Download AsefCh01b(Probability)

Continuous Probability Distributions
Uniform Distribution
Exponential Distribution
b
a

Triangular Distribution
Normal Distribution
s
a
m
b

Continuous Probability Distributions
The probability of the continuous random
variable assuming a specific value is 0.
The probability of the random variable
assuming a value within a given interval from x1
to x2 is defined to be the area under the graph
of the probability density function between x1
and x2.
Uniform Probability Density Distribution
b
a
Uniform Probability Density Function
f (x) = 1/(b - a) for a < x < b
= 0 elsewhere
a = smallest value the variable can assume
b = largest value the variable can assume
µ = (a + b)/2
s 2 = (b - a)2/12
Continuous Probability Distributions
a
x1 x2
P(x1 ≤ x≤ x2)

b
a
x1

P(x≤ x1)
P(x≥ x1)= 1- P(x<x1)
a
x1

P(x≥ x1)
b
P(x =x1)= 0
b
Example
Slater customers are charged for the amount of salad
they take. Sampling suggests that the amount of salad
taken is uniformly distributed between 5 ounces and 15
ounces.
Probability Density Function
f (x ) = 1/10 for 5≤ x ≤ 15
= 0 elsewhere
where x = salad plate filling weight
Example
What is the probability that a customer will
take between 12 and 15 ounces of salad?
F (x )
P(12 < x < 15) = 1/10(3) = .3
1/10
x
5
10 12
15
Salad Weight (oz.)
The Uniform Probability Distribution
P(8 ≤ x ≤ 12) = ?
f (x )
1/10
5
8
12
15
x
P(8 ≤ x ≤ 12) = (1/10)(12-8) = .4
The Uniform Probability Distribution
P(0<x < 12) = ?
f (x )
1/10
5
12
15
P(0<x < 12) = P(5<x < 12)=
= (1/10)(12-5) = .7
x
Generate a Uniform Random Number
We want to select a student, randomly, to come
to the white board.
How?
Simulation is based on random number
generation
Crystal Ball is a simulation software
AutoMod is another simulation software
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Show Some AutoMod Examples
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Triangular Probability Distributions
a x1
m
b
a = smallest value the variable can assume
b = largest value the variable can assume
m = most likely value the variable can assume
Compute the height of this triangle
If a ≤ x1 ≤ m, compute the height associated with x1.
You can compute it. But you need time.
Triangular Probability Distributions
a
x1 m
b
It is equal to 2(x1-a)/[(m-a)(b-a)].
Compute P (x ≤ x1). It is equal to the area from a to x1
It is equal to (x1-a)2/[(m-a)(b-a)].
It is not difficult to compute the area if m ≤ x1 ≤ b,
Triangular Probability Distribution
Expected Value of x
µ = (a +m+ b)/3
Variance of x
s 2 = (a2 + m2 +b2 – ab- am-bm)/18
Exponential Probability Distribution

1 x / 
f ( x)  e

 = mean
s =
e = 2.71828182845904, the base of the natural logarithm.
Exponential Probability Distribution
1 x / 
f ( x)  e

F (x )
.4
.3
P(x < x1 ) = area from 0 to x1
.2
.1
x
1 2 3 4 5 6 7 8 9 10
P( x  x1 )  1  e
 x1 / 
Example
The time between arrivals of cars at Al’s Carwash follows an
exponential probability distribution with a mean time between
arrivals of 3 minutes. Al would like to know the probability that the
time between two successive arrivals will be 2 minutes or less.
F (x )
.4
.3
P(x < 2) = 1 - e-2/3 = 1 - .5134 = .4866
.2
.1
x
1
2
3
4
5 6
7
8
9 10
Generate an Exponential l Random
Variable
Parameters: Mean and StDev.
How can I generate an exponential random
variable?
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Normal Distribution
s
μ
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z
1-18
Normal Probability Density Function
1
(x μ)2 / 2 σ 2
f(x) 
e
2π σ
where

s

e
= mean
= standard deviation
= 3.14159
= 2.71828
Standard Normal Probability Density Function
where
 =0
s =1
 = 3.14159265358979
Normal and Standard Normal
A Normal Variable x is associated with a standard Normal
Variable z
x is Normal (Average x , Standard Deviation x)  z is Normal (0,1)
μ = Average x
σ = Standard Deviation of x
x
s
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is z  a Standard Normal variable.
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Standard Normal Table
s

x
0
z
x-scale
z-scale
There is a table for z which tells us
a) Given any probability of not exceeding z, what the value of z is
b) Given any value for z, what the probability of not exceeding z is
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Given z, Find the Probability
Table returns probability
0.05
z
Given z
1.6
z = 1.65
Up to
the first
digit
after
decimal
Second
digit
after
decimal
Probability
Given Probability, Find z
Given a 95% Probability
0.05
z
The table will give you z
Z = 1.65
1.6
Up to
the first
digit
after
decimal
Second
digit
after
decimal
Probability
The Standard Normal Distribution F(z)
z
F(z) = Prob( N(0,1) < z)
F(z)
0
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
0.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.9032
0.9192
0.9332
0.9452
0.9554
0.9641
0.9713
0.9772
0.9821
0.9861
0.9893
0.9918
0.9938
0.9953
0.9965
0.9974
0.9981
0.9987
0.9990
0.9993
0.9995
0.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.9049
0.9207
0.9345
0.9463
0.9564
0.9649
0.9719
0.9778
0.9826
0.9864
0.9896
0.9920
0.9940
0.9955
0.9966
0.9975
0.9982
0.9987
0.9991
0.9993
0.9995
0.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.9066
0.9222
0.9357
0.9474
0.9573
0.9656
0.9726
0.9783
0.9830
0.9868
0.9898
0.9922
0.9941
0.9956
0.9967
0.9976
0.9982
0.9987
0.9991
0.9994
0.9995
0.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.9082
0.9236
0.9370
0.9484
0.9582
0.9664
0.9732
0.9788
0.9834
0.9871
0.9901
0.9925
0.9943
0.9957
0.9968
0.9977
0.9983
0.9988
0.9991
0.9994
0.9996
0.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.9099
0.9251
0.9382
0.9495
0.9591
0.9671
0.9738
0.9793
0.9838
0.9875
0.9904
0.9927
0.9945
0.9959
0.9969
0.9977
0.9984
0.9988
0.9992
0.9994
0.9996
0.05
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.9115
0.9265
0.9394
0.9505
0.9599
0.9678
0.9744
0.9798
0.9842
0.9878
0.9906
0.9929
0.9946
0.9960
0.9970
0.9978
0.9984
0.9989
0.9992
0.9994
0.9996
0.06
0.5239
0.5636
0.6026
0.6406
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.9131
0.9279
0.9406
0.9515
0.9608
0.9686
0.9750
0.9803
0.9846
0.9881
0.9909
0.9931
0.9948
0.9961
0.9971
0.9979
0.9985
0.9989
0.9992
0.9994
0.9996
0.07
0.5279
0.5675
0.6064
0.6443
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.9147
0.9292
0.9418
0.9525
0.9616
0.9693
0.9756
0.9808
0.9850
0.9884
0.9911
0.9932
0.9949
0.9962
0.9972
0.9979
0.9985
0.9989
0.9992
0.9995
0.9996
0.08
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.9429
0.9535
0.9625
0.9699
0.9761
0.9812
0.9854
0.9887
0.9913
0.9934
0.9951
0.9963
0.9973
0.9980
0.9986
0.9990
0.9993
0.9995
0.9996
0.09
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
0.9441
0.9545
0.9633
0.9706
0.9767
0.9817
0.9857
0.9890
0.9916
0.9936
0.9952
0.9964
0.9974
0.9981
0.9986
0.9990
0.9993
0.9995
0.9997
Excel: Given z, Compute Probability
Excel: Given Probability, Compute z
Example
Pep Zone sells auto parts and supplies including multigrade motor oil. When the stock of this oil drops to
20 gallons, a replenishment order is placed.
The store manager is concerned that sales are being
lost due to stockouts while waiting for an order.
It has been determined that lead time demand is
normally distributed with a mean of 15 gallons and a
standard deviation of 6 gallons.
In Summary; we have a N (15, 6): A normal random
variable with mean of 15 and std of 6.
The manager would like to know the probability of a
stockout, P(x > 20).
Example
If the manager of Pep Zone wants the probability of a
stockout to be no more than .05, what should the
reorder point be?
Area = .05
Area = .95
0
z.05
Generate Normal Random Varaiable
Parameters: Mean and StDev.
How can I generate a standard Normal random
variable?
How can I generate a Normal random variable?
Central Limit Theorem
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A Random Variable = Summation of
Several Random Variables
We have a random variable y which is equal to
summation of n random variables x: N(x, σx)
y = x1+x2+x3+…….+xn
Find average and standard deviation of the random
variable y in terms of x, σx
Note: on the basis of CLT, if n is large, we do not need N, and other
distribution also leads to Normal distribution for y
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A Random Variable = Summation of
Several Random Variables
Mean (y) = Mean(x1)+ Mean(x2)+ ….. +Mean(xn)
y = mx1+…….xn
Var(y )= Var(x1)+Var(x2)+ …+Var(xn)
sy2 = sx12 +…….s2 xn
StdDev (y) = SQRT(Var(y))
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A Random Variable = Another Random
Variable Multiplied by a Constant
We have a random variable y which is equal to random
variables x: N(x, σx) multiplied by a constant k
y = kx
Find Average and Standard deviation of the random
variable y in terms of x, σx
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A Random Variable = Another Random
Variable Multiplied by a Constant
Mean (y) = kMean(x)
y = kmx
Var(y )= k2Var(x)
sy2 = k2sx2
StdDev (y) = SQRT(Var(y))
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Additional Assignment for next week
Problem 1. Initial investment at the end of year 0 is 100, cash
inflow at the end of year 1 and 2 each N(100,20).
Compute mean and standard deviation of the NPV.
Compute Probability of NPV ≥ 0.
2. A project has three sequential tasks. The first task follows
Exponential distribution with parameter of 5 days. The second task
follows Uniform distribution with parameters of 5 and 10 days, the
last task follows Normal distribution with parameters of 9 and 2
days. The numbers on the next page were generated using RAND()
function in excel. Starting from the first row and column, moving
through the first row first, and not using any given element twice,
generate a random instance of duration of this project.
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Example
0.1
0.5
0.7
0.9
0.8
0.1
0.3
0.2
0.4
0.1
0.4
0.4
0.0
0.2
0.6
By not using any element twice we mean not using a specific
element – in a specific row and column twice.
For example you can not use the number in row 1 and column 1
twice. Obviously you can use all the three 0.1s as soon as you
reach their corresponding box in the table.
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