ECON 214 Elements of Statistics for Economists · 2017-02-22 · Elements of Statistics for...
Transcript of ECON 214 Elements of Statistics for Economists · 2017-02-22 · Elements of Statistics for...
College of Education
School of Continuing and Distance Education 2014/2015 – 2016/2017
ECON 214
Elements of Statistics for
Economists
Session 6 – The Binomial and Poisson Distributions
Lecturer: Dr. Bernardin Senadza, Dept. of Economics Contact Information: [email protected]
Session Overview
• In this session, we discuss two standard discrete probability distributions; the Binomial and Poisson distributions.
• The Binomial distribution arises whenever the underlying probability experiment has just two possible outcome, e.g. heads or tails from the toss of a coin.
• On the other hand, the Poisson distribution describes rare events, when the probability of occurrence is low, e.g. the number of accidents on the Accra-Tema motorway in a day.
• We shall discuss these two discrete probabilities and compute probabilities for a possible number of occurrences of events.
Slide 2
Session Overview
• At the end of the session, the student will
– Be able to describe the characteristics and compute probabilities using the binomial distribution
– Be able to describe the characteristics and compute probabilities using the Poisson distribution
– Use the Poisson distribution to approximate the binomial distribution
Slide 3
Session Outline
The key topics to be covered in the session are as follows:
• Binomial probability distribution
• Poisson probability distribution
• Poisson approximation of the binomial probability distribution
Slide 4
Reading List
• Michael Barrow, “Statistics for Economics, Accounting and Business Studies”, 4th Edition, Pearson
• R.D. Mason , D.A. Lind, and W.G. Marchal, “Statistical Techniques in Business and Economics”, 10th Edition, McGraw-Hill
Slide 5
The Binomial distribution
• The binomial distribution (BD) is applied when;
– Only two mutually exclusive outcomes are possible in each trial (success or failure)
– The outcomes in the series of trials are independent
– The probability of success, denoted P, in each trial remains constant from trial to trial
Slide 7
The Binomial distribution
• The objective of using the BD is to determine the probability values for various possible number of successes (X);
– given the number of trails (n)
– and the known (and constant) probability of success (p).
Slide 8
The Binomial distribution
• Consider five tosses of a coin.
• Recall from Session 5, we can write the probability of 1 Head in 2 tosses as the probability of a head and a tail (in that order) times the number of possible orderings (# of times that event occurs). – P (1 Head) = ½ ½ 2C1 = ¼ 2 = ½
• We can apply same technique to calculate the probability for the number of heads in five tosses of a coin.
Slide 9
The Binomial distribution
• P (X Heads in five tosses of a coin) – P(X = 0) = (½)0 (½)5 5C0 = 1/32 1 = 1/32
– P(X = 1) = (½)1 (½)4 5C1 = 1/32 5 = 5/32
– P(X = 2) = (½)2 (½)3 5C2 = 1/32 10 = 10/32
– P(X = 3) = (½)3 (½)2 5C3 = 1/32 10 = 10/32
– P(X = 4) = (½)4 (½)1 5C4 = 1/32 5 = 5/32
– P(X = 5) = (½)5 (½)0 5C5 = 1/32 1 = 1/32
Slide 10
The Binomial distribution
• To construct a binomial distribution, let – n be the number of trials – x be the number of observed successes – p be the probability of success on each trial
• the formula for the binomial probability distribution is:
Slide 11
!( ) (1 )
!( )!
n x n x
x
nP x C p p
x n x
The Binomial distribution
• When the BD is used, it is typically because we wish to determine the probability of:
• “X or more” successes [ P(X ≥ xi) ]
• or
• “ X or fewer successes [P (X ≤ xi)]
• If the individual probabilities to be summed is large, it is easier to use the complement rule.
• Example: P (X ≥ xi) = 1 - P(X < xi)
Slide 12
The Binomial distribution
• Suppose the probability is .05 that a randomly selected student of the University of Ghana owns a car. What is the probability of observing two or more student car-owners in a random sample of 20 students?
• P(X≥2) = P(X=2) + P(X=3) + …..+ P(X=20)
• P(X≥2) = 1 – P(X<2) = 1- P(X=0, 1)
• = 1 – [P(X=0) +P(X=1)]
• Now
– P(X=0) = 20C0 (.05)0(.95)20 = .3585
– P(X=1) = 20C1 (.05)(.95)19 = .3774
• So P(X≥2) = 1 – (.3585 + .3774) = .2641
Slide 13
The Binomial distribution
• The mean of the binomial random variable is
E(X) = np
• And the variance is
σ2 = np(1-p)
• The standard deviation is the square root of the variance.
Slide 14
The Poisson distribution
• It is a sampling process in which events occur over time or space.
• The Poisson distribution is used to describe a number of processes or events such as
– The distribution of telephone calls going through a switch board
– The demand of patients for service at a health facility
– The arrival of vehicles at a tollbooth
– The number of accidents occurring at a road intersection, etc.
Slide 16
The Poisson distribution
• Characteristics defining a Poisson random variable are
– The experiment consists of counting the number of times a particular event occurs during a given time interval
– The probability that the event occurs in one time interval is independent of the probability of the event occurring in another time interval
– The mean number of events in each unit of time is proportional to the length of the time interval.
Slide 17
The Poisson distribution
• The probability that the Poisson random variable will assume the value X is given by
• For X = 0, 1, 2,3 ……. • λ is the mean of the distribution (mean number of
events occurring in a given unit of time) • e is approximately 2.7183 and is the base of the
natural logarithms. Slide 18
P!
xex
x
The Poisson distribution
• Suppose an average of 2 calls per minute are received at a switchboard during a designated time interval, find the probability that exactly 3 calls are received in a randomly sampled minute.
• Here, X = 3 and λ = 2. So
Slide 19
3 22 8(.1353)
P 3 .18043! 6
ex
The Poisson distribution
• As with the BD, the PD typically involves determining the probability of
• “X or more” number of events • or • “X or fewer” number of events. • To calculate, we sum the appropriate
probability values. • The use of the complement rule may also come
in handy.
Slide 20
The Poisson distribution
• For example, we may want to calculate the probability of receiving 3 or more calls in a three-minute interval.
• That is P(X≥3) = P(X=3) + P(X=4) + … • Using the complement rule, we have • P(X≥3) = 1 - P(X<3) = 1 – [P(x=0) + P(x=1) +
P(x=2)] • From our proposition 3, we know the mean
number of occurrences is proportional to the length of the time interval.
Slide 21
The Poisson distribution
• So if we expect a mean of 2 calls per minute, in three minutes we must expect 6 calls.
• For an interval of 30 seconds, the mean number of calls is one.
• The probability of 5 calls in a three-minute interval implies λ = 6, so P(X=5 / λ=6) = .1606
• The probability of no calls in an interval of 30 seconds implies that λ = 1, so P(X=0 / λ=1) = .3679
Slide 22
The Poisson distribution
• The expected value and variance for a Poisson random variable are both equal to the mean number of events for the time interval of interest
• E (X) = λ
• Var (X) = λ
• The standard deviation is the square root of the variance.
Slide 23
Poisson approximation of Binomial
• When the probability of occurrence (success) is very small (P<.05)
• And the number of trials is large (n>20)
• So that λ = nP and we apply the Poisson formula
• Some say use the Poisson in place of the Binomial when nP < 5
Slide 25
Poisson approximation of Binomial
• A manufacturer claims a failure rate of 0.2% for its hard disk drives. In an assignment of 500 drives, what is the probability that, none are faulty, one is faulty, two are faulty?
• On average, 1 drive (0.2% of 500) should be faulty, so λ = nP = 1.
Slide 26
Poisson approximation of Binomial
• The probability of no faulty drives is
• The probability of one faulty drive is
• The probability of two faulty drives is
Slide 27
0 11
P 0 0.3680!
ex
1 11
P 1 0.3681!
ex
2 11
P 2 0.1842!
ex