CVE 472 3. NORMAL DISTRIBUTIONusers.metu.edu.tr/bertug/CVE472/CVE472 Statistical...CVE 472...

CVE 472 Statistical Techniques in Hydrology. 1/47

Assist. Prof. Dr. Bertuğ Akıntuğ

Civil Engineering ProgramMiddle East Technical University

Northern Cyprus Campus

CVE 472CVE 472

3. NORMAL DISTRIBUTION3. NORMAL DISTRIBUTION



Outline

General Normal Distribution

Per-productive Properties

Standard Normal Distribution

Central Limit Theorem

Constructing Normal Curves for Data



The Normal Distribution

‘Bell Shaped’SymmetricalMean, Median and Mode

are EqualLocation is determined by the mean, µSpread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + ∞ to − ∞ Mean = Median= Mode

X

px(x)

µ

σ



By varying the parameters µ and σ, we obtain different normal distributions


Many Normal Distributions




X

f(X)

µ

σ

Changing µ shifts the distribution left or right.

Changing σ increases or decreases the spread.

(continued)

The Normal Distribution Shape




Small standard deviation

Large standard deviation

The Normal Distribution Shape




The formula for the normal probability density function is

Where e = the mathematical constant approximated by 2.71828π = the mathematical constant approximated by 3.14159µ = the population meanσ = the population standard deviationX = any value of the continuous variable

2µ)/σ](1/2)[(Xx e

2π1(x)p −−=

σ

The Normal Probability Density Function



Outline








Perproductive Properties

If Xi for i=1, 2, … n are independently and normally distributed with mean, µ and variance, σ2

X ~ N(µ,σ2)

Then Yi = a + bXi is normally distributed with

Y ~ N(a+bµ, b2σ2)

Solve Example 5.1 p.86



The Sum of Two Independent Normal Random Variables

The sum of two independent Normal random variables is also normally distributed.

If X1 ~ N(µ1, σ12) and X2 ~ N(µ2, σ2

2) are independent random variables, then

( )22

212121 σσ,µµN~XXY +++=



Linear Combinations of Independent Normal Random Variables

The two results presented so far can be synthesized into the following general result.If Xi ~ N(µi, σi

2), 1≤ i ≤ n, are independent random variables and if ai, 1≤ i ≤ n, and b are constants, then

whereµ = a + b1 µ 1 + … + bn µ n

and σ2 = b1

2 σ12 + … + bn

2 σn2

( )2nn11 σµ,N~XbXbaY +++= K



Outline








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



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:

σµXZ −

=

The Z distribution always has mean = 0 and standard deviation = 1



The Standardized Normal Probability Density Function

The formula for the standardized normal probability density function is

Where e = the mathematical constant approximated by 2.71828π = the mathematical constant approximated by 3.14159z = any value of the standardized normal distribution

2(1/2)zz e

2π1(z)p −=



The StandardizedNormal Distribution

Also known as the “Z” distributionMean is 0Standard Deviation is 1

z

gz (z)

0

1

Values above the mean have positive Z-values, values below the mean have negative Z-values



Example

If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is

This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.

2.050

100200σ

µXZ =−

=−

=



Comparing X and Z units

Z100

2.00200 X

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)

(µ = 100, σ = 50)

(µ = 0, σ = 1)



Finding Normal Probabilities

Probability is the area under thecurve!

a b X

f(X) P a X b( )≤

Probability is measured by the area under the curve

≤P a X b( )<<=

(Note that the probability of any individual value is zero)



f(X)

Xµ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)XP( =∞<<−∞

0.5)XP(µ =∞<<0.5µ)XP( =<<−∞



Empirical Rules

µ ± 1σ encloses about 68% of X’s

f(X)

Xµ µ+1σµ-1σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%



The Empirical Rule

µ ± 2σ covers about 95% of X’s

µ ± 3σ covers about 99.7% of X’s

xµ2σ 2σ

xµ3σ 3σ

95.44% 99.73%

(continued)



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)

Z0 2.00

0.9772Example:P(Z < 2.00) = 0.9772



The Standardized Normal Table

The value within the table gives the probability from Z = − ∞ up to the desired Z value.9772

2.0P(Z < 2.00) = 0.9772

The row shows the value of Z to the first decimal point

The column gives the value of Z to the second decimal point

2.0

...

(continued)

Z 0.00 0.01 0.02 …

0.0

0.1



General Procedure for Finding Probabilities

Draw the normal curve for the problem in terms of X

Translate X-values to Z-values

Use the Standardized Normal Table

To find P(a < X < b) when X is distributed normally:




Suppose X is normal with mean 8.0 and standard deviation 5.0Find P(X < 8.6)

X

8.68.0



Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)

Z0.120X8.68

µ = 8σ = 10

µ = 0σ = 1

(continued)


0.125.0

8.08.6σ

µXZ =−

=−

=

P(X < 8.6) P(Z < 0.12)



Z

0.12

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

Solution: Finding P(Z < 0.12)

.5478.02

0.1 .5478

Standardized Normal Probability Table (Portion)

0.00

= P(Z < 0.12)P(X < 8.6)



Upper Tail Probabilities

Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(X > 8.6)

X

8.68.0



Now Find P(X > 8.6)…(continued)

Z

0.120

Z

0.12

0.5478

0

1.000 1.0 - 0.5478 = 0.4522

P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)

= 1.0 - 0.5478 = 0.4522

Upper Tail Probabilities



Probability Between Two Values

Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6)

P(8 < X < 8.6)

= P(0 < Z < 0.12)

Z0.120X8.68

05

88σ

µXZ =−

=−

=

0.125

88.6σ

µXZ =−

=−

=

Calculate Z-values:



Z

0.12

Solution: Finding P(0 < Z < 0.12)

0.0478

0.00

= P(0 < Z < 0.12)P(8 < X < 8.6)

= P(Z < 0.12) – P(Z ≤ 0)= 0.5478 - .5000 = 0.0478

0.5000

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.02

0.1 .5478




Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(7.4 < X < 8)

X

7.48.0

Probabilities in the Lower Tail



Probabilities in the Lower Tail

Now Find P(7.4 < X < 8)…

X7.4 8.0

P(7.4 < X < 8)

= P(-0.12 < Z < 0)

= P(Z < 0) – P(Z ≤ -0.12)

= 0.5000 - 0.4522 = 0.0478

(continued)

0.0478

0.4522

Z-0.12 0

The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)



Steps to find the X value for a known probability:1. Find the Z value for the known probability2. Convert to X units using the formula:

Finding the X value for a Known Probability

ZσµX +=



Finding the X value for a Known Probability

Example: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

X? 8.0

0.2000

Z? 0

(continued)



Find the Z value for 20% in the Lower Tail

20% area in the lower tail is consistent with a Z value of -0.84

Z .03

-0.9 .1762 .1736

.2033

-0.7 .2327 .2296

.04

-0.8 .2005


.05

.1711

.1977

.2266

…

…

…

…X? 8.0

0.2000

Z-0.84 0

1. Find the Z value for the known probability



2. Convert to X units using the formula:

Finding the X value

80.3

0.5)84.0(0.8

ZσµX

=

−+=

+=

So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80




Solve Example 5.3 p.87

Example 5.4 p.88

Example 5.5 p.88



Outline








Averaging Independent Normal Random Variables

If Xi ~ N(µ, σ2), 1≤ i ≤ n, are independent random variables, then their average is distributed X

nσµ,N~X

2




Outline








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.




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()




ExampleData: Kentucky River peak flows 1895-1960Xmin= 20600 cfsXmax= 115000 cfsµ = sample mean = 67509 cfsσ = sample standard deviation = 20952 cfsTo find expected relative frequency for class mark 25000 with 10000 class interval (class limits 20000 – 30000) using EXCEL:f25000 =normdist(30000, µ, σ,TRUE)-normdist(20000, µ, σ,TRUE)

= 0.0250MATLAB:f25000 =normcdf(30000, µ, σ)-normcdf(20000, µ, σ)

= 0.0250



Distribution Comparison

Comparison of normal distribution with the observed distribution of peak flows in the Kentucky River.

2 3 4 5 6 7 8 9 10 11 12

x 104

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

PEAK FLOW (cfs)

RE

LATI

VE

FR

EQ

UE

NC

Y

Normal DistributionObserved Distribution



Excersises

Page 94 and 95:5.1 (a) and (b) only5.25.35.65.75.85.145.18

CVE 472 3. NORMAL DISTRIBUTIONusers.metu.edu.tr/bertug/CVE472/CVE472 Statistical...CVE 472...

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