STATISTICS FOR BUSINESS AND ECONOMICS

Statistics for Business and Economics Anderson

Sweeney

Williams

Slides by

John Loucks St. Edward’s University

© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 1

Chapter 5 Discrete Probability Distributions n n n n n n

Random Variables Discrete Probability Distributions Expected Value and Variance Binomial Probability Distribution Poisson Probability Distribution .40 Hypergeometric Probability .30 Distribution .20

.10

0

1


2

3

4

Slide 2

Random Variables A A random random variable variable is is aa numerical numerical description description of of the the outcome outcome of of an an experiment. experiment. A A discrete discrete random random variable variable may may assume assume either either aa finite finite number number of of values values or or an an infinite infinite sequence sequence of of values. values. A A continuous continuous random random variable variable may may assume assume any any numerical numerical value value in in an an interval interval or or collection collection of of intervals. intervals.


Slide 3

Discrete Random Variable with a Finite Number of Values n

Example: JSL Appliances Let Let xx == number number of of TVs TVs sold sold at at the the store store in in one one day, day, where where xx can can take take on on 55 values values (0, (0, 1, 1, 2, 2, 3, 3, 4) 4) We can count the TVs sold, and there is a finite upper limit on the number that might be sold (which is the number of TVs in stock).


Slide 4

Discrete Random Variable with an Infinite Sequence of Values n

Example: JSL Appliances Let Let xx == number number of of customers customers arriving arriving in in one one day, day, where where xx can can take take on on the the values values 0, 0, 1, 1, 2, 2, .. .. .. We can count the customers arriving, but there is no finite upper limit on the number that might arrive.


Slide 5

Random Variables Question Family size

Type

Random Variable x x = Number of dependents reported on tax return

Discrete

Distance from x = Distance in miles from home to the store site home to store

Continuous

Own dog or cat

Discrete

x = 1 if own no pet; = 2 if own dog(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s)


Slide 6

Discrete Probability Distributions The The probability probability distribution distribution for for aa random random variable variable describes describes how how probabilities probabilities are are distributed distributed over over the the values values of of the the random random variable. variable. We We can can describe describe aa discrete discrete probability probability distribution distribution with with aa table, table, graph, graph, or or formula. formula.


Slide 7

Discrete Probability Distributions The The probability probability distribution distribution is is defined defined by by aa probability (x), which probability function function,, denoted denoted by by ff(x), which provides provides the the probability probability for for each each value value of of the the random random variable. variable. The The required required conditions conditions for for aa discrete discrete probability probability function function are: are: ff(x) (x) > 0  f(x) = 1 f(x)


Slide 8

Discrete Probability Distributions n

Example: JSL Appliances • Using past data on TV sales, … • a tabular representation of the probability distribution for TV sales was developed. Units Sold 0 1 2 3 4

Number of Days 80 50 40 10 20 200

x 0 1 2 3 4

f(x) .40 .25 .20 .05 .10 1.00


80/200

Slide 9

Discrete Probability Distributions n

Example: JSL Appliances Graphical representation of probability distribution

Probability

.50 .40

.30 .20 .10 0

1

2

3

4

Values of Random Variable x (TV sales) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 10

Discrete Uniform Probability Distribution The The discrete discrete uniform uniform probability probability distribution distribution is is the the simplest simplest example example of of aa discrete discrete probability probability distribution distribution given given by by aa formula. formula. The The discrete discrete uniform uniform probability probability function function is is ff(x) (x) = 1/ n 1/n

the values of the random variable are equally likely

where: n = the number of values the random variable may assume


Slide 11

Expected Value The The expected expected value value,, or or mean, mean, of of aa random random variable variable is is aa measure measure of of its its central central location. location. E (x) =  =  xf(x) E(x) xf(x) The The expected expected value value is is aa weighted weighted average average of of the the values values the the random random variable variable may may assume. assume. The The weights weights are are the the probabilities. probabilities. The The expected expected value value does does not not have have to to be be aa value value the the random random variable variable can can assume. assume. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 12

Variance and Standard Deviation The The variance variance summarizes summarizes the the variability variability in in the the values values of of aa random random variable. variable. Var( x) =  2 =  (x - )2ff(x) (x) Var(x) (x The The variance variance is is aa weighted weighted average average of of the the squared squared deviations deviations of of aa random random variable variable from from its its mean. mean. The The weights weights are are the the probabilities. probabilities. The The standard standard deviation deviation,, ,, is is defined defined as as the the positive positive square square root root of of the the variance. variance. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 13

Expected Value n

Example: JSL Appliances x 0 1 2 3 4

f(x) xf(x) .40 .00 .25 .25 .20 .40 .05 .15 .10 .40 E(x) = 1.20

expected number of TVs sold in a day © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 14

Variance n

Example: JSL Appliances x

x-

0 1 2 3 4

-1.2 -0.2 0.8 1.8 2.8

(x - )2

f(x)

(x - )2f(x)

1.44 0.04 0.64 3.24 7.84

.40 .25 .20 .05 .10

.576 .010 .128 .162 .784

TVs squared

Variance of daily sales = 2 = 1.660 Standard deviation of daily sales = 1.2884 TVs


Slide 15

Binomial Probability Distribution n

Four Properties of a Binomial Experiment 1. The experiment consists of a sequence of n identical trials. 2. Two outcomes, success and failure failure,, are possible on each trial. 3. The probability of a success, denoted by pp,, does not change from trial to trial. stationarity assumption 4. The trials are independent.


Slide 16

Binomial Probability Distribution Our interest is in the number of successes occurring in the n trials. We let x denote the number of successes occurring in the n trials.


Slide 17


Binomial Probability Function

n! f (x)  p x (1  p )( n  x ) x !(n  x )! where: x = the number of successes p = the probability of a success on one trial n = the number of trials f(x) = the probability of x successes in n trials


Slide 18


Binomial Probability Function

n! f (x)  p x (1  p )( n  x ) x !(n  x )!

Number of experimental outcomes providing exactly x successes in n trials

Probability of a particular sequence of trial outcomes with x successes in n trials


Slide 19


Example: Evans Electronics Evans Electronics is concerned about a low retention rate for its employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year?


Slide 20


Example: Evans Electronics The probability of the first employee leaving and the second and third employees staying, denoted (S, F, F), is given by p(1 – p)(1 – p) With a .10 probability of an employee leaving on any one trial, the probability of an employee leaving on the first trial and not on the second and third trials is given by (.10)(.90)(.90) = (.10)(.90)2 = .081


Slide 21


Example: Evans Electronics Two other experimental outcomes also result in one success and two failures. The probabilities for all three experimental outcomes involving one success follow. Experimental Outcome

Probability of Experimental Outcome

(S, F, F) (F, S, F) (F, F, S)

p(1 – p)(1 – p) = (.1)(.9)(.9) = .081 (1 – p)p(1 – p) = (.9)(.1)(.9) = .081 (1 – p)(1 – p)p = (.9)(.9)(.1) = .081 Total = .243


Slide 22


Example: Evans Electronics Let: p = .10, n = 3, x = 1

Using the probability function

n! f ( x)  p x (1  p ) ( n  x ) x !( n  x )! 3! f (1)  (0.1)1 (0.9)2  3(.1)(.81)  .243 1!(3  1)!


Slide 23


Example: Evans Electronics 1st Worker

2nd Worker Leaves (.1)

Leaves (.1)

Using a tree diagram

3rd Worker L (.1)

x 3

Prob. .0010

S (.9)

2

.0090

L (.1)

2

.0090

S (.9)

1

.0810

L (.1)

2

.0090

S (.9)

1

.0810

1

.0810

0

.7290

Stays (.9)

Leaves (.1) Stays (.9)

L (.1) Stays (.9) S (.9)


Slide 24

Binomial Probabilities and Cumulative Probabilities Statisticians have developed tables that give probabilities and cumulative probabilities for a binomial random variable. These tables can be found in some statistics textbooks. With modern calculators and the capability of statistical software packages, such tables are almost unnecessary.


Slide 25


Expected Value E(x) =  = np

n

Variance Var(x) =  2 = np(1  p)

n

Standard Deviation

  np(1  p )


Slide 26


Example: Evans Electronics

•

Expected Value E(x) = np = 3(.1) = .3 employees out of 3

•

Variance Var(x) = np(1 – p) = 3(.1)(.9) = .27

•

Standard Deviation

  3(.1)(.9)  .52 employees © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 27

Poisson Probability Distribution A A Poisson Poisson distributed distributed random random variable variable is is often often useful useful in in estimating estimating the the number number of of occurrences occurrences over over aa specified specified interval interval of of time time or or space space It It is is aa discrete discrete random random variable variable that that may may assume assume an an infinite infinite sequence sequence of of values values (x (x == 0, 0, 1, 1, 2, 2, .. .. .. ).).


Slide 28

Poisson Probability Distribution Examples Examples of of aa Poisson Poisson distributed distributed random random variable: variable: the the number number of of knotholes knotholes in in 14 14 linear linear feet feet of of pine pine board board the the number number of of vehicles vehicles arriving arriving at at aa toll toll booth booth in in one one hour hour Bell Bell Labs Labs used used the the Poisson Poisson distribution distribution to to model model the the arrival arrival of of phone phone calls. calls.


Slide 29

Poisson Probability Distribution n

Two Properties of a Poisson Experiment

1. 1. The The probability probability of of an an occurrence occurrence is is the the same same for for any any two two intervals intervals of of equal equal length. length.

2. 2. The The occurrence occurrence or or nonoccurrence nonoccurrence in in any any

interval interval is is independent independent of of the the occurrence occurrence or or nonoccurrence nonoccurrence in in any any other other interval. interval.


Slide 30


Poisson Probability Function

f ( x) 

 x e x!

where: x = the number of occurrences in an interval f(x) = the probability of x occurrences in an interval  = mean number of occurrences in an interval e = 2.71828


Slide 31


Poisson Probability Function Since there is no stated upper limit for the number of occurrences, the probability function ff(x) (x) is applicable for values x = 0, 1, 2, … without limit. In practical applications, x will eventually become large enough so that ff(x) (x) is approximately zero and the probability of any larger values of x becomes negligible.


Slide 32


Example: Mercy Hospital Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening?


Slide 33


Example: Mercy Hospital

 = 6/hour = 3/half-hour, x = 4


34 (2.71828)3 f (4)   .1680 4!


Slide 34


Example: Mercy Hospital Poisson Probabilities

Probability

0.25 0.20

actually, the sequence continues: 11, 12, …

0.15 0.10 0.05 0.00 0

1

2

3

4

5

6

7

8

9

10

Number of Arrivals in 30 Minutes © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 35

Poisson Probability Distribution A A property property of of the the Poisson Poisson distribution distribution is is that that the the mean mean and and variance variance are are equal. equal.

 = 2


Slide 36


Example: Mercy Hospital Variance for Number of Arrivals During 30-Minute Periods

 =  22 = 3


Slide 37

Hypergeometric Probability Distribution The hypergeometric distribution is closely related to the binomial distribution. However, However, for for the the hypergeometric hypergeometric distribution: distribution: the the trials trials are are not not independent, independent, and and the the probability probability of of success success changes changes from from trial trial to to trial. trial.


Slide 38

Hypergeometric Probability Distribution n

Hypergeometric Probability Function

 r  N  r     x  n  x   f ( x)  N   n where:

x = number of successes n = number of trials f(x) = probability of x successes in n trials N = number of elements in the population r = number of elements in the population labeled success


Slide 39


Hypergeometric Probability Function

f (x) 

r  N r x  n  x      N n  

number of ways x successes can be selected from a total of r successes in the population

for 0 < x < r

number of ways n – x failures can be selected from a total of N – r failures in the population

number of ways n elements can be selected from a population of size N


Slide 40


Hypergeometric Probability Function The probability function ff(x) (x) on the previous slide is usually applicable for values of x = 0, 1, 2, … nn.. However, only values of x where: 1) x < r and 2) n – x < N – r are valid. If these two conditions do not hold for a value of xx,, the corresponding ff(x) (x) equals 0.


Slide 41


Example: Neveready’s Batteries Bob Neveready has removed two dead batteries from a flashlight and inadvertently mingled them with the two good batteries he intended as replacements. The four batteries look identical. Bob now randomly selects two of the four batteries. What is the probability he selects the two good batteries?


Slide 42


Example: Neveready’s Batteries


 r  N  r   2  2   2!  2!   x  n  x   2  0   2!0!  0!2!  1              .167 f (x)  6 N  4  4!  n 2  2!2!       

where: x = 2 = number of good batteries selected n = 2 = number of batteries selected N = 4 = number of batteries in total r = 2 = number of good batteries in total © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 43


Mean

 r  E ( x)    n   N n

Variance

r  N  n   r  Var ( x)    n  1     N  N  N  1  2


Slide 44


Example: Neveready’s Batteries • Mean

 r  2   n   2   1 N 4

•

Variance

 2  2  4  2  1   2   1       .333  4   4   4 1  3 2


Slide 45

Hypergeometric Probability Distribution Consider Consider aa hypergeometric hypergeometric distribution distribution with with nn trials trials and r/n) denote and let let pp == ((r/n) denote the the probability probability of of aa success success on on the the first first trial. trial. If N –– nn)/(N )/(N –– 1) If the the population population size size is is large, large, the the term term ((N 1) approaches approaches 1. 1. TThe he expected expected value value and and variance variance can can be be written written EE(x) (x) == np (x) == np (1 –– pp). ). np and and Var Var(x) np(1 Note he expected Note that that these these are are the the expressions expressions for for tthe expected value value and and variance variance of of aa binomial binomial distribution. distribution. continued © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Slide 46

Hypergeometric Probability Distribution When When the the population population size size is is large, large, aa hypergeometric hypergeometric distribution distribution can can be be approximated approximated by by aa binomial binomial distribution distribution with with nn trials trials and and aa probability probability of of success success pp == ((r/N). r/N).


Slide 47

End of Chapter 5


Slide 48

STATISTICS FOR BUSINESS AND ECONOMICS

STATISTICS FOR BUSINESS AND ECONOMICS

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