 8 1.39
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Probability Distributions CDF is the most useful form for analysis
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Moments of a Distribution First Moment about the Origin Discrete Continuous
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Var(x) = Variance Second moment about mean
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Estimates of Moments from Data Std Dev. S x = (S x 2 ) 1/2
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Skewness Coefficient Used to evaluate high or low data points - flood or drought data
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Mean, Median, Mode Positive Skew moves mean to right Negative Skew moves mean to left Normal Dist’n has mean = median = mode Median has highest prob. of occurrence
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Skewed PDF - Long Right Tail
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01 '98 76 83
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Skewed Data
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Climate Change Data
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Siletz River Data Stationary Data Showing No Obvious Trends
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Data with Trends
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Frequency Histogram Probability that Q is 10,000 to 15, 000 = 17.3% Prob that Q < 20,000 = 1.3 + 17.3 + 36 = 54.6% 17.3 36 1.3 27 8 1.39
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Cumulative Histogram Probability that Q < 20,000 is 54.6 % Probability that Q > 25,000 is 19 %
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PDF - Gamma Dist
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Major Distributions Binomial - P (x successes in n trials) Exponential - decays rapidly to low probability - event arrival times Normal - Symmetric based on and Lognormal - Log data are normally dist’d Gamma - skewed distribution - hydro data Log Pearson III -skewed logs -recommended by the IAC on water data - most often used
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Binomial Distribution The probability of getting x successes followed by n-x failures is the product of prob of n independent events: p x (1-p) n-x This represents only one possible outcome. The number of ways of choosing x successes out of n events is the binomial coeff. The resulting distribution is the Binomial or B(n,p). Bin. Coeff for single success in 3 years = 3(2)(1) / 2(1) = 3
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Binomial Dist’n B(n,p)
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Risk and Reliability The probability of at least one success in n years, where the probability of success in any year is 1/T, is called the RISK. Prob success = p = 1/T and Prob failure = 1-p RISK = 1 - P(0) = 1 - Prob(no success in n years) = 1 - (1-p) n = 1 - (1 - 1/T) n Reliability = (1 - 1/T) n
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Design Periods vs RISK and Design Life Risk % 5102550100 754.17.718.536.672.6 507.714.936.672.6144.8 2022.945.3112.5224.6448.6 104895.4237.8475.1949.6 Expected Design Life (Years) x 2 x 3
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Risk Example What is the probability of at least one 50 yr flood in a 30 year mortgage period, where the probability of success in any year is 1/T = 1.50 = 0.02 RISK = 1 - (1 - 1/T) n = 1 - (1 - 0.02) 30 = 1 - (0.98) 30 = 0.455 or 46% If this is too large a risk, then increase design level to the 100 year where p = 0.01 RISK = 1 - (0.99) 30 = 0.26 or 26%
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Exponential Dist’n Poisson Process where k is average no. of events per time and 1/k is the average time between arrivals f(t) = k e - kt for t > 0 Traffic flow Flood arrivals Telephone calls
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Exponential Dist’n f(t) = k e - kt for t > 0 F(t) = 1 - e - kt Avg Time Between Events
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Gamma Dist’n n =1 n =2 n =3 Unit Hydrographs
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Parameters of Dist’n DistributionNormalxLogN Y =logx GammaxExpt Mean xxxx yyyynk1/k Variance xxxx yyyy nk 1/k 2 Skewnesszerozero 2/n 0.5 2
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Normal, LogN, LPIII Normal Data in bins
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Normal Prob Paper Normal Prob Paper converts the Normal CDF S curve into a straight line on a prob scale
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Normal Prob Paper Mean = 5200 cfs Std Dev = +1000 cfs Std Dev = –1000 cfs Place mean at F = 50% Place one S x at 15.9 and 84.1% Connect points with st. line Plot data with plotting position formula P = m/n+1
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Normal Dist’n Fit Mean
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Frequency Analysis of Peak Flow DataYearRank Ordered cfs 1940142,700 1925231,100 1932320,700 1966419,300 1969514,200 1982614,200 1988712,100 1995810,300 2000………….
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Frequency Analysis of Peak Flow Data Take Mean and Variance (S.D.) of ranked data Take Skewness C s of data (3rd moment about mean) If C s near zero, assume normal dist’n If C s large, convert Y = Log x - (Mean and Var of Y) Take Skewness of Log data - C s (Y) If C s near zero, then fits Lognormal If C s not zero, fit data to Log Pearson III
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Siletz River Example 75 data points - Excel Tools Mean20,4524.2921 Std Dev 60890.129 Skew0.7889 - 0.1565 Coef of Variation 0.2980.03 Original QY = Log Q
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Siletz River Example - Fit Normal and LogN Normal Distribution Q = Q m + z S Q Q 100 = 20452 + 2.326(6089) = 34,620 cfs Mean + z (S.D.) Where z = std normal variate - tables Log N Distribution Y = Y m + k S Y Y 100 = 4.29209 + 2.326(0.129) = 4.5923 k = freq factor and Q = 10 Y = 39,100 cfs
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Log Pearson Type III Log Pearson Type III Y = Y m + k S Y K is a function of Cs and Recurrence Interval Table 3.4 lists values for pos and neg skews For Cs = -0.15, thus K = 2.15 from Table 3.4 Y 100 = 4.29209 + 2.15(0.129) = 4.567 Q = 10 Y = 36,927 cfs for LP III Plot several points on Log Prob paper
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LogN Prob Paper for CDF What is the prob that flow exceeds some given value - 100 yr value Plot data with plotting position formula P = m/n+1, m = rank, n = # Log N dist’n plots as straight line
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LogN Plot of Siletz R. Straight Line Fits Data Well Mean
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Siletz River Flow Data Various Fits of CDFs LP3 has curvature LN is straight line
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Flow Duration Curves
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01 '98 98 92 Trends in data have to be removed before any Frequency Analysis
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