Transcript ECON1003: Analysis of Economic Data - Ka

Lesson 4:
Discrete Probability
Distributions
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-1
Outline
Random variables and probability distribution
Features of univariate probability distribution
Features of bivariate probability distribution
Marginal distribution and Conditional distribution
Expectation and conditional expectation
Variance, Covariance and Correlation Coefficient
Binomial Probability Distribution
Hypergeometric Probability Distribution
Poisson Probability Distribution
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ECON1003: Analysis of Economic Data
Lesson4-2
Random Variables and probability
distribution
 A random variable is a numerical value determined by the
outcome of an experiment. A random variable is often
denoted by a capital letter, e.g., X or Y.
 A probability distribution is the listing of all possible
outcomes of an experiment and the corresponding
probability.
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Lesson4-3
Types of Probability Distributions
 A discrete probability distribution can assume only
certain outcomes (need not be finite) – for random
variables that take discrete values.
 The number of students in a class.
 The number of children in a family.
 A continuous probability distribution can assume an
infinite number of values within a given range – for
random variables that take continuous values.
 The distance students travel to class.
 The time it takes an executive to drive to work.
 The amount of money spent on your last haircut.
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Lesson4-4
Types of Probability Distributions
Probability distribution may be classified according to the
number of random variables it describes.
Number of random variables
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Joint distribution
1
Univariate probability distribution
2
Bivariate probability distribution
3
Trivariate probability distribution
…
…
n
Multivariate probability distribution
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Lesson4-5
Features of a Univariate Discrete
Distribution
 Let x1,…,xN be the list of all possible outcomes (N of them).
 The main features of a discrete probability distribution are:
 The probability of a particular outcome, P(xi), is
between 0 and 1.00.
 The sum of the probabilities of the various outcomes is
1.00. That is,
P(x1) + … + P(xN) = 1
 The outcomes are mutually exclusive. That is,
P(x1and x2) = 0 and
Outcome
Prob.
P(x1or x2) = P(x1)+ P(x2)
x1
P(x1)
Generally, for all i not equal to k.
x2
P(x2)
P(xi and xk) = 0.
…
…
P(xi or xk) = P(xi)+ P(xk)
xN
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P(xN)
Lesson4-6
Features of a Univariate Discrete
Distribution
Can the following be a probability distribution of a
random variable?
x
Prob.
x
Prob.
1
0.2
2
0.3
1
0.6
3
0.1
2
0.3
1
0.4
3
0.1
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event
Prob.
1 or 2
0.6
2 or 3
0.3
3 or 1
0.1
ECON1003: Analysis of Economic Data
Lesson4-7
EXAMPLE:
Univariate probability distribution
 Consider a random experiment in which a coin is tossed three
times. Let x be the number of heads. Let H represent the
outcome of a head and T the outcome of a tail.
 The possible outcomes for such an experiment will be:
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
 Thus the possible values of x (number of heads) are
x=0:
x=1:
x=2:
x=3:
TTT
TTH, THT, HTT
THH, HTH, HHT
HHH
P(x=0) =1/8
P(x=1) =3/8
P(x=2) =3/8
P(x=3) =1/8
If the
coin is
fair
 From the definition of a random variable, x as defined in this
experiment, is a random variable.
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Features of a Bivariate Discrete
Distribution
 If X and Y are discrete random variables, we may define their
joint probability function as PXY(xi,yi)
 Let (x1,…,xR) and (y1,…,yS) be the list of all possible outcomes for
X and Y respectively.
 The main features of a bivariate discrete probability distribution
are:
 The probability of a particular outcome, PXY(xi,yi) is between 0
and 1.
 The sum of the probabilities of the various outcomes is 1.00.
That is,
PXY(x1,y1) + PXY(x2,y1) +…+ PXY(xR,y1) + + … + PXY(xR,yS) = 1
 The outcomes are mutually exclusive. That is,
if xi not equal to xk, or yi not equal to yk
PXY((xi,yi) and (xk,yk)) = 0 and
PXY((xi,yi) or (xk,yk)) = PXY(xi,yi) + PXY(xk,yk)
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Example: Bivariate Discrete Distribution
X takes 3 possible values and Y takes 4 possible values.
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y1
y2
y3
y4
x1
P(x1,y1)
P(x1,y2)
P(x1,y3)
P(x1,y4)
x2
P(x2,y1)
P(x2,y2)
P(x2,y3)
P(x2,y4)
x3
P(x3,y1)
P(x3,y2)
P(x3,y3)
P(x3,y4)
ECON1003: Analysis of Economic Data
Lesson4-10
EXAMPLE: Bivariate distribution
The joint distribution of the movement of Hang Seng
Index (HSI) and weather is shown in the following table.
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
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ECON1003: Analysis of Economic Data
Lesson4-11
EXAMPLE: Bivariate distribution
The joint distribution of the movement of Hang Seng
Index (HSI) and weather is shown in the following table.
Rainy
Not Rainy
HSI falls
0
a
HSI rises
0
b
Totals
0
Totals
Suppose …..
PX|Y(x | y) = P(X = x | Y = y)=P(x,y)/P(y) if P(Y = y) > 0
PX|Y(x | y) =0
if P(Y = y) = 0
P(HSI falls|Rainy) = P(HSI falls, Rainy) / P(Rainy)= 0/0
Forcing P(HSI falls|Rainy) in the definition eliminates the
difficulty in interpreting 0/0.
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Lesson4-12
Marginal Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The marginal probability function of X.
PX(x) = yPXY(x, y) = PXY(x, y1) +PXY(x, y2) +…+ PXY(x, yn)
P(HSI falls)= P(HSI falls and rainy) + P(HSI falls and not rainy)
P(HSI rises)= P(HSI rises and rainy) + P(HSI rises and not rainy)
 The double sum
xyPXY(x, y)
= P(HSI falls and rainy) + P(HSI falls and not rainy)+
P(HSI rises and rainy) + P(HSI rises and not rainy)
= P(HSI falls)+P(HSI rises)= 1
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Marginal Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The marginal probability function of X.
yPXY(x, y) = PX(x).
 The marginal probability function of Y.
xPXY(x, y) = PY(y).
 The double sum
yxPXY(x, y) = 1
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Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The conditional probability function of X given Y:
PX|Y(x | y) = P(X = x | Y = y) = PXY(x,y)/PY(y) if P(Y = y) > 0
PX|Y(x | y) =0
if P(Y = y) = 0
Note that PX|Y(x | y) when P(Y = y) = 0 is undefined
using the top formula.
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Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 For each fixed y this is a probability function for X, i.e. the
conditional probability function is non-negative and
XPX|Y(x | y) = PX|Y(x1 | y)+ PX|Y(x2 | y)
= PX,Y(x1, y)/ PY(y) + PX,Y(x2, y)/ PY(y)
=[PX,Y(x1, y) + PX,Y(x2, y)]/ PY(y)
=1.
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ECON1003: Analysis of Economic Data
Lesson4-16
Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The conditional probability function of X given Y:
PX|Y(x | y) = P(X = x | Y = y) if P(Y = y) > 0
PX|Y(x | y) =0
if P(Y = y) = 0
 For each fixed y this is a probability function for X, i.e. the
conditional probability function is non-negative and
XPX|Y(x | y) = 1.
 By the definition of conditional probability,
PX|Y(x | y) = PX,Y(x, y)/ PY(y).
E.g., P(HSI rises| Rainy) = 0.2/0.35.
 When X and Y are independent,
PX|Y(x | y) is equal to PX(x).
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-17
Example: Conditional Distributions
Y
X HSI falls
HSI rises
Totals
Rainy
Not Rainy
Totals
0.15
0.4
0.55
0.15/0.35
0.4/0.65
0.2
0.25
0.45
0.2/0.35
0.25/0.65
0.35
0.65
1.0
1.0
1.0
0.4/0.55
1.0
0.25/0.45
1.0
P(Y|HSI falls) 0.15/0.55
P(Y|HSI rises)
PX|Y(x | y) = PX,Y(x, y)/ PY(y).
0.2/0.45
P(X|Rainy) P(X| Not Rainy)
PY|X(y | x) = PX,Y(x, y)/ PX(x).
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Lesson4-18
Transformation of Random variables
 A transformation of random variable(s) results in a new
random variable.
 For example, if X and Y are random variables, the following
are also random variables:
 Z=2X
 Z=3+2X
 Z=X2
 Z=log(X)
 Z=X+Y
 Z=X2+Y2
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The Expectation (mean) of a Discrete
Probability Distribution
 The expectation (mean):
 reports the central location of the data.
 is the long-run average value of the random variable.
That is, the average of the outcomes of many
experiments.
 is also referred to as its expected value, E(X), in a
probability distribution.
 Is also known as first moment of a random variable.
 is a weighted average.
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Moments of a random variable
The n-th moment is defined as the expectation of
the n-th power of a random variable: E(Xn)
E(X)
First moment
E(X2)
Second moment
The n-th centralized moment is defined as:
E[X-E(X)]n
E(X-m)2
Second centralized moment
E(X-m)3
Third centralized moment
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The Expectation (Mean) of Discrete
Probability Distribution
 For univariate probability distribution, the expectation or
mean E(X) is computed by the formula:
E(x)  Σ[xP(x)]
 x 1P(x1 )  x 2P(x 2 )  ...  x nP(x n )
 For bivariate probability distribution, the the expectation
or mean E(X) is computed by the formula:
E(x)  Σ Y Σ X [xPX, Y (x, y)]
 x 1PX, Y (x 1 , y1 )  x 2PX, Y (x 2 , y1 )  ...  x nPX, Y (x n , y1 )
 ...
 x1PX, Y (x 1 , y n )  x 2PX, Y (x 2 , y n )  ...  x nPX, Y (x n , y n )
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Conditional Mean of Bivariate Discrete
Probability Distribution
 For bivariate probability distribution, the conditional
expectation or conditional mean E(X|Y) is computed by
the formula:
E(X | Y  y i )  Σ X [xPX|Y (x | y i )]
 x 1PX|Y (x 1 | y i )  x 2PX|Y (x 2 | y i )  ...  x nPX|Y (x n | y i )
 Unconditional expectation or mean of X, E(X)
E(X)  Σ YE(X | Y  y i )PY (y i )
 E[E(X | Y)]
 E[μX ]
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Expectation of a linear transformed
random variable
 If a and b are constants and X is a random variable, then
E(a) = a
E(bX) = bE(X)
E(a+bX) = a+bE(X)
E(a  bx)  Σ[(a  bx)P(a  bx)]
 Σ[(a  bx)P(x)]
 (a  bx1 )P(x 1 )  (a  bx 2 )P(x 2 )  ...  (a  bx n )P(x n )
 aP(x1 )  bx1P(x1 )  aP(x 2 )  bx 2P(x 2 )  ...  aP(x n )  bx nP(x n )
 a[P(x 1 )  P(x 2 )  ...  P(x n )]  b[x1P(x1 )  x 2P(x 2 )  ...  x nP(x n ) ]
 a  bE(x)
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Lesson4-24
The Variance of a Discrete
Probability Distribution
 The variance measures the amount of spread (variation) of
a distribution.
 The variance of a discrete distribution is denoted by the
Greek letter 2 (sigma squared).
 The standard deviation is the square root of  2.
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Lesson4-25
The Variance of a Discrete
Probability Distribution
 For univariate discrete probability distribution
V( X )  E[(X  μ)2 ]
 Σ[(x  μ)2 P(x)]
 (x 1  μ)2 P(x1 )  (x 2  μ)2 P(x 2 )  ...  (x n  μ)2 P(x n )
 For bivariate discrete probability distribution
V( X )  E[(X  μX )2 ]
 Σ Y Σ X [(x - μX )2 PX, Y (x, y)]
 (x 1 - μX )2 PX, Y (x 1 , y1 )  ( x 2 - μX )2 PX, Y (x 2 , y1 )  ...  ( x n - μX )2 PX, Y (x n , y1 )
 ...
 (x 1 - μX )2 PX, Y (x 1 , y n )  ( x 2 - μX )2 PX, Y (x 2 , y n )  ...  ( x n - μX )2 PX, Y (x n , y n )
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ECON1003: Analysis of Economic Data
Lesson4-26
Variance of a linear transformed random
variable
 If a and b are constants and X is a random variable, then
V(a) = 0
V(bX) = b2V(X)
V(a+bX) = b2V(X)
V( a  bX )  E[ a  bX  ( a  bμ ) ]2
 E[ bX  bμ ]2
 E[ b(X  μ) ]2
 E[ b 2 (X  μ) 2 ]
 b 2 E[ (X  μ) 2 ]
 b 2 V(X)
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Lesson4-27
The Covariance of a Bivariate Discrete
Probability Distribution
Covariance measures how two random variables co-vary.
C( X ,Y )  E[(X  μX )(Y  μ Y )]
 Σ Y Σ X [(x - μX )(Y  μ Y )PX,Y (x, y)]
C( X ,Y )  E[(X  μX )(Y  μ Y )]
 E[XY  μX Y  μ Y X  μXμ Y ]
 E[XY]  μXE[Y]  μ YE[X]  μXμ Y
 E[XY]  μXμ Y  μ Y μX  μXμ Y
 E[XY]  μXμ Y
 E[XY]  E[X]E[Y]
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Lesson4-28
Covariance of linear transformed random
variables
 If a and b are constants and X is a random variable, then
C(a,b) = 0
C(a,bX) = 0
C(a+bX,Y) = bC(X,Y)
C(a  bX, Y )  E[ a  bX  (a  bμX ) ](Y  μ Y )
 E[ bX  bμX ](Y  μ Y )
 E[ b(X  μX ) ](Y  μ Y )
 b E(X  μ)(Y  μ Y )
 bC(X, Y)
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Lesson4-29
Variance of a sum of random variables
 If a and b are constants and X and Y are random variables,
then
V(X+Y) = V(X) + V(Y) + 2C(X,Y)
V(aX+bY) =a2V(X) + b2V(Y) + 2abC(X,Y)
V( X  Y )  E[ X  Y  (μX  μ Y ) ]2
 E[ (X  μX )  (Y  μ Y )]2
 E[ (X  μX ) 2  (Y  μ Y )2  2(X  μX )(Y  μ Y )]
 E[ (X  μX ) 2 ]  E[(Y  μ Y )2  2E[(X  μX )(Y  μ Y )]
 V ( X )  V (Y )  2C(X, Y)
V(aX  bY )  E[ aX  bY  (aμX  bμ Y ) ]2
 E[ (aX  aμX )  (bY  bμ Y )]2
 E[ a 2 (X  μX ) 2  b 2 (Y  μ Y )2  2(aX  aμX )(bY  bμ Y )]
 a 2 E[ (X  μX ) 2 ]  b 2 E[(Y  μ Y )2  2abE[(X  μX )(Y  μ Y )]
 a 2V ( X )  b 2V (Y )  2abC(X, Y)
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ECON1003: Analysis of Economic Data
Lesson4-30
Correlation coefficient
 The strength of the dependence between X and Y is
measured by the correlation coefficient:
C(X, Y)
Corr(X, Y) 
V(X)V(Y)
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ECON1003: Analysis of Economic Data
Lesson4-31
EXAMPLE
 Dan Desch, owner of College Painters, studied his records
for the past 20 weeks and reports the following number of
houses painted per week:
Number of houses painted, x
Weeks
Probability, P(x)
10
5
.25
11
6
.30
12
7
.35
13
2
.10
Total
20
1.00
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ECON1003: Analysis of Economic Data
Lesson4-32
EXAMPLE
x
P(x)
10
11
12
13
Total
.25
.30
.35
.10
1.00
 Compute the mean and variance of the
number of houses painted per week and:
μ  E(x)  Σ[xP(x)]
 (10)(.25)  (11)(.30)  (12)(.35)  (13)(.10)
 11.3
σ 2  Σ[(x  μ) 2 P(x)]
 (10  11.3)2 (.25)  ...  (13  11.3)2 (.10)
 0.4225  0.0270  0.1715  0.2890
 0.91
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ECON1003: Analysis of Economic Data
Lesson4-33
Binomial Probability Distribution
 The binomial distribution has the following characteristics:
 An outcome of an experiment is classified into one of two
mutually exclusive categories, such as a success or failure.
 The data collected are the results of counts in a series of trials.
 The probability of success stays the same for each trial.
 The trials are independent.
 For example, tossing an unfair coin three times.
 H is labeled success and T is labeled failure.
 The data collected are number of H in the three tosses.
 The probability of H stays the same for each toss.
 The results of the tosses are independent.
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ECON1003: Analysis of Economic Data
Lesson4-34
Binomial Probability Distribution
 To construct a binomial distribution, let
 n be the number of trials
 x be the number of observed successes
  be the probability of success on each trial
 The formula for the binomial probability distribution is:
P(x) = nCx  x(1- )n-x
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Lesson4-35
The density functions of binomial distributions
with n=20 and different success rates p
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Lesson4-36
EXAMPLE
x = number of patients who will experience
nausea following treatment with Phe-Mycin
n = 4 , p = 0.1 , q = 1 – p = 1 - 0.1 = 0.9
Find the probability that 2 of the 4 patients
treated will experience nausea.
4!
p(2)  P(x=2)=
(0.1)2 (0.9)4-2 =6(0.1)2 (0.9)2 =0.0486
2!(4-2)!
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ECON1003: Analysis of Economic Data
Lesson4-37
Binomial Probability Distribution
 The formula for the binomial probability distribution is:
P(x) = nCx  x(1- )n-x
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
n!
n Cr 
r!(n  r)!
 X=number of heads
 The coin is fair, i.e., P(head) = 1/2.
 P(x=0) = 3C0 0.5 0(1- 0.5)3-0 =3!/(0!3!) (1) (1/8)=1/8
 P(x=1) = 3C1 0.5 1(1- 0.5)3-1 =3!/(1!2!) (1) (1/8)= 3/8
 P(x=2) = 3C2 0.5 2(1- 0.5)3-2 =3!/(2!1!) (1) (1/8)= 3/8
 P(x=3) = 3C3 0.5 3(1- 0.5)3-3 =3!/(3!0!) (1) (1/8)= 1/8
When the coin is not fair, simple counting rule will not work.
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ECON1003: Analysis of Economic Data
Lesson4-38
Mean & Variance of the Binomial
Distribution
 The mean is found by:
m  n
 The variance is found by:
  n (1  )
2
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ECON1003: Analysis of Economic Data
Lesson4-39
EXAMPLE
The Alabama Department of Labor reports that 20% of the
workforce in Mobile is unemployed. From a sample of 14
workers, calculate the following probabilities:
 Exactly three are unemployed.
 At least three are unemployed.
 At least one are unemployed.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-40
EXAMPLE
 The probability of exactly 3:
P (3)14 C 3 (.20)3 (1  .20)11
 (364)(.0080)(.0859)  .2501
 The probability of at least 3 is:
P ( x  3)14 C3 (.20)3 (.80)11  ... 14 C14 (.20)14 (.80)0
 .250  .172  ...  .000  .551
 The probability of at least one being unemployed:
P(x  1)  1  P(0)
 114 C 0 (.20)0 (1  .20)14
 1  .044  .956
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-41
EXAMPLE
 Since  =.2 and n=14.
 Hence, the mean is:
m= n  = 14(.2) = 2.8.
 The variance is:
2 = n  (1-  ) = (14)(.2)(.8) =2.24.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-42
Finite Population
 A finite population is a population consisting of a fixed
number of known individuals, objects, or measurements.
Examples include:
 The number of students in this class.
 The number of cars in the parking lot.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-43
Hypergeometric Distribution
 The hypergeometric distribution has the following
characteristics:
 There are only 2 possible outcomes.
 The probability of a success is not the same on each trial.
 It results from a count of the number of successes in a
fixed number of trials.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-44
EXAMPLE
In a bag containing 7 red chips and 5 blue chips you
select 2 chips one after the other without replacement.
6/11
7/12
5/12
R2
R1
5/11
B2
7/11
R2
B1
4/11
B2
The probability of a success (red chip) is not the same on each trial.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-45
Hypergeometric Distribution
 The formula for finding a probability using the
hypergeometric distribution is:
( S C x )( N S Cn  x )
P( x ) 
N Cn
where N is the size of the population, S is the number
of successes in the population, x is the number of
successes in a sample of n observations.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-46
Hypergeometric Distribution
 Use the hypergeometric distribution to find the probability
of a specified number of successes or failures if:
 the sample is selected from a finite population without
replacement (recall that a criteria for the binomial
distribution is that the probability of success remains
the same from trial to trial)
 the size of the sample n is greater than 5% of the size
of the population N .
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-47
The density functions of hypergeometric
distributions with N=100, n=20 and different
success rates p (=S/N).
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-48
EXAMPLE: Hypergeometric Distribution
 The National Air Safety Board has a list of 10 reported
safety violations. Suppose only 4 of the reported violations
are actual violations and the Safety Board will only be able
to investigate five of the violations. What is the probability
that three of five violations randomly selected to be
investigated are actually violations?
( 4 C3 )(10  4 C5 2 )
P (3 ) 
10 C5

Ka-fu Wong © 2004
( 4 C3 )( 6 C2 ) 4(15)

 .238
252
10 C5
ECON1003: Analysis of Economic Data
Lesson4-49
Poisson Probability Distribution
The formula for the binomial probability distribution is:
P(x) = nCx  x(1- )n-x
 The binomial distribution becomes more skewed to the
right (positive) as the probability of success become
smaller.
 The limiting form of the binomial distribution where the
probability of success  is small and n is large is called
the Poisson probability distribution.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-50
Poisson Probability Distribution
The Poisson distribution can be described
mathematically using the formula:
P( x ) 
m xem
x!
where m is the mean number of successes in a particular
interval of time, e is the constant 2.71828, and x is the
number of successes.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-51
Poisson Probability Distribution
 The mean number of successes m can be determined in
binomial situations by n , where n is the number of trials
and  the probability of a success.
 The variance of the Poisson distribution is also equal to n .
 X, the number of success generally has no specific upper
limit.
 Probability distribution always skewed to the right.
 Becomes symmetrical when m gets large.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-52
EXAMPLE: Poisson Probability Distribution
 The Sylvania Urgent Care facility specializes in caring for
minor injuries, colds, and flu. For the evening hours of 610 PM the mean number of arrivals is 4.0 per hour. What is
the probability of 2 arrivals in an hour?
P( x ) 
Ka-fu Wong © 2004
m x e m
x!
42 e 4

 .1465
2!
ECON1003: Analysis of Economic Data
Lesson4-53
Example: Poisson Probabilities
x = number of Cleveland air traffic control errors
during one week
m = 0.4 (expected number of errors per week)
Find the probability that 3 errors will occur in a
week.
p(3)  P(x = 3) =
Ka-fu Wong © 2004
e
-0.4
3
(0.4)
= .0072
3!
ECON1003: Analysis of Economic Data
Lesson4-54
Mean and Variance of a Poisson Random
Variable
If x is a Poisson random variable with parameter m, then
Mean
mX = m
Variance
 x2 = m
Standard Deviation
Ka-fu Wong © 2004
 x =  x2  m
ECON1003: Analysis of Economic Data
Lesson4-55
Several Poisson Distributions
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-56
What distributions to use?
 Poisson considers the number of times an event occurs
over an INTERVAL of TIME or SPACE. Note that we are not
considering a sample of given number of observations.
 Thus, if we are considering a sample of 10 observations
and we are asked to compute the probability of having
6 successes, we should not use Poisson. Instead, we
should consider Binomial or Hypergeometric.
 Hypergeometric consider the number of successes in a
sample when the probability of success varies across trials
due to “without replacement” sampling strategy. To
compute the Hypergeometric probability, one will need to
know N and S separately.
 Suppose we know that the probability of success is 0.3.
We are considering a sample of 10 observations and we
are asked to compute the probability of having 6
successes. We cannot use Hypergeometric because we
do not have N and S separately. Instead, we have to
use Binomial.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-57
What distributions to use?
Example
In a shipment of 15 hard disks, 5 are defective. If 4 of the
disks are inspected, what is the probability that exactly 1 is
defective?
 First, we recognize that it is not Poisson because "4 of the disks
are inspected" (i.e., sample size =4).
 Second, it is sampling without replacement because if we were
to inspect four disks for defects, we will not want to sample with
replacement.
 Third, both N (15 hard disks) and S (5 are defective) are
given. Hence we will use Hypergeometric.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-58
What distributions to use?
Example
From an inventory of 48 cars being shipped to local
automobile dealers, 12 have had defective radios installed.
What is the probability that one particular dealership
receiving 8 cars obtains all with defective radios?
 First, we recognize that it is not Poisson because 8 cars are
“inspected" (i.e., sample size =8).
 Second, it is sampling without replacement because if we were
to inspect all 8 cars for defects, we will not want to sample with
replacement.
 Third, both N (48 cars) and S (12 have defective radio) are
given. Hence we will use Hypergeometric.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-59
What distributions to use?
Example
The number of claims for missing baggage for a well-known
airline in a small city averages nine per day. What is the
probability that, on a given day, there will be fewer than
three claims made?
 First, we recognize that it is likely Poisson because “on a given
day”.
 Second, we are asked to compute the probability of the number
of claims larger than some number. There is no limit on the
number of claims that can arrive in a given day.
 Third, “average per day” is given. Hence we will use Poisson.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-60
What distributions to use?
Example
When a customer places an order with Rudy’s on-Line Office
Supplies, a computerized accounting information system (AIS)
automatically checks to see if the customer has exceeded his or her
credit limit. Past records indicate that the probability of customers
exceeding their credit limit is 0.05. Suppose that, on a given day,
20 customers place orders. What is the probability that zero
customers will exceed their limits?
 First, we recognize that it is not Poisson because 20 customers
place orders (i.e., sample size =20).
 Second, the probability of drawing a particular type of customers
appears the same across trials because “the probability of
customers exceeding their credit limit is 0.05”.
 Hence we will use Binomial.
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-61
Lesson 4:
Discrete Probability Distributions
- END -
Ka-fu Wong © 2004
ECON1003: Analysis of Economic Data
Lesson4-62