Anderson_Ch5_updated From the book .pptx

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Statistics for Business and Economics (13e)
Slides by
John
Loucks
St. Edward’s
University
1
Statistics for
Business and Economics (13e)
Anderson, Sweeney, Williams, Camm, Cochran
© 2017 Cengage Learning
Slides by John Loucks
St. Edwards University

Chapter 5: Discrete Probability Distributions
.10
.20
.30
.40
0 1 2 3 4
• Random Variables
• Developing Discrete Probability Distributions
• Expected Value and Variance
• Binomial Probability Distribution
• Poisson Probability Distribution
• Hypergeometric Probability
Distribution
2

• A random variable is a numerical description of the outcome of an
experiment.
Random Variables
• A discrete random variable may assume either a finite number of values
or an infinite sequence of values.
• A continuous random variable may assume any numerical value in an
interval or collection of intervals.
3

Let x = number of TVs sold at the store in one day,
where x can take on 5 values (0, 1, 2, 3, 4)
• Example: JSL Appliances
Discrete Random Variable
with a Finite Number of Values
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).
4

Let x = number of customers arriving in one day,
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is
no finite upper limit on the number that might arrive.
5
Discrete Random Variable
with a Finite Number of Values

Random Variables
Illustration Random Variable x Type
Family
size
x = Number of dependents
reported on tax return
Discrete
Distance from
home to stores on a
highway
x = Distance in miles from
home to the store site
Continuous
Own dog
or cat
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)
Discrete
6

• The probability distribution for a random variable describes how
probabilities are distributed over the values of the random variable.
• We can describe a discrete probability distribution with a table, graph,
or formula.
Discrete Probability Distributions
7

Types of discrete probability distributions:
• First type: uses the rules of assigning probabilities to experimental
outcomes to determine probabilities for each value of the random variable.
• Second type: uses a special mathematical formula to compute the
probabilities for each value of the random variable.
8

• The probability distribution is defined by a probability function, denoted by
f(x), that provides the probability for each value of the random variable.
• The required conditions for a discrete probability function are:
P (x) > 0 and ∑P (x) = 1
9

• There are three methods for assigning probabilities to random variables:
classical method, subjective method, and relative frequency method.
• The use of the relative frequency method to develop discrete probability
distributions leads to what is called an empirical discrete distribution.
10

Using past data on TV sales, a tabular representation
of the probability distribution for sales was developed.
Number
Units Sold of Days
0 80
1 50
2 40
3 10
4 20
200
x f(x)
0 .40 = 80/200
1 .25
2 .20
3 .05
4 .10
1.00
11

.10
.20
.30
.40
.50
0 1 2 3 4
Values of Random Variable x (TV sales)
Probability
Graphical
representation
of probability
distribution
12

• In addition to tables and graphs, a formula that gives the probability
function, P(x) or f(x), for every value of x is often used to describe the
probability distributions.
• Several discrete probability distributions specified by formulas are the
discrete-uniform, binomial, Poisson, and hypergeometric distributions.
13

• The discrete uniform probability distribution is the simplest example of a
discrete probability distribution given by a formula.
• The discrete uniform probability function is
P (x) = 1/n
where: n = the number of values the
random variable may assume
14
• The values of the random variable are equally likely.

Expected Value
• The expected value, or mean, of a random variable is a measure of its
central location.
• The expected value is a weighted average of the values the random
variable may assume. The weights are the probabilities.
• The expected value does not have to be a value the random variable can
assume.
E (x) =  = ∑x.P (x)
15

Variance and Standard Deviation
• The variance summarizes the variability in the values of a random variable.
• The variance is a weighted average of the squared deviations of a random
variable from its mean. The weights are the probabilities.
Var(x) =  2 = (x - )2P (x)
• The standard deviation, , is defined as the positive square root of the
variance.
16

x f(x) xf(x)
0 .40 .00
1 .25 .25
2 .20 .40
3 .05 .15
4 .10 .40
E(x) = 1.20 = expected number of TVs sold in a day
Expected Value
17

Standard deviation of daily sales = 1.2884 TVs
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
1.44
0.04
0.64
3.24
7.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
x -  (x - )2 P (x) (x - )2P (x)
Variance of daily sales =  2 = 1.660
x
Variance
18

Binomial Probability Distribution
• Four Properties of a Binomial Experiment
3. The probability of a success, denoted by p, does not change from trial to
trial. (This is referred to as the stationarity assumption.)
4. The trials are independent.
2. Two outcomes, success and failure, are possible on each trial.
1. The experiment consists of a sequence of n identical trials.
19

• Our interest is in the number of successes occurring in the n trials.
• Let x denote the number of successes occurring in the n trials.
20

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
n! = n(n – 1)(n – 2) ….. (2)(1)
• Binomial Probability Function
𝑃 𝑥 =
𝑛!
𝑥! 𝑛 − 𝑥 !
𝑝𝑥
(1 − 𝑝)(𝑛−𝑥)
21

Probability of a particular
sequence of trial outcomes
with x successes in n trials
Number of experimental
outcomes providing exactly
x successes in n trials
𝑃 𝑥 =
𝑛!
𝑥! 𝑛 − 𝑥 !
𝑝𝑥 (1 − 𝑝)(𝑛−𝑥)
22
• Binomial Probability Function

• 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.
Choosing 3 hourly employees at random, what is the probability that 1 of
them will leave the company this year?
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.
23

• 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
24

• Two other experimental outcomes result in one success and two
failures. The probabilities for all three experimental outcomes involving
one success follow.
Experimental
Outcome
(S, F, F)
(F, S, F)
(F, F, S)
Probability of
Experimental Outcome
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
25

Let: p = .10, n = 3, x = 1
Using the probability function:
𝑃 𝑥 =
𝑛!
𝑥! 𝑛 − 𝑥 !
𝑝𝑥(1 − 𝑝)(𝑛−𝑥)
𝑃 1 =
3!
1! 3−1 !
0.1 1
(0.9)2
= .243
26

1st Worker 2nd Worker 3rd Worker x Prob.
Leaves
(.1)
Stays
(.9)
3
2
0
2
2
Leaves (.1)
Leaves (.1)
S (.9)
Stays (.9)
Stays (.9)
S (.9)
S (.9)
S (.9)
L (.1)
L (.1)
L (.1)
L (.1) .0010
.0090
.0090
.7290
.0090
1
1
.0810
.0810
.0810
1
27

Binomial Probabilities and Cumulative Probabilities
• With modern calculators and the capability of statistical software
packages, such tables are almost unnecessary.
• These tables can be found in some statistics textbooks.
• Statisticians have developed tables that give probabilities and
cumulative probabilities for a binomial experiment random variable .
28

• Using Tables of Binomial Probabilities
n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50
3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250
1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750
2 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750
3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
p
29

E(x) =  = np
• Expected Value
• Variance
• Standard Deviation
𝜎 = 𝑛𝑝(1 − 𝑝)
Var(x) =  2 = np(1 – p)
30

E(x) = np = 3(.1) = .3 employees out of 3
Var(x) = np(1 – p) = 3(.1)(.9) = .27
• Expected Value
• Variance
• Standard Deviation
𝜎 = 3 .1 . 9) = .52 employees
31

• A Poisson distributed random variable is often useful in estimating
the number of occurrences over a specified interval of time or space.
• It is a discrete random variable that may assume an infinite sequence
of values (x = 0, 1, 2, . . . ).
Poisson Probability Distribution
32

• Examples of Poisson distributed random variables:
• number of knotholes in 14 linear feet of pine board
• number of vehicles arriving at a toll booth in one hour
• Bell Labs used the Poisson distribution to model the arrival of phone calls.
33

• Two Properties of a Poisson Experiment
2. The occurrence or nonoccurrence in any interval is independent
of the occurrence or nonoccurrence in any other interval.
1. The probability of an occurrence is the same for any two intervals
of equal length.
34

• Poisson Probability Function
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
x! = x(x – 1)(x – 2) . . . (2)(1)
𝑃 𝑥 =
𝜇𝑥
𝑒−𝜇
𝑥!
35

• Poisson Probability Function–
• In practical applications, x will eventually become large enough so that
P(x) is approximately zero and the probability of any larger values of x
becomes negligible.
• Since there is no stated upper limit for the number of occurrences, the
probability function P(x) is applicable for values x = 0, 1, 2, … without
limit.
36

• 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?
37

 = 6/hour = 3/half-hour, x = 4
𝑃 4 =
34(2.71828)−3
4!
= .1680
38

Poisson Probabilities
0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5 6 7 8 9 10
Number of Arrivals in 30 Minutes
Probability
39

• A property of the Poisson distribution is that the mean and variance
are equal.
 =  2
40
Variance for Number of Arrivals during 30-Minute
periods
 =  2 = 3

41
𝑓 𝑥 =
𝜇𝑥𝑒−𝜇
𝑥!
a) x = 3
𝜇 𝑓𝑜𝑟 1 ℎ𝑜𝑢𝑟 = 48 𝑐𝑎𝑙𝑙𝑠 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟
𝜇 for 5 minutes:
Since there are 12 intervals of 5 min each in an
hour, 𝜇 = 48/12 = 4 for 5 min interval
P 3 𝑐𝑎𝑙𝑙𝑠 𝑖𝑛 5 𝑚𝑖𝑛 =
43𝑒−4
3!
= 0.1954 = 19.54%

42
b) x = 10
𝜇 for 15 minutes:
Since there are 4 intervals of 15 min each in an
hour, 𝜇 = 48/4 = 12 for 15 min interval
P 10 𝑐𝑎𝑙𝑙𝑠 𝑖𝑛 15 𝑚𝑖𝑛 =
1210𝑒−12
10!
=0.1048=10.48 %
c) x = 1
𝜇 for 5 minutes:
Since there are 12 intervals of 5 min each in an hour, 𝜇 = 48/12 = 4 for
5 min interval.
Apart from the current call, there is an expectation that there will be 3
more callers (because the avg is 4 calls per 5 min).
P 1 𝑐𝑎𝑙𝑙 𝑖𝑛 5 𝑚𝑖𝑛 =
41𝑒−4
1!
= 0.0732 = 7.32%

43
d) x = 0 in 3 min interval
𝜇 for 3 minutes:
Since there are 20 intervals of 3 min each in an hour,
𝜇 = 48/20 = 2.4 calls in a 3 min interval. (because 60/3 = 20)
P 0 𝑐𝑎𝑙𝑙 𝑖𝑛 3 𝑚𝑖𝑛 =
2.40𝑒−2.4
0!
= 0.0907 = 9.07%

Hypergeometric Probability Distribution
• The hypergeometric distribution is closely related to the binomial
distribution.
• n number of fixed trials
• However, for the hypergeometric distribution:
• the trials are not independent, and
• the probability of success changes from trial to trial.
• Generally, without replacement experiment!
44

45

• Hypergeometric Probability Function
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
𝑟
𝑥
= rCx
𝑃 𝑥 =
𝑟
𝑥
𝑁 − 𝑟
𝑛 − 𝑥
𝑁
𝑛
46

for 0 < x < r
number of ways
x successes can be selected
from a total of r successes
in the population
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
𝑃 𝑥 =
𝑟
𝑥
𝑁 − 𝑟
𝑛 − 𝑥
𝑁
𝑛
47

• If these two conditions do not hold for a value of x, the corresponding
P(x) equals 0.
• However, only values of x where: 1) x < r and 2) n – x < N – r are
valid.
• The probability function P(x) on the is usually applicable for values of
x = 0, 1, 2, … n.
48

49
• A deck of cards
• Total 52 cards
• Draw 4 cards in succession, without replacement
• Find all the probabilities of getting spades in 4 trials.
• X = Number of spades in 4 trials without replacement.
• X= 0, 1, 2, 3, 4
• Two outcomes: Spades (success) and Non-spades (failure)
• N = 52, r = 13, N - r = 39
• n = 4,

50
Find all the probabilities of getting spades in 4 trials.
P(0 spades):
x = 0, n-x = 4
𝑃 𝑋 = 0 =
13
0
39
4
52
4
=
13𝐶0 .39𝐶4
52𝐶4
= 0.3038 = 30.38%
𝑃 𝑋 = 1 =
13
1
39
3
52
4
=
13𝐶1 .39𝐶3
52𝐶4
= 0.4388 = 43.88%
𝑃 𝑋 = 2 =
13
2
39
2
52
4
=
13𝐶2 .39𝐶2
52𝐶4
= 0.2135 = 21.35%
𝑃 𝑋 = 3 =
13
3
39
1
52
4
=
13𝐶3 .39𝐶1
52𝐶4
= 0.0412 = 4.12%
P(1 spades):
x = 1, n-x = 3
P(2 spades):
x = 2, n-x = 2
P(3 spades):
x = 3, n-x = 1
P(4 spades):
x = 4, n-x = 0
𝑃 𝑋 = 4 =
13
4
39
0
52
4
=
13𝐶4 .39𝐶0
52𝐶4
= 0.00264 = 0.26%

51

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.
• Example: Neveready’s Batteries
Bob now randomly selects two of the four batteries. What is the
probability he selects the two good batteries?
52

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
𝑃 𝑥 =
𝑟
𝑥
𝑁−𝑟
𝑛−𝑥
𝑁
𝑛
=
2
2
2
0
4
2
=
2!
2!0!
2!
0!2!
4!
2!2!
=
1
6
= .167
53

• Mean
• Variance
𝐸 𝑥 = 𝜇 = 𝑛
𝑟
𝑁
𝑉𝑎𝑟 𝑥 = 𝜎2 = 𝑛
𝑟
𝑁
1 −
𝑟
𝑁
𝑁 − 𝑛
𝑁 − 1
54

• Mean
• Variance
𝜇 = 𝑛
𝑟
𝑁
= 2
2
4
= 1
𝜎2 = 2
2
4
1 −
2
4
4 − 2
4 − 1
=
1
3
= .333
55

• Consider a hypergeometric distribution with n trials and let p = (r/N)
denote the probability of a success on the first trial.
• If the population size is large, the term (N – n)/(N – 1) approaches 1.
• The expected value and variance can be written E(x) = np and Var(x) =
np(1 – p).
• Note that these are the expressions for the expected value and variance of
a binomial distribution.
continued
56

• When the population size is large, a hypergeometric distribution can be
approximated by a binomial distribution with n trials and a probability of success
p = (r/N).
• Population size = 1000
• Number of successes in population = 100
• Number of trials = 5 without replacement (hypergeometric)
• Probability of success for first trial = 100/1000
• Probability of success for second trial after getting a success in first = 99/999
• Hypergeometric experiments are sometimes approximated using Binomial, when
the population is very large.
57

End of Chapter 5
58

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