Statistic and Probabilities in Hydrology

22.1 Introduction

2.2 Statistical Parameters

2.3 Theoretical Probability Distribution

2.4 Frequency Analysis

32.5 Graphic Method Using Probability Paper

2.6 Confidence Limits

2.7 Regression and Correlation

2.8 Multivariate Linear Regression and Correlation

4**2.9 Analysis of Time Series

**2.10 Dependence Models

2.11 Chi-square Test of Goodness of Fit

52.1 Introduction

:

62.1 Introduction

: , ., , .,.

72.2 Statistical ParametersMean

for a grouped series with N data points Arithmetic Mean

Geometric Mean

1

1 Nav i

iX X

N = =

( )1/1 2 3... Ngm nX X X X X=

82.2 Statistical ParametersMean

for a grouped series with N data points Harmonic Mean

Weighted Mean1 1 2 3

1 1 1 1 1...hm N

i i n

N NX

X X X X X=

= = + + + +

1 1 2 2

1 2

......

n nw

n

w X w X w XXw w w+ + += + + +

92.2 Statistical ParametersExample :

Station A B C D E F G H IRainfall(cm) 51.8 32.0 28.7 43.4 38.6 50.5 59.6 31.5 31.7

Area(sq. km) 31.0 58.0 31.5 31.0 86.0 71.0 27.0 43.5 66.0

Arithmetic Mean51.8 32.0 28.7 ... 31.7 367.8 40.87

9 9i

av

XX

N+ + + += = = =

Geometric Mean( ) ( )1/ 1/91 2 3... 51.8 32.0 ... 31.7 39.6Ngm nX X X X X= = =

10


Harmonic Mean

1 1 2 3

9 9 38.421 1 1 0.2342251 1 1 1 1 ...... 51.8 32 31.7

hm N

i i n

N NX

X X X X X=

= = = = = + + ++ + + +

Weighted Mean

1 1 2 2 9 9

1 2 9

... 51.8 31 32 58 ... 31.7 66 17688 39.75... 31 58 ... 66 445w

w X w X w XXw w w+ + + + + + = = = =+ + + + + +

11

2.2 Statistical ParametersMedian

, . , .

Mode , . , .

( 1/ 2): nMe X +=( / 2) ( / 2 1):

2n nX XMe +

+=

12

2.2 Statistical ParametersStandard Deviation

, .

( )21

1

n

ii

x x

n =

= ( )2

1

N

ii

x

N

=

=

13


Coefficient of Variation, C.V., , , C.V..

. .C Vx=

2

14

2.2 Statistical ParametersCoefficient of Skewness

( )33 1 i avX XN =

( )( ) ( )2

33 1 2 i av

N X XN N

= 33. .C S

=

15

2.2 Statistical ParametersCoefficient of Skewness

0sC

Coefficient of Kurtosis

18

2.3 Theoretical Probability Distribution

Biominal Distribution 1. n2. , ,

3. , P4. x

( ) ( )1 n xn xxf x C P P =

19

2.3.1.1 Biominal DistributionExample, 20%. .

(1) ?

(2) ?

5 0 50(0) (0.2) (0.8) 0.328p C= =

5 1 41(1) (0.2) (0.8) 0.41p C= =

5, 0.2, 0.8n p q= = =

20

2.3.1.1 Biominal DistributionExample, 5.

(1) ?

(2) ?

10, 0.2, 0.8n p q= = =10 0 100(0) (0.2) (0.8) 0.107p C= =

10 2 82(0) (0.2) (0.8) 0.302p C= =

21

Poisson Distribution 1. n 2. p 3. p n ( finite )

( ) 0 0,1,2,...

!

xef x xx

= > =

22

2.3.1.2 Poisson DistributionExample

80, 300 mm 0.02.

(1) 300 mm ?

( ) ( )3 0.2

0.0210

0.02 10 0.20

0.23 0.0011

3!

pnnp

ep

=== = =

= =

23

2.3.2 Continuous DistributionNormal Distribution

1. ,2. 3.,,

( )

= 2)(

21exp

21

xxf x

24

2.3.2 Continuous DistributionLog-Normal Distribution z = log x

2

ln121( ) 0,

2

z

z

x

z avf x e x zx

= > =

25

2.3.2 Continuous DistributionExtreme Value Distribution

1. 2.

3.

26

2.4 Frequency Analysis

4.5.

1.2.3.

27

(i) Frequency factor

Dr. Ven Te Chow, 1951()

(ii)Plotting position 2.5

28

Return periodTx

1( )

TP X x

=

X ()x (200cms)

29

( ) 11 1PT

=

1( )P X xT

=

n

n

11n

T

11 1n

T RiskRisk

30

Q. 40()(Risk)RR=5%T

11 1n

RT

= 4010.05 1 1 780T year

T = =

Answer

31

(i)Frequency factor

(XT)

T av

T

X x x

kX =

= av: T

::

TXx

x

:k frenquncy factor

32

Frequency factorkTXT

TX k = +k

1.2.3.k-TTk

33

2.4.1 Gumbels Distribution

( Extreme-value type I distribution )

[ ] ( )( )

( ) ( ) ( )( )

ln ln ( ) ln ln 1

ln ln ln 1

ln ln ln 1

6 6 ln ln ln 1

6 ln ln1

6 0.57721 ln ln1

T

a x F x Pc

X a c T T

c c T T where a c

T T

TT

TT

+ = = = = =

= = + = +

34

6 0.57721 ln ln 1TTX

T

= + TX k = +

{ }6 0.57721 ln ln

1

0.45005 0.7797 ln ln ( 1)

TkT

T T

= + = +

35

2.4.2 Pearson Type-III Distribution

(Three parameters gamma distribution)

(skewed data)

10 01( ) exp

( )

0

x x x xf x

x

=

36

TXT1.Cs ()2.NN100

1+8.5/N* Cs3.k-Ttables 2.8a & 2.8bk4.ChowsXT

TX k = +

37

2.4.3 Log-Pearson Type-III Distribution

x10

10logx x

37

38

1.2.yyCs ()3.NN1001+8.5/N* Cs

4.k-Ttables 2 .8a & 2.8b k5.ChowsYT

10logy x=

10 TYTX =

TXT

39

2.4.3 Normal Distribution

Cs 02.8a 2.8bk Cs 0

Cs 0(2.30)

tx2.6k

xt x t = = +

40

2.4.3 Log Normal Distribution

Cs 02.8a 2.8bk Cs 0

xy=ln(x)xx

lnx x

ln( )1 1( ) exp22

y

yy

xf x

x

=

41

10 TYTX =k 2.9

ChowsYT

k( )

2

2

0.5

exp 2 1

1

y y ykk

e =

CvCs 33S V VC C C= +

42

Example 2.8

Q.T50100

Table 2.10 annual maximum runoff recordsYear 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959

Runoff(mm) 133 94.5 76 87.5 92.7 71.3 77.3 85.1 122.8 69.4

Year 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969

Year 1970 1971 1972 1973 1974Runoff(mm) 91 106.8 102.2 87 84

Runoff(mm) 81 94.5 86.3 68.6 82.5 90.7 99.8 74.4 66.6 65

43

Ans. Example 2.8

For the seriesFor the series

( )

( )

1/ 2 1/ 2

3 3

2170 86.8 25

5156.7 = 14.6

Mean

Standard deviatio 6 1 25 1

14.66 0.1689

n

Coefficient of variation

Coefficien

86.81 t of s

1kewness

iav

i av

v

s

xx mm

n

x xmm

n

C

NCN N

= = = =

= =

= =

( ) ( )

( ) ( )

3

3

2

1 25 = 41793.7 0.55314.66 25 1 25 2

i avx x

=

44

Example 2.8

For the log transferred seriesFor the log transferred series

( ) ( )

1/ 2

3

Mean

Standard deviation

Coefficient of

48.381 1.9327 25

0.124358 0.072 25 1

0.072 0.03721.9327

1 25 0.

variation

Coefficient of skewn 001762 =0.072 25 1 25

ess2

av

v

s

y mm

mm

C

C

= = = = = =

0.197 =

45

Example 2.8

Extreme Value Type-IExtreme Value Type-I

2.7 T50100k

50

50

100

3.088 and 3.729

86.8 3.088 14.66 132.086.8 3.729 14.66 141.

'

5

T av

100

From Chow s relation X x kk k

X mmX mm

= =

= + == + =

= +

46

Example 2.8

Pearson Type-IIIPearson Type-III

Cs=0.5532.8(a) Cs=0.553T50100k

50

50

100

2.315 and 2.70

86.8 3.088 2.315 120.786.8 3.729 2.70 126.

'

4

T av

100

From Chow s relation X x kk k

X mmX mm

= =

= + == + =

= +

47

Example 2.8

Log Pearson Type-IIILog Pearson Type-III

Cs=0.1972.8(a) Cs=0.197T50100k

50

50

1002.0875005

502.110535

100

2.15 and 2.70

1.932737 0.071983 2.15 2.08750051.932737 0.071983 2.70 2.110535

1

'

22.3129.0

T av

100

From Chow s relk k

Y mmY mmX e mmX

ation Y

e

k

m

y

m

== =

= + == + =

= == =

+

48

Example 2.9Q. 280 m3/sec40

m3/sec 10400 m3/sec

Given thatGiven that

xav=280 m3/sec40 m3/sec

XT=400 m3/sec

TP

49

Example 2.9Assume Gumbel distribution for data.

XTxavk

400 = 280 40 k k3

From Chows relationFrom Chows relation

0.45005+0.7797 ln ln1

TkT

=

( ) ( )4.42483 0.45-0.7797 ln ln

1

exp exp 0.01197 1.0120491

84

TT

T eT

T year

= = = =

=

For Gumbel For Gumbel

50

Example 2.9

The Probability of the eventThe Probability of the event

P = 1 / T = 1/84 = 0.0119

occurring in next 10 yearsoccurring in next 10 years10 101 11 1 1 1 0.1128 11.3%

84T = = =

51

Graphical Method Using Probability Paper2.5

Plotting position

1. 2.

1950 2851951 150

52

3. m

14. x

P[Xxm]m/(N+1) N( )

1950 2851951 1501952 501953 121954 4451955 135

xm1954 4451950 2851951 1501955 1351952 501953 12 6

54321

P [Xxm]1 0.14287 =2 0.28577 =3 0.42887 =4 0.57147 =5 0.71437 =6 0.85717 =

53

5. TT 1/P (N+1)/m

515

[ ]1[ ] 0.25

m

m

if T

P X x

P X x

= = = =

121953501952

1351955150195128519504451954

xm

654321

P [Xxm]1 0.14287 =2 0.28577 =3 0.42887 =4 0.57147 =5 0.71437 =6 0.85717 =

285445

19501954

21 1 0.14287 =

2 0.28577 =

5X294.6 cms

54

6. PX

7. (Frequency curve)

55

2.5.1 Construction of Probability Paper

kT

1. ()2. ()

58

2.5.2 Selection of Type of Distribution

(Extreme value distribution)

Extreme value Type-III distribution

Log normal distribution

Exponential distribution

59

2.6 Confidence Limits

Confidence Limits

TU T

TL T

TU T TL

X X S xX X S xX X X

= + = > >

1

2 1/ 2(1 1.3 1.1 ) :

nwhere x an

a k kk

== + +

60

S

Confidence limit(%)

50 68 80 90 95 99

Value of S 0.674 1.00 1.282 1.645 1.96 2.58

61

Regression and Correlation 2.7

EX

(regression line)

y = f (x)

(coefficient of correlation)

62

2.7.1 Graphical Method

63

2.7.2 Analytical Method

( Method of Least squares ) xy

y = a + bx

Se

( )( )

2

01

2

1

SN

e i eii

N

i ii

y y

y a bx

=

=

=

=

64

y = a + bx

Se

2

2

0

0

i i

i i

i i i

i i i

y Na b x

y Na b x

xy a x b x

xy a x b x

== + == +

( )21

S 0N

e i ii

y a bx=

= =

( )( )

2

22

22

i i i i i i i

i i

i i i i

i i

y x x x y y b xa

NN x x

N x y x yb

N x x

= =

=

65

2.7.3 Correlation

(correlation coefficient , )

66

( ),,

i i

x y

Cov x y =

( ) ( ) ( )

( )( ) ( )

1

2 2 2 2

2 22 2

1 ,

=

=

N

i i i x i yi

i i x y

i x i y

i i i i

i i i i

Cov x y x yN

x y N

x N y N

x y x y N

x x N y y N

=

=

( )yi y i xx

y x =

67

2.7.4 Significance of Parameters

abFor testing b

For testing a

2(1 )r

rSN=

1/ 2

2

( ) ( 2)(1 )

b Nt rb r =

1/ 2

2 2 2

( ) ( 2)(1 )( )n av

a Nt rb r x

= +

68

2.7.5 Standard Form of Bivariate Equations

LinearExponentialParabolaHigher order equationOther forms of equations:

y a bx= +axy be=by ax=

21 2 3 1

nny a a x a x a x+= + + + +K

( ) ( )a xby a y yx b x a bx

= + = =+ +

69

Example 2.13Q. 2.14

y=a+bx

where y is runoff and x is rainfall of July, respectively

see Table2.15

Assume a linear relationAssume a linear relation

( )( )

2

22

22

94.95

0.704

i i i i i

i i

i i i i

i i

y x x x ya

N x x

N x y x yb

N x x

= =

= =

70

Example 2.13

y=-94.95+0.704x =0.855

Test for

= 0.081

the value of is significantly different from zero

2(1 )

rrSN=

3 0.855 3 0.081 1.0983 0.855 3 0.081 0.6124 0.855 4 0.081 1.1794 0.855 4 0.081 0.531

r

r

r

r

SSSS

+ = + = + = = ++ = + = + = = +

71

Example 2.13

Assume a non-linear relationAssume a non-linear relation y=axb

log => log(y) = log(a)+blog(x)Y = A+BX log(a) = -1.839 => a = 0.0145

b = 1.563 see Table2.16

y=0.0145x1.563

72

2.8 Multivariate Linear Regression and Correlation

x1x2xx =b0 + b1x1 + b2x2

=> 0 1 1 2 22

1 0 1 1 1 2 1 2

22 0 2 1 1 2 2 2

x Nb b x b x

xx b x b x b x x

xx b x b x x b x

= + += + += + +

:

73

2.11 Chi-Square Test of Goodness of Fit

(chi-square test)

2

222

2

1 1

( )i ei ii iei ei

Q P ZP P

= =

= =

74

vOiPei

75

Example 2.19Q. 2.3 2

Mean = 74.27

Variance = {5(45-74.27)2++2(115-74.27)2} / 55 = 315.85

Standard Deviation = (315.85)0.5 = 17.78

zi of col. (6) (for row 4)=

For zi = 0.329, F(xi) = 0.5 + 0.129 = 0.629 and so on

80 74.27 0.32917.78

x = =

76

The chi-square test statistics is calculated in col. (9)

For example, 0.141 in row(4) is

For degrees of freedom of (8-2-1) = 5,

22 {55(0.2 0.224) } 0.141

0.224 = =

20.95 11.1 =

Since 4.459

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Statistic and Probabilities in Hydrology