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PROBABILITY AND STATISTICS
ENGR 351 Numerical Methods for Engineers
Southern Illinois University Carbondale
College of EngineeringDr. L.R. Chevalier
Copyright© 1999 by Lizette R. Chevalier
Permission is granted to students at Southern Illinois University at Carbondaleto make one copy of this material for use in the class ENGR 351, NumericalMethods for Engineers. No other permission is granted.
All other rights are reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means,electronic, mechanical, photocopying, recording, or otherwise, withoutthe prior written permission of the copyright owner.
Why statistics in a course on numerical methods?
• Inputs on models are typically variable and uncertain– media specific
• hydraulic conductivity
• organic carbon content
• wind speed
– receptor specific
• body weight
• ingestion rate
– chemical specific
• Henry’s law constant
• decay rate
• toxicity values
– landuse specific
• industrial
• commercial
Why statistics in a course on numerical methods?
• Inputs on models are typically variable and uncertain
• Best represented by probability distributions• Tools for estimating values • Quantifying uncertainty• Tool for risk assessment
Random Variables
• Discrete – variable can only attain a finite number of
values.
• Continuous – variable can be from zero to infinity.
Probability
Probability of Occurrencenumber of successestotal number of trials
mn
nmn
lim
The following properties are associated with probability.
1. Probability is a nonnegative number.
2. If an event is certain, then m and n are equal and p(x) = 1
3. Mutually exclusive
p(A+B) = p(A) + p(B)
In other words, the probability of A or B is the sum of the probability of either event. For example, the probability of 2 or 4 on the throw of a dice is:
p(2) + p(4) = 1/6 + 1/6 = 1/3
4. If events are independent
p(AB) = p(A) . p(B)
The occurrence of one does not affect the occurrence of the other. For example, the probability that 2 and 4 will occur in two dice simultaneously thrown is:
p(2,4) = p(2) . p(4) . 2
= (1/6) . (1/6) . (2)
= 1/18
Density FunctionThe density function is defined as the function which yields the probability that the random variable takes on any one of its values.
Density function for a continuous random variable
Prob
abil
ity
of O
ccur
renc
ep(
x) =
f(x
)
x
# on dice
f(x)
1/6
Density function for a discrete random variable
Cumulative Distribution Function, F(x)
F(x
)
x
Continuousfunction
Allows us to determinethe probabilitythat x is less thanor equal to a
For a continuous variable
F x a F a p x dx f x dx P x aa a
( ) ( ) ( ) ( ) ( )
For a discrete variableF x a p x p x p a( ) ( ) ( ) ( ) 1 2
F(x)
1
5/6
2/3
1/2
1/3
1/6
0 1 2 3 4 5 6
# on dice
Cumulative distance: Discrete function
f x dxa
( )0F x a can be as( ) : expressed
F x a p x
where x a
i
a
i
( ) ( )
which is the area underthe densityfunction
Prob
abil
ity
of O
ccur
renc
ep(
x) =
f(x
)
xa
Samples and Populations
• Sample- a random selection of items from a lot or population in order to evaluate the characteristics of the lot or population– mean– expected value– variance
Mean or Expected Value
E x xf x dx( ) ( ) .
E x x p xii
n
i( ) ( )1
continuous
discrete
In the case of a discrete sample, is thisthe mean?
Example:Expected value on dice
E x x p xii
( ) ( )
.
1
6
116
216
316
416
516
616
35
...end of problem
Example: Let x = # of hours of a light bulb. Find the expected life.
E x xx
dx( ) (,
)20 000
3100
f x xx
elsewhere( )
,
20 000100
0
3
E x xf x dx( ) ( ) .
Example (cont.)
E x xx
dx( ) (,
)20 000
200
3100
...end of problem
f x xx
elsewhere( )
,
20 000100
0
3
VarianceDescribes the “spread” or shape of thedistribution
0 1 2 3 41 2 3
2 2 2 E x x f xx
[( ) ] ( ) ( ) discrete
( ) ( )x f x dx 2continuous
The following equations describe the discrete and continuous cases for variance.
Example: Let x = number of orders received per day
Probability density function: Company A
x 1 2 3
f(x) 0.3 0.4 0.3
Probability density function: Company B
x 0 1 2 3 4
f(x) 0.2 0.1 0.3 0.3 0.1
Probability density function: Company A
E x( ) ( . ) ( . ) ( . )
.
1 0 3 2 0 4 3 0 3
2 0
x 1 2 3
f(x) 0.3 0.4 0.3
Probability density function: Company B
x 0 1 2 3 4
f(x) 0.2 0.1 0.3 0.3 0.1
E x( ) ( . ) ( . ) ( . ) ( . ) ( . )
.
0 0 2 1 01 2 0 3 3 0 3 4 01
2 0
VarianceCompany A
Company B
2 2 2 21 2 0 3 2 2 0 4 3 2 0 3
0 6
( ) ( . ) ( ) ( . ) ( ) ( . )
.
2 2 2 2
2 2
0 2 0 2 1 2 01 2 2 0 3
3 2 0 3 4 2 01
16
( ) ( . ) ( ) ( . ) ( ) ( . )
( ) ( . ) ( ) ( . )
.
0 1 2 3 41 2 3
...end of problem
2 2
( ) ( )x f x dx
Third momentmeasure of asymmetry
+ -
If symmetricS= 0
Often referred to as skewness, s
( ) ( )x f x dx
3
Fourth moment, measure of flatness
small kurtosis
large kurtosis
( ) ( )x f x dx
4
Monte Carlo Technique A technique for modeling processes that involve random variables.
You need:
• a random variable and its probability distribution• a sequence of random numbers
Suppose x is a random number that describes the demand per day of a commodity where:
f x
x
x
x
x
( )
.
.
.
.
01 0
0 2 1
0 4 2
0 3 3
Cumulative distribution function
F x
x
x
x
x
x
( )
.
.
.
.
0 0
0 1 0 1
0 3 1 2
0 7 2 3
10 3
The task is to generate values of x (demand) such that the relative frequency of each value of k will be equal to its probability.
Need a sequence of random numbers
Will use the “MIDSQUARE” technique
Monte Carlo Method
1. Take a 4 digit number (preferably selected at random)
2. Square the number
3. Take 4 digits starting at the third from the left.
4. Record
5. Square -------- etc
ie. Select 1653
(1653)2 = 53640976
Select 6409
1653, 6409, 0752, ......... 0 9999 x
ENGR 351 Numerical Methods for EngineersRandom numbers
Monte Carlo ExcelUsing the random number function =rand()
6922 47914084 0.6922 0.351035 0.506976 0.0623719140 83539600 0.914 0.61144 0.223697 0.7550615396 29116816 0.5396 0.492761 0.143406 0.6614361168 1364224 0.1168 0.964795 0.597764 0.1696093642 13264164 0.3642 0.269986 0.851376 0.9441912641 6974881 0.2641 0.718226 0.571283 0.1097749748 95023504 0.9748 0.401943 0.77973 0.305236
235 55225 0.0235 0.953922 0.934942 0.937367552 304704 0.0552 0.301322 0.279323 0.095127
3047 9284209 0.3047 0.543647 0.44348 0.4534432842 8076964 0.2842 0.365798 0.632637 0.476239
769 591361 0.0769 0.909206 0.338302 0.3541585913 34963569 0.5913 0.851988 0.67051 0.3984659635 92833225 0.9635 0.238579 0.286329 0.5363368332 69422224 0.8332 0.585925 0.248756 0.5658834222 17825284 0.4222 0.454564 0.244167 0.7452738252 68095504 0.8252 0.375883 0.179672 0.540868
955 912025 0.0955 0.749491 0.766341 0.0751059120 83174400 0.912 0.25066 0.450009 0.8184041744 3041536 0.1744 0.172349 0.745362 0.753676
415 172225 0.0415 0.111162 0.407399 0.0645571722 2965284 0.1722 0.461135 0.274751 0.2277849652 93161104 0.9652 0.361817 0.543096 0.2820721611 2595321 0.1611 0.06993 0.956483 0.4380415953 35438209 0.5953 0.282377 0.829095 0.211325
average 0.454512 0.473998 0.516195 0.439272standard deviation 0.343788 0.260592 0.251293 0.280533
Recall that we are studying the following system:Suppose x is a random number that describes the demand per day of a commodity where:
f x
x
x
x
x
( )
.
.
.
.
0 1 0
0 2 1
0 4 2
0 3 3
The “middle” digits are considered random. Since the study of demand requires 4 subsets.Using:
x Random0 0 - 999 (10%)1 1000 - 2999 (20%)2 3000 - 6999 (40%)3 7000 - 9999 (30%)
Number 7324 x = 3 6409 x = 2
x Random0 0 - 9991 1000 - 29992 3000 - 69993 7000 - 9999 Random Ct.
#9140 35396 21168 13642 2
Total: 8
Example: Generate numbers with class
Probability of hydraulic conductivity
v Kdhdx
Soil samples at site:
K(cm/s) f(x)
10-5(silty sand)0.410-4 0.3510-3(sand) 0.1510-2 0.0710-1(gravel) 0.03
dept
h
Ground surface
Soil samples at site:
K(cm/s) f(x) Random #
10-5(silty sand)0.4 0-399910-4 0.35 4000 - 749910-3(sand) 0.15 7500 - 899910-2 0.07 9000 - 969910-1(gravel) 0.03 9700 - 9999
Soil samples at site:
K(cm/s) f(x) Random #
10-5(silty sand) 0.4 0-399910-4 0.35 4000 - 749910-3(sand) 0.15 7500 - 899910-2 0.07 9000 - 969910-1(gravel) 0.03 9700 - 9999
Node Random # K (cm/s)(1,1)(2,1)(3,1)(4,1)(5,1)
end of lecture
