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
l
1.
2.
3.
4.
5.
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.
6.
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 Let H outcome of a tail.
x be the number of heads. represent the outcome of a head and T the
The possible outcomes for such an experiment will be: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH.
Thus the possible values of are 0,1,2,3 x (number of heads)
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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, in this experiment, is a random variable .
x as defined
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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, in a probability distribution.
E ( X ),
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 probability of the various outcomes ( x x .
) is the 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:
σ
2
Σ[(x
μ)
2
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:
# o f H o u s e s P a i n t e d 10 11 12 13 W e e k s 5 6 7 2 Ka-fu Wong © 2003
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 5 6 7 2 20 .25 .30 .35 .10 1.00 Ka-fu Wong © 2003
Chap 6- 15
EXAMPLE 2
continued
Compute the mean number of houses painted per week:
μ E(x) 11.3
Σ[xP(x)] (10)(.25) (11)(.30) (12)(.35) (13)(.10) Ka-fu Wong © 2003
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 characteristics: has the following
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 x
be the number of trials be the number of observed successes be the probability of success on each trial
The formula for the binomial probability distribution is: P(x) = n C x
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) = n C x
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
.
250 3 ) 14
C
3 (.
20 ) 3 (.
80 ) 11 .
172 ...
.
000 ...
14
C
14 .
551 (.
20 ) 14 (.
80 ) 0 Ka-fu Wong © 2003
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)
1
14
C
0
(.20)
0
(1
.20)
14
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:
2
n
( 1 ) Ka-fu Wong © 2003
Chap 6- 25
EXAMPLE 4
From EXAMPLE 3 , recall that
=.2 and n =14.
Hence, the mean
= n
is: = 14(.2) = 2.8.
The variance
2 is: = 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 following characteristics: has the
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 R2 7/12 R1 5/11 7/11 B2 R2 5/12 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:
P
(
x
) (
S C x
)(
N
S C n
x
)
N C n
where N is the size of the population, S number of successes in the population, the number of successes in a sample of observations. is the x n is
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Chap 6- 30
Hypergeometric Distribution
Use the hypergeometric distribution probability of a specified number of successes or failures if: to find the
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?
P
( 3 ) ( 4
C
3 )( 10 4
C
5 2 ) 10
C
5 ( 4
C
3 )( 6
C
2 ) 10
C
5 4 ( 15 ) .
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Chap 6- 33
Poisson Probability Distribution The formula for the binomial probability distribution is: P(x) = n C x
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 n is large is called the Poisson probability distribution .
is small and
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Chap 6- 34
Poisson Probability Distribution The Poisson distribution can be described mathematically using the formula:
P
(
x
)
x e x
!
where
is the mean number of successes in a particular interval of time, 2.71828, and x e is the constant is the number of successes.
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Chap 6- 35
Poisson Probability Distribution
The mean number of successes determined in binomial situations by n is the number of trials and a success.
can be n
, where the probability of 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
)
x e
x
!
4 2
e
4 2 !
.
1465 Ka-fu Wong © 2003
Chap 6- 37
Ka-fu Wong © 2003
Chapter Six
Discrete Probability Distributions
- END -
Chap 6- 38