calculus

Outline Discrete probability in general Mean and variance Binomial (or Bernoulli) Poisson Hypergeometric

ECON 4630 Research Methods

Discrete Probability Distributions

Kwon, Economics, UNT

Binomial, Poisson

1 Discrete probability in general

2 Mean and variance

3 Binomial (or Bernoulli)

4 Poisson

5 Hypergeometric

Probability Distribution

1 Random Variable: a numerical description of the outcome of anexperiment.

2 A random variable can be classified as either discrete or continuous1 Discrete Random Variable: a random variable that can assume only

certain clearly separated values.(Counting)2 Continuous Random Variable: a random variable that can assume

one of an infinitely large numbers of values, within certainrange.(Measuring-Weight, Height, Commuting time)

3 Probability Distribution: A listing of all the outcomes of anexperiment and the probability associated with each outcome.

1 The probability of a particular outcome is between 0 and 1 inclusive.2 The outcomes are mutually exclusive events3 The list is exhaustive, so the sum of the probabilities of all the

possible outcomes is equal to 1.

Mean and variance

The population mean of a random variable X:

E(X) = µX =∑all x

[xP (x)] ,

The population variance of a random variable X:

Var(X) = σ2X =∑all x

[(x− µx)

2P (x)]

The population standard deviation of a random variable X:σX

Mean and variance

[xP (x)] ,

[(x− µx)

2P (x)]

Mean and variance

[xP (x)] ,

[(x− µx)

2P (x)]

Mean and variance

[xP (x)] ,

[(x− µx)

2P (x)]

Mean and variance

[xP (x)] ,

[(x− µx)

2P (x)]

Exercise 5-Probability Distribution

The information below is the number of daily emergency service callsmade by the volunteer ambulance service of Walterboro for the last 50days.

Number of calls (x) Frequency Relative Frequency0 81 102 223 94 1

Sum 50

Is this a discrete or continuous random variable?

Convert this frequency distribution to a probability distribution.

What is the mean number of calls per day?

What is the standard deviation of the number of calls made perday?

Sum 50

Number of calls (x) Frequency Relative Frequency

0 8 850

= .161 102 223 94 1

Sum 50

0 8 850

= .161 10 .202 22 .443 9 .184 1 .02

Sum 50

0 8 850

= .161 10 .202 22 .443 9 .184 1 .02

Sum 50 1.00

Exercise 5-Mean

µX = E(X) =∑all x

xP (x)

Number of Calls (x) P (x) x× P (x)

0 0.161 0.202 0.443 0.184 0.02

Total 1

Exercise 5-Mean

xP (x)

0 0.161 0.202 0.443 0.184 0.02

Total 1

Exercise 5-Mean

xP (x)

0 0.161 0.202 0.443 0.184 0.02

Total 1

Exercise 5-Mean

xP (x)

0 0.16 0.001 0.202 0.443 0.184 0.02

Total 1

Exercise 5-Mean

xP (x)

0 0.16 0.001 0.20 0.202 0.443 0.184 0.02

Total 1

Exercise 5-Mean

xP (x)

0 0.16 0.001 0.20 0.202 0.44 0.883 0.18 0.544 0.02 0.08

Total 1

Exercise 5-Mean

xP (x)

0 0.16 0.001 0.20 0.202 0.44 0.883 0.18 0.544 0.02 0.08

Total 1 1.7

Exercise 5-Variance

What is the mean number of calls per day? 1.7

What is the variance of the number of calls made per day?σ2X = Var(X) =

∑all x(x− µx)

2P (x)

Number of Calls (x) P (x) x− µx (x− µx)2 (x− µx)

2 × P (x)

0 0.161 0.202 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.161 0.202 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.161 0.202 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.161 0.202 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.701 0.202 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.701 0.20 -0.702 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.70 2.891 0.20 -0.70 0.492 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.70 2.89 0.46241 0.20 -0.70 0.492 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.70 2.89 0.46241 0.20 -0.70 0.49 0.0982 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1

Exercise 5-Variance

∑all x(x− µx)

2P (x)

2 × P (x)

0 0.16 -1.70 2.89 0.46241 0.20 -0.70 0.49 0.0982 0.44 0.30 0.09 0.03963 0.18 1.30 1.69 0.30424 0.02 2.30 5.29 0.1058

Total 1 1.01

The Binomial Distribution

How to calculate a probability of obtaining 33 heads when you tossa coin 100 times?

When a telemarketer makes 10 calls per hour and is able to make asale on 30 percent of phone contacts, what is the probability ofmaking 7 sales next hour?

The binomial random variable is based on a Bernoulli trial. We candefine a Bernoulli trial with the following properties:

1 The result of each trial may be either S or F: only two outcomes!!!.2 The probability of success, π is the same in every trial.3 The trials are independent.

Probability density function of a binomial random variable:

P (X = x) =

)πx(1− π)n−x

where1(nx

)is the binomial coefficient,

2 n is the number of trials,3 x is the number of successes,4 and π is the probability of success.

Binomial coefficient

Binomial coefficient,(nx

)combination of n items selected x at a

time.(nx

)is denoted as nCx (reads Cnx) counts the number of x

combinations from a set of n objects. The binomial coefficientcounts all possible ways of getting x successes in n trials.

One possible outcome with x successes and (n− x) failures hasprob. of πx(1− π)n−x. Then, the question is how to count suchoutcomes?

Suppose that when a telemarketer makes 10 calls (to 10 differentcustomers). Count all the possible ways of making 7 sales nexthour.

time.(nx

Binomial coefficient, factorial

x!(n− x)!

n! = n× (n− 1)× · · · × 1 reads “n-factorial”.

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

Table for binomial coefficient

x!(n− x)!

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

x!(n− x)!

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

x!(n− x)!

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

x!(n− x)!

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

x!(n− x)!

8! = 8 ∗ 7 ∗ 6 ∗ · · · ∗ 2 ∗ 10! = 1

Calculate(42

Example: Consider a family planning to have four children. It isknown that P (B) = 0.6: in fact, it was 51.20% in 2003.

1 What is the probability of having exactly 1 boy in a 4-child family?

How would you define “success” to solve the problem?What is n?What is π?Does the answer change if we define “success” differently?

2 What is the probability of having exactly 3 girls in an 8-child family?

How would you define “success” to solve the problem?What is n?What is π?(Cumulative probability) What about the probability of having lessthan or equal to 3 girls?

Mean and variance of a binomial random variable, X:

µ =∑all x

xP (x) =∑all x

)πx(1− π)n−x

]= nπ

σ2 =∑all x

(x− µx)2P (x)

=∑all x

[(x− µx)

)πx(1− π)n−x

]= nπ(1− π)

The Binomial Distribution: Exercise 42

The Bank of Hawaii reports that 20 percent of its credit card holderswill default at sometime in their life. The Hilo Branch just mailed out12 new cards today.

How many of these new cardholders would you expect to default?

What is the standard deviation?

What is the probability that exactly 3 of the cardholders willdefault?

What is the probability that at least 4 will default?

The Poisson Distribution

Description: the number of times some event occurs during aspecified interval if these events occur with a known average rate,µ. The poisson probability distribution provides a good model forthe probability distribution of the number of rare events that occurin space, time, or volume etc.

Examples are:

the number of defective parts in outgoing shipments in every threeweek.the number of fatal car accidents on I-35 of the Denton county in ayear.

We make the following assumptions:1 The probability of the event is proportional to the size of the

interval.2 The intervals (spaces, times, ...) which do not overlap are

independent.

Examples are:

independent.

Examples are:

independent.

Probability density function

P (x) =µxe−µ

x!=µx exp(−µ)

µ = the known average number of “successes” in a particularinterval of time or space, µ > 0,The constant, e ≡ lima→∞

(1 + 1

)a= 2.71828 · · · , Or, e can be

defined by the infinite series,∑∞

a=01a! .

x = the number of “successes” = 0, 1, 2, 3, · · · .Example: the number of typos on a page is distributed Poissonwith an average of 0.8.

What is the probability distribution of typos?What is the probability of having 2 typos on a single page?What is the probability that there are at least 2 typos on a singlepage?

P (x) =µxe−µ

x!=µx exp(−µ)

(1 + 1

)a= 2.71828 · · · , Or, e can be

a=01a! .

P (x) =µxe−µ

x!=µx exp(−µ)

(1 + 1

)a= 2.71828 · · · , Or, e can be

a=01a! .

P (x) =µxe−µ

x!=µx exp(−µ)

(1 + 1

)a= 2.71828 · · · , Or, e can be

a=01a! .

P (x) =µxe−µ

x!=µx exp(−µ)

(1 + 1

)a= 2.71828 · · · , Or, e can be

a=01a! .

Mean and variance of a Poisson random variable

µ =∑all x

xP (x) =∑all x

[xµxe−µ

σ2 =∑all x

(x− µx)2P (x)

=∑all x

[(x− µx)

2µxe−µ

µ =∑all x

xP (x) =∑all x

[xµxe−µ

σ2 =∑all x

(x− µx)2P (x)

=∑all x

[(x− µx)

2µxe−µ

µ =∑all x

xP (x) =∑all x

[xµxe−µ

σ2 =∑all x

(x− µx)2P (x)

=∑all x

[(x− µx)

2µxe−µ

The Poisson Distribution: Example

Suppose 1.5 percent of the antennas on new Nokia cell phones aredefective. For a random sample of 200 antennas:

What is the probability distribution of the number of defectiveantennas?

What is the probability that none of the antennas is defective?

What is the probability that three or more of the antennas aredefective?

Description: For use in a population with two distinct, dichotomousgroups. Hypergeometric is sampling without the replacement, butbinomial can be regarded as sampling with replacement.

P (x) =

)(N−Sn−x

N = the population size,S = the number of “successes” in the population,N − S = the number of “failures” in the population,n = the number of trials,x = the number of “successes” in the sample,n− x = the number of “failures” in the sample.

P (x) =

)(N−Sn−x

P (x) =

)(N−Sn−x

Example: according to an industry publication(?), Subway andJimmyJones are ranked one and two in sales of sandwiches.Assume that in a group of 11 individuals, 7 preferred Subway and 4preferred JimmyJones. A random sample of 3 of these individualsis selected.

What is the probability that exactly 2 preferred Subway?What is the probability that the majority preferred JimmyJones?

Mean and variance of a hypergeometric random variable

µX = nS

σ2X = n

)(N − S

)(N − n

N − 1

µX = nS

σ2X = n

)(N − S

)(N − n

N − 1

µX = nS

σ2X = n

)(N − S

)(N − n

N − 1

µX = nS

σ2X = n

)(N − S

)(N − n

N − 1

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This is without replacement, which is not the case for the binomialdistribution. The hypergeometric distribution is withoutreplacement.

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