Transcript No Slide Title

Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data
Ka-fu Wong © 2003
Chap 6- 1
Chapter Six
Discrete Probability Distributions
GOALS
1.
2.
3.
4.
5.
6.
l
Define the terms random variable and probability
distribution.
Distinguish between a discrete and continuous
probability distributions.
Calculate the mean, variance, and standard deviation
of a discrete probability distribution.
Describe the characteristics and compute probabilities
using the binomial probability distribution.
Describe the characteristics and compute probabilities
using the hypergeometric distribution.
Describe the characteristics and compute the
probabilities using the Poisson distribution.
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Chap 6- 2
Random Variables
 A random variable is a numerical value
determined by the outcome of an experiment.
 A probability distribution is the listing of all
possible outcomes of an experiment and the
corresponding probability.
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Chap 6- 3
Types of Probability Distributions
 A discrete probability distribution can assume
only certain outcomes.
 A continuous probability distribution can
assume an infinite number of values within a
given range.
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Chap 6- 4
Types of Probability Distributions
 Examples of a discrete distribution are:
 The number of students in a class.
 The number of children in a family.
 The number of cars entering a carwash in a
hour.
 Number of home mortgages approved by
Coastal Federal Bank last week.
 Number of CDs you own.
 Number of t rips made outside Hong Kong in
the past one year.
 The number of ten-cents coins in your pocket.
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Chap 6- 5
Types of Probability Distributions
 Examples of a continuous distribution include:
 The distance students travel to class.
 The time it takes an executive to drive to work.
 The length of an afternoon nap.
 The length of time of a particular phone call.
 The amount of money spent on your last
haircut.
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Chap 6- 6
Features of a Discrete Distribution
 The main features of a discrete probability
distribution are:
 The sum of the probabilities of the various
outcomes is 1.00.
 The probability of a particular outcome is
between 0 and 1.00.
 The outcomes are mutually exclusive.
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Chap 6- 7
Example 1
 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 0,1,2,3
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Chap 6- 8
EXAMPLE 1 continued
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
 The outcome of zero heads occurred once.
 TTT
 The outcome of one head occurred three times.
 TTH, THT, HTT
 The outcome of two heads occurred three times.
 THH, HTH, HHT
 The outcome of three heads occurred once.
 HHH
 From the definition of a random variable, x as defined
in this experiment, is a random variable.
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Chap 6- 9
The Mean of a Discrete Probability
Distribution
 The mean:
 reports the central location of the data.
 is the long-run average value of the random
variable.
 is also referred to as its expected value, E(X),
in a probability distribution.
 is a weighted average.
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Chap 6- 10
The Mean of a Discrete Probability
Distribution
The mean is computed by the formula:
μ  Σ[xP(x)]
where  represents the mean and P(x) is the
probability of the various outcomes x.
Similar to the formula for computing grouped mean
where P(x) is replaced by relative frequency.
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Chap 6- 11
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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Chap 6- 12
The Variance of a Discrete
Probability Distribution
 The variance of a discrete probability distribution
is computed from the formula:
2
2
σ  Σ[(x  μ) P(x)]
Similar to the formula for computing grouped variance
where P(x) is replaced by relative frequency.
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Chap 6- 13
EXAMPLE 2
 Dan Desch, owner of College Painters, studied
his records for the past 20 weeks and reports
the following number of houses painted per
week:
# of Houses
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Painted
Weeks
10
5
11
6
12
7
13
2
Chap 6- 14
EXAMPLE 2 continued
 Probability Distribution:
Number of houses painted, x W e e k s Probability, P(x)
10
11
12
13
Total
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5
6
7
2
20
.25
.30
.35
.10
1.00
Chap 6- 15
EXAMPLE 2 continued
 Compute the mean number of houses painted
per week:
μ  E(x)  Σ[xP(x)]
 (10)(.25)  (11)(.30)  (12)(.35)  (13)(.10)
 11.3
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Chap 6- 16
EXAMPLE 2 continued
 Compute the variance of the number of houses
painted per week:
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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Chap 6- 17
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.
 The probability of success stays the same for
each trial.
 The trials are independent.
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Chap 6- 18
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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Chap 6- 19
The density functions of binomial distributions
with n=20 and different success rates p
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Chap 6- 20
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.
 X=number of heads
 The coin is fair, i.e., P(head) = 1/2.
 P(x=0) = 1/8
 P(x=1) = 3/8
 P(x=2) = 3/8
 P(x=3) = 1/8
When the coin is not fair, simple counting rule will not work.
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Chap 6- 21
EXAMPLE 3
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.
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Chap 6- 22
EXAMPLE 3
continued
The Alabama Department of Labor reports that 20% of the
workforce in Mobile is unemployed. From a sample of 14 workers
 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
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Chap 6- 23
Example 3 continued
The Alabama Department of Labor reports that 20% of the
workforce in Mobile is unemployed. From a sample of 14 workers
 The probability of at least one being unemployed.
P(x  1)  1  P(0)
0
14
 114 C0 (.20) (1  .20)
 1  .044  .956
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Chap 6- 24
Mean & Variance of the Binomial
Distribution
 The mean is found by:
  n
 The variance is found by:
  n (1  )
2
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Chap 6- 25
EXAMPLE 4
 From EXAMPLE 3, recall that  =.2 and n=14.
 Hence, the mean is:
= n  = 14(.2) = 2.8.
 The variance is:
2 = n (1-  ) = (14)(.2)(.8) =2.24.
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Chap 6- 26
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.
 The number of homes built in Blackmoor.
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Chap 6- 27
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.
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Chap 6- 28
EXAMPLE 8 of last lecture
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
R1
R2
5/11
B2
7/11
R2
B1
4/11
B2
The probability of a success (red chip) is not the same on each trial.
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Chap 6- 29
Hypergeometric Distribution
 The formula for finding a probability using
the hypergeometric distribution is:
( S Cx )(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.
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Chap 6- 30
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 .
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Chap 6- 31
The density functions of hypergeometric distributions with
N=100, n=20 and different success rates p (=S/N).
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Chap 6- 32
EXAMPLE 5
 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
( 4 C3 )(6 C2 ) 4(15)


 .238
252
10 C5
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Chap 6- 33
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.
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Chap 6- 34
Poisson Probability Distribution
The Poisson distribution can be described
mathematically using the formula:
P( x ) 
 xe
x!
where  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.
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Chap 6- 35
Poisson Probability Distribution
 The mean number of successes  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  gets large.
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Chap 6- 36
EXAMPLE 6
 The Sylvania Urgent Care facility specializes in
caring for minor injuries, colds, and flu. For the
evening hours of 6-10 PM the mean number of
arrivals is 4.0 per hour. What is the probability of
2 arrivals in an hour?
P( x ) 
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 e
x

x!
2
4 e

2!
4
 .1465
Chap 6- 37
Chapter Six
Discrete Probability Distributions
- END -
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Chap 6- 38