CVE 472 Assist. Prof. Dr. Bertuğ Akıntuğ Civil Engineering Program Middle East Technical University Northern Cyprus Campus CVE 472 Statistical Techniques in Hydrology. 1/47
Outline General Normal Distribution Per-productive Properties Standard Normal Distribution Central Limit Theorem Constructing Normal Curves for Data CVE 472 Statistical Techniques in Hydrology. 2/47
The Normal Distribution Bell Shaped Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, µ Spread is determined by the standard deviation, σ p x (x) The random variable has an infinite theoretical range: + to Mean = Median= Mode µ σ X CVE 472 Statistical Techniques in Hydrology. 3/47
The Normal Distribution Many Normal Distributions By varying the parameters µ and σ, we obtain different normal distributions CVE 472 Statistical Techniques in Hydrology. 4/47
The Normal Distribution The Normal Distribution Shape (continued) f(x) Changing µ shifts the distribution left or right. σ Changing σ increases or decreases the spread. µ X CVE 472 Statistical Techniques in Hydrology. 5/47
The Normal Distribution The Normal Distribution Shape Small standard deviation Large standard deviation CVE 472 Statistical Techniques in Hydrology. 6/47
The Normal Distribution The Normal Probability Density Function The formula for the normal probability density function is p x (x) = 1 e 2πσ (1/2)[(X µ)/σ] 2 Where e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 µ = the population mean σ = the population standard deviation X = any value of the continuous variable CVE 472 Statistical Techniques in Hydrology. 7/47
Outline General Normal Distribution Per-productive Properties Standard Normal Distribution Central Limit Theorem Constructing Normal Curves for Data CVE 472 Statistical Techniques in Hydrology. 8/47
Perproductive Properties If X i for i=1, 2, n are independently and normally distributed with mean, µ and variance, σ 2 X ~ N(µ,σ 2 ) Then Y i = a + bx i is normally distributed with Y ~ N(a+bµ, b 2 σ 2 ) Solve Example 5.1 p.86 CVE 472 Statistical Techniques in Hydrology. 9/47
The Sum of Two Independent Normal Random Variables The sum of two independent Normal random variables is also normally distributed. If X 1 ~ N(µ 1, σ 12 ) and X 2 ~ N(µ 2, σ 22 ) are independent random variables, then ( 2 2 µ + µ,σ ) Y = X + 1 + X2 ~ N 1 2 1 σ2 CVE 472 Statistical Techniques in Hydrology. 10/47
Linear Combinations of Independent Normal Random Variables The two results presented so far can be synthesized into the following general result. If X i ~ N(µ i, σ i2 ), 1 i n, are independent random variables and if a i, 1 i n, and b are constants, then where and Y = a + b K + µ= a + b 1 µ 1 + + b n µ n σ 2 = b 12 σ 12 + + b n2 σ 2 n ( 2 ) 1X1 + bn Xn ~ N µ, σ CVE 472 Statistical Techniques in Hydrology. 11/47
Outline General Normal Distribution Per-productive Properties Standard Normal Distribution Central Limit Theorem Constructing Normal Curves for Data CVE 472 Statistical Techniques in Hydrology. 12/47
The Standardized Normal Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z) Need to transform X units into Z units CVE 472 Statistical Techniques in Hydrology. 13/47
Translation to the Standardized Normal Distribution Translate from X to the standardized normal (the Z distribution) by subtracting the mean of X and dividing by its standard deviation: Z = X σ µ The Z distribution always has mean = 0 and standard deviation = 1 CVE 472 Statistical Techniques in Hydrology. 14/47
The Standardized Normal Probability Density Function The formula for the standardized normal probability density function is p z (z) = 1 2π e (1/2)z 2 Where e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 z = any value of the standardized normal distribution CVE 472 Statistical Techniques in Hydrology. 15/47
The Standardized Normal Distribution Also known as the Z distribution Mean is 0 Standard Deviation is 1 g z (z) 1 0 z Values above the mean have positive Z-values, values below the mean have negative Z-values CVE 472 Statistical Techniques in Hydrology. 16/47
Example If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is X µ 200 100 Z = = = σ 50 2.0 This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. CVE 472 Statistical Techniques in Hydrology. 17/47
Comparing X and Z units 100 0 200 X 2.0 Z (µ = 100, σ = 50) (µ = 0, σ = 1) Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z) CVE 472 Statistical Techniques in Hydrology. 18/47
Finding Normal Probabilities Probability is the area under the curve! Probability is measured by the area under the curve f(x) P ( a X b) = P ( a < X < b) (Note that the probability of any individual value is zero) a b X CVE 472 Statistical Techniques in Hydrology. 19/47
Probability as Area Under the Curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(x) P( < X < µ) = 0.5 P(µ < X < ) = 0.5 0.5 0.5 µ P( < X < ) = 1.0 X CVE 472 Statistical Techniques in Hydrology. 20/47
Empirical Rules What can we say about the distribution of values around the mean? There are some general rules: f(x) σ σ µ±1σ encloses about 68% of X s µ-1σ µ µ+1σ 68.26% X CVE 472 Statistical Techniques in Hydrology. 21/47
The Empirical Rule (continued) µ±2σ covers about 95% of X s µ±3σ covers about 99.7% of X s 2σ µ 2σ x 3σ µ 3σ x 95.44% 99.73% CVE 472 Statistical Techniques in Hydrology. 22/47
The Standardized Normal Table The Cumulative Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value for Z (i.e., from negative infinity to Z) Example: P(Z < 2.00) = 0.9772 0.9772 0 2.00 Z CVE 472 Statistical Techniques in Hydrology. 23/47
The Standardized Normal Table The column gives the value of Z to the second decimal point Z 0.00 0.01 0.02 (continued) The row shows the value of Z to the first decimal point 0.0 0.1. 2.0.9772 The value within the table gives the probability from Z = up to the desired Z value P(Z < 2.00) = 0.9772 2.0 CVE 472 Statistical Techniques in Hydrology. 24/47
General Procedure for Finding Probabilities To find P(a < X < b) when X is distributed normally: Draw the normal curve for the problem in terms of X Translate X-values to Z-values Use the Standardized Normal Table CVE 472 Statistical Techniques in Hydrology. 25/47
Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0 Find P(X < 8.6) 8.0 X CVE 472 Statistical Techniques in Hydrology. 26/47 8.6
Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6) (continued) X µ 8.6 8.0 Z = = = σ 5.0 0.12 µ= 8 σ = 10 µ = 0 σ = 1 8 8.6 X 0 0.12 Z P(X < 8.6) P(Z < 0.12) CVE 472 Statistical Techniques in Hydrology. 27/47
Solution: Finding P(Z < 0.12) Standardized Normal Probability Table (Portion) Z.00.01.02 P(X < 8.6) = P(Z < 0.12).5478 0.0.5000.5040.5080 0.1.5398.5438.5478 0.2.5793.5832.5871 0.3.6179.6217.6255 0.00 0.12 Z CVE 472 Statistical Techniques in Hydrology. 28/47
Upper Tail Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(X > 8.6) 8.0 X CVE 472 Statistical Techniques in Hydrology. 29/47 8.6
Upper Tail Probabilities Now Find P(X > 8.6) P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z 0.12) = 1.0-0.5478 = 0.4522 (continued) 1.000 0.5478 1.0-0.5478 = 0.4522 0 Z 0 Z 0.12 0.12 CVE 472 Statistical Techniques in Hydrology. 30/47
Probability Between Two Values Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6) Calculate Z-values: X µ 8 8 Z = = = 0 σ 5 X µ 8.6 8 Z = = = 0.12 σ 5 8 8.6 0 0.12 P(8 < X < 8.6) = P(0 < Z < 0.12) X Z CVE 472 Statistical Techniques in Hydrology. 31/47
Solution: Finding P(0 < Z < 0.12) Standardized Normal Probability Table (Portion) Z.00.01.02 P(8 < X < 8.6) = P(0 < Z < 0.12) = P(Z < 0.12) P(Z 0) = 0.5478 -.5000 = 0.0478 0.0.5000.5040.5080 0.1.5398.5438.5478 0.2.5793.5832.5871 0.5000 0.0478 0.3.6179.6217.6255 Z 0.00 0.12 CVE 472 Statistical Techniques in Hydrology. 32/47
Probabilities in the Lower Tail Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(7.4 < X < 8) 7.4 8.0 X CVE 472 Statistical Techniques in Hydrology. 33/47
Probabilities in the Lower Tail (continued) Now Find P(7.4 < X < 8) P(7.4 < X < 8) = P(-0.12 < Z < 0) = P(Z < 0) P(Z -0.12) = 0.5000-0.4522 = 0.0478 0.4522 0.0478 The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12) 7.4 8.0-0.12 0 X Z CVE 472 Statistical Techniques in Hydrology. 34/47
Finding the X value for a Known Probability Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: X = µ + Zσ CVE 472 Statistical Techniques in Hydrology. 35/47
Example: Finding the X value for a Known Probability (continued) Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X 0.2000? 8.0? 0 CVE 472 Statistical Techniques in Hydrology. 36/47 X Z
Find the Z value for 20% in the Lower Tail 1. Find the Z value for the known probability Standardized Normal Probability Table (Portion) Z.03.04.05 20% area in the lower tail is consistent with a Z value of -0.84-0.9.1762.1736-0.8.2033.2005-0.7.2327.2296.1711.1977.2266 0.2000? 8.0-0.84 0 X Z CVE 472 Statistical Techniques in Hydrology. 37/47
Finding the X value 2. Convert to X units using the formula: X = µ + Zσ = 8.0 + ( 0.84)5.0 = 3.80 So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 CVE 472 Statistical Techniques in Hydrology. 38/47
Standard Normal Distribution Solve Example 5.3 p.87 Example 5.4 p.88 Example 5.5 p.88 CVE 472 Statistical Techniques in Hydrology. 39/47
Outline General Normal Distribution Per-productive Properties Standard Normal Distribution Central Limit Theorem Constructing Normal Curves for Data CVE 472 Statistical Techniques in Hydrology. 40/47
Averaging Independent Normal Random Variables If X i ~ N(µ, σ 2 ), 1 i n, are independent random variables, then their average is distributed σ X ~ N µ, n 2 X Central Limit Theorem CVE 472 Statistical Techniques in Hydrology. 41/47
Outline General Normal Distribution Per-productive Properties Standard Normal Distribution Central Limit Theorem Constructing Normal Curves for Data CVE 472 Statistical Techniques in Hydrology. 42/47
Constructing Normal Curves For Data Frequently the histogram of a set of observed data suggests that the data may be approximated by a normal distribution. One way to investigate the goodness of this approximation is by superimposing a normal curve on the frequency histogram and visually compare the two distributions. CVE 472 Statistical Techniques in Hydrology. 43/47
Constructing Normal Curves For Data This is possible by plotting both observed relative frequency and expected (according to normal distribution) relative frequency and visually compare two. In order to find expected relative frequency for any class interval, standard normal table or software such as EXCELL or MATLAB can be used. EXCEL normdist() MATLAB normcdf() CVE 472 Statistical Techniques in Hydrology. 44/47
Constructing Normal Curves For Data Example Data: Kentucky River peak flows 1895-1960 X min = 20600 cfs X max = 115000 cfs µ = sample mean = 67509 cfs σ = sample standard deviation = 20952 cfs To find expected relative frequency for class mark 25000 with 10000 class interval (class limits 20000 30000) using EXCEL: f 25000 =normdist(30000, µ, σ,true)-normdist(20000, µ, σ,true) = 0.0250 MATLAB: f 25000 =normcdf(30000, µ, σ)-normcdf(20000, µ, σ) = 0.0250 CVE 472 Statistical Techniques in Hydrology. 45/47
Distribution Comparison 0.18 0.16 Normal Distribution Observed Distribution 0.14 RELATIVE FREQUENCY 0.12 0.1 0.08 0.06 0.04 0.02 0 2 3 4 5 6 7 8 9 10 11 12 PEAK FLOW (cfs) x 10 4 Comparison of normal distribution with the observed distribution of peak flows in the Kentucky River. CVE 472 Statistical Techniques in Hydrology. 46/47
Excersises Page 94 and 95: 5.1 (a) and (b) only 5.2 5.3 5.6 5.7 5.8 5.14 5.18 CVE 472 Statistical Techniques in Hydrology. 47/47