How can it be that mathematics being after all a product of human thought independent of experience is so admirably adapted to the objects of reality Albert Einstein Some parts of these slides were prepared based on ID: 781074
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Slide1
Slide2Basic Probability Distributions
How
can it be that mathematics, being after all a product of human thought independent of experience, is so admirably adapted to the objects of realityAlbert Einstein
Some parts of these slides were prepared based on
Essentials of Modern
Busines
Statistics, Anderson et al.
2012, Cengage
.
Managing Business Process Flow, Anupindi et al. 2012, Pearson.
Project Management in Practice,
Meredith et al. 2014, Wiley
Slide3Continuous Probability Distributions
Slide4Continuous Probability Distributions
Uniform
Normal
Exponential
Slide5Exponential and Poisson Relationship
Slide6Probability Distribution
The
probability distribution for a random variable describes the probabilities associated with the values of the random variable
.A probability distribution can be represented by a table, a graph, or a function (equation).
f(x):
Probability distribution function (pdf).
f(x)
represents the probability associated with each
value
(or the range of values) of
the random variable
.
Two requirements for all pdfs:
Slide7Continuous Probability Distribution
A
continuous random variable can assume any value over a real interval.
It is not possible to talk about the probability of the random variable assuming a particular
value. Why?
How many values exist between 0 and 1?
In continuous distributions the probability of the random variable to be equal to a specific value is?
0
In continuous distributions
,
we talk about the probability of the random variable
as
P(x ≤X1),
P(x ≥
X1), P(X1
≤
x
≤
X1).
Probability? The area under the pdf
Slide8
a
b
a
b
Continuous RV. The Area under the pdf
a
b
a
b
x
1
x
2
x
1
x
1
P(X1
≤
x
≤
X2)
P(x
≤
X1)
P(x
≥ X
1)
P(x
≥
X1)=
1- P(x
<
X1)
x
2
x
1
P(X1
≤
x
≤
X2)
Slide9Uniform Probability Distribution
Watch
the following repository lecture on youtube
Uniform Random Variables
Slide10Uniform Probability Distribution
a
= lowest value
b
=
highest value
f (x) = 1/(b – a) for a < x < b
= 0 elsewhere
A random variable is
uniformly distributed
whenever the probability is proportional to the interval’s length.
Var(x) = (b - a)2/12
E(x) = (a + b)/2
f
(
x
)
x
1
/(b-a)
a
b
Slide11Example: Costco Gas Station
Sampling
suggests that the amount of gas the consumers put in their car in a Costco gas station is uniformly
distributed between 10 gallons and 20 gallons.
f
(
x
)
x
1/10
Gas Volume
10
15
20
f(x
) = 1/10 for
10
< x <
20
= 0 elsewhere
E(
x
) = (
a
+
b
)/2
=
(10
+
20)/
2
=
15
Var(
x
) = (
b
-
a
)
2
/12
=
(20
–
10)
2
/12
= 8.33
x
= volume of gas filled
Slide12What if I have forgotten the formulas?
x = 10+10Rand()
Generate 1000
random values from U(10,20)
Compute AVERAGE and STDEV.S
E(
x
) = (
a
+
b
)/2
=
(10
+
20)/
2
=
15
Var(
x
) = (
b
-
a
)
2
/12
=
(20
–
10)
2
/12
= 8.33
Slide13f
(
x
)
x
1/10
Gas Volume
10
15
20
P(12
<
x
<
20)
= 1/10(3) = .3
What
is the probability that a
customer will
take between
17
and
20 gallons of gas?
Uniform Probability Distribution
17
Slide14Uniform Probability Distribution
P(13
<
x
<
17)
= ?
x
f
(
x
)
10
20
17
1/10
13
P(13
<
x
<
17)
= (1/10)(
17-13)
=
0.4
x
f
(
x
)
10
20
17
1/10
P(0
<
x
<
17)
= ?
P(0
<
x
<
17)
=
P(10
<
x
<
17)=
= (1/10)(
17-10)
=
0.7
P(15
<
x
<
22
) = ?
P(15
<
x
<
22
) =
P(15
<
x
<
20)=
= (1/10
)(20-15)
=
0.5
Slide15Most computer languages include a function that can be used to generate random numbers. In Excel, the RAND() function can be used to generate random numbers between 0 and 1. If we let x denote a random number generated using RAND(), then x is a continuous random variable
w
ith the following probability density function.
f(x) = 1 for 0 ≤
x ≤
1
f(x) =
0, elsewhere
a
) Graph the density function.
Uniform Distribution Problem 4
x
1
0
1
b) Compute P( .25 ≤ x
≤
.75)
c
)
Compute
P( x
≤
.3)
d)
Compute P( x
> .6)
e
) Generate 100 uniform random variables between 20 and 80 using RAND() function.
f) Compute Mean and StdDev for part (e)
Slide16U-Distribution- Random Problem Generator
Suppose
this is the probability distribution of the sales price
of a piece
of antique in
thousand dollars
. What price (in
thousand
dollars) do you offer to maximize
the probability of
getting this
antique.
Slide17A Non-Trivial Problem – Curve, Solver, Data table
x = price offered.
Probability of wining = 0.2(x-4)
Profit = (12-x)
E(Profit) = 0.2(x-4)(12-x)
E(profit) = -0.2x2+3.2x-9.6
Slide18The News Vendor Problem
Swell Productions is sponsoring an outdoor conclave for owners of collectible and classic Fords. The concession stand in the T-Bird area will sell clothing such as official Thunderbird racing jerseys. Suppose the probability of jerseys sales quantities is uniformly (and continuously) distributed between 100 and
400. Suppose sales price is $80 per jersey, purchase cost is $40, and unsold jerseys are returned to the manufacturer for $20 per unit. How many Jerseys Swell Production orders?
100
4
00
Q
Cu: Underage Cost
Cu = 80-40 =
40
Co: Overage
Cost
White /(White + Gray)
White/(
White+White
)
Cu/(
Cu+Cp
) = 40/(40+40)
= 0.5
40
40
40
20
Co
= 40-20 = 20
If Co was also 40
40/(20+60
) =2/3
Slide19The News Vendor Problem
100
4
00
Q
SL* = Cu/(Cu+Co)
SL* = 40/(40+20) = 2/3
(Q-100)/(400-100) = 2/3
Q= 300
Slide20The expected number of participants in a conference is uniformly distributed between 100 and 700. The participants spend one night in the hotel and the cost is paid by the conference. The hotel has offered a rate of $200 per room if a block of rooms is reserved (non-refundable) in advance. The rate in the conference day is
$
300. All rooms will be single occupied. How many rooms should we reserve in the non-refundable block to minimize our expected total cost.
100
7
00
The News Vendor Problem
Service level (Probability of demand not exceeding what we have ordered) SL* = Cu/(Cu+Co)
Co: Overage cost
Co = 200.
Cu: Underage cost
Cu = 300-200 = 100
Slide21The News Vendor Problem
a
=100
b
=700
?
B-a
=600
1/600
0.3333
SL* = Cu/(Cu+Co)
SL* = 100/(100+200) = 1/3
SL* = (Q-a)/(b-a) = (Q-100)/600 = 1/3
Q= 300
Slide22Simulation of Project Management Network
URV Generation
x= a+(b-a)Rand()
x= 20+(60-40)Rand()
https://
youtu.be/wqjGsLsadOo
Slide23Central Limit Theorem
Given
certain conditions, the arithmetic mean of a sufficiently large number of
independent
random variables, each with
a
well-defined expected
value
and well-defined
variance
, will be approximately normally distributed, regardless of the underlying distribution
The distribution of each of the activity was uniform. Summation of them moves towards normal distribution.
Slide24Simulation of Project Management Network
Slide25Simulation of Project Management Network
Slide26Simulation of Project Management Network
Slide27Simulation of Project Management Network