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```					Doing Statistics for Business
Data, Inference, and Decision Making
Marilyn K. Pelosi
Theresa M. Sandifer
Chapter 6
Probability

1
Chapter 6 Objectives

 Basic Probability Rules
 Random Variables and Probability
Distributions
 The Binomial Probability Distribution
 The Normal Probability Distribution

2

Probability is measure of how likely it
is that something will occur.

An Experiment is any action whose
outcomes are recordable data.

3

The Sample Space is the set of all
possible outcomes of an experiment.

4
TRY IT NOW!
The Spinner Problem
Writing out the Sample Space
An experiment consists of spinning the different spinners pictured below:

Write down the sample space for this experiment.

5

An event, A, is an outcome or a set of outcomes
that are of interest to the experimenter.

The probability of an event A, P(A), is a
measure of the likelihood that an event A will
occur.

6
TRY IT NOW!
The Spinner Problem
Classical Definition of Probability
In the previous exercise you found the sample space
for the spinner example to be:
S = {1A, 1B, 1C, 2A, 2B, 2C, 3A, 3B, 3C}
Let A be the event that the first spinner lands on an odd number. Find
P (A).
Let B be the event that the second spinner is a vowel. Find P (B).
7

The complement of an event A, denoted A´, is
the set of all outcomes in the sample space, S,
that do not correspond to the event A.

8

The event A OR B describes when either A
happens or B happens or they both happen.

The event A and B is the event that A and B
both occur.

Two events A and B are said to be mutually
exclusive if they have no outcomes in common.
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TRY IT NOW!
The Spinner Problem
Calculating the Probability of A OR B
The sample space for the experiment of spinning the two spinners is:
S = {1A, 1B, 1C, 2A, 2B, 2C, 3A, 3B, 3C}
Let A be the event that the first spinner comes up a 1 and let B be the
event that the first spinner comes up a 3.
Find the probability that A OR B occurs using the sample space.

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TRY IT NOW!
The Spinner Problem
Calculating the Probability of A OR B
(con’t)
Now find the same probability using the simple addition rule.

Why are the two answers the same?

11
TRY IT NOW!
Quality Problems
The company that manufactures cardboard boxes collected
data on the defect type and production shift. The data are
summarized in the contingency table below:
Shift
Defect      1      2     3   Total
Color       8      4     3    15
Printing    6      5     2    13
Skewness    0      2     0    2
Total       14    11     5    30

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TRY IT NOW!
Quality Problems
Using the General Addition Rule (con’t)
If a box has more than one defect, then it is classified by the more serious
of the defects only. Suppose that a box from the sample is selected at
random and examined more closely. What is the probability that the box
has a color defect?

What is the probability that the box was produced during the second shift?

13
TRY IT NOW!
Quality Problems
Using the General Addition Rule (con’t)

Is it possible for the selected box to have a color defect and to have
been produced on the second shift? If so, what is the probability?

What is the probability that the selected box will have a color defect or
will have been produced on the second shift?

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The conditional probability of an event A
given an event B is

P(A|B) = P(A and B)
P(B)

15
Two events are independent if the
probability that one event occurs on any
given trial of an experiment is not affected
or changed by the occurrence of the other
event.

16
A Random Variable, X, is a quantitative
variable whose value varies according to
the rules of probability.

The Probability Distribution of a random
variable, X, written as p (x), gives the
probability that the random variable will
take on each of its possible values.

17
TRY IT NOW!
Defective Diskettes
Finding Interval Probabilities
A company that sells computer diskettes in bulk packages for
a warehouse club outlet knows that the number of defective diskettes
in a package is a random variable with the probability distribution given
below:
x       0      1      2      3      4      5     6
p(x)   0.30   0.21   0.12   0.10   0.10   0.09   0.08

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TRY IT NOW!
Defective Diskettes
Finding Interval Probabilities (con’t)
Find the probability that a package of the diskettes will contain
at least 3 defective disks.

Find the probability that the package will contain between 2 and 5
defective diskettes.
Find the probability that the number of defective diskettes will be at most
2.
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TRY IT NOW!
Defective Diskettes
Creating a Probability Histogram
The company that sells computer diskettes in bulk packages for
warehouse clubs would like to have a picture of how the number of
defective diskettes in a package behaves. The probability distribution
is given below:

x       0      1      2      3      4      5     6
p(x)   0.30   0.21   0.12   0.10   0.10   0.09   0.08

20
TRY IT NOW!
Defective Diskettes
Creating a Probability Histogram
(con’t)

Create a probability histogram for the number of defective diskettes.

Use the probability histogram to describe the distribution of the number
of defective diskettes in a package.
21

A Binomial Random Variable is the
number of successes in n trials or in a
sample of size of n.

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TRY IT NOW!
Loan Defaults
Recognizing a Binomial Random Variable
While the Chamber of Commerce is concerned about the problems
of small businesses, it must also be sensitive to the problems that the
lending institutions have when issuing credit. One of the problems that
banks have with small businesses is default on loan payments. It is
estimated that approximately 20% of all small business with less than
50 employees are at least six months behind in loan payments.

23
TRY IT NOW!
Loan Defaults
Recognizing a Binomial Random Variable
The Chamber of Commerce that surveyed the small businesses of
a city wants to look at this problem in more detail. It finds that of the
1536 small businesses in the city, 965 have less than 50 employees. It
randomly selects 25 of these small businesses, checks their credit histories
and counts the number of companies in the sample of 25 that are at least
six months behind in loan payments. Does this qualify as a binomial
probability distribution?

24
TRY IT NOW!
Loan Defaults
Solving Binomial Probability Problems

The Chamber of Commerce that is checking credit problems of small
businesses estimated that 20% of all small businesses were at least six
months behind in load payments. The Chamber of commerce took a
random sample of 25 small businesses and counted that number of the
businesses that were at least six months behind in loan payments.

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TRY IT NOW!
Loan Defaults
Solving Binomial Probability Problems
(con’t)

Define a success for this problem.

Describe the random variable, X, in words.

Find the parameters of the binomial distribution for this problem.

26
TRY IT NOW!
Loan Defaults
Solving Binomial Probability Problems
(con’t)

Find the probability that in the sample of 25 businesses less than 6 were
at least 6 months behind in loan payments.

Find the probability that between 4 and 9 inclusive were at least 6 months
behind in loan payments.

27
TRY IT NOW!
Loan Defaults
Calculating the Mean & Standard
Deviation of a Binomial Random Variable
The Chamber of Commerce that was looking at the loan defaults
for small businesses wants to know the mean and standard deviation for
the binomial random variable with n = 25 and  = 0.20.

Find the mean and standard deviation of the number of small businesses in
25 that will default on their loans.
28
Discovery Exercise 6.1
Exploring the Binomial Distribution
Dear Mom and Dad: Send Cash
According to USA Today, 70% of college students receive
spending money from their parents when at school.

For this exercise, you will need to simulate selecting 30 samples of 5
students from this population of college students and observe whether
they receive spending money from their parents. Consider the successful
outcome to be “receives money” with  = 0.70 and the failure outcome to
29
Discovery Exercise 6.1
Exploring the Binomial Distribution
(con’t)
If your instructor does not provide you with a method, you can take ten
pieces (small) of paper and write an S on 7 of them and an F on 3 of them.
Put the papers in a bag or other container and select one at random to
simulate an observation. Note: Be sure to replace the the paper each time
or  will not always be 0.70.
Record an S when you select a student who receives money from his/her
parents and a F when you select a student who does not receive spending
money from his/her parents.

30
Discovery Exercise 6.1
Exploring the Binomial Distribution
(con’t)
For each sample, record the number of successes you sampled.
In the last outcome compute a running estimate of . Remember that  is
the probability of a successful outcome. In this case,  is known to be
0.70. Let’s see how close the estimate gets to 0.70 as the sample size
increases. So, after the first sample is selected your estimate of  is
simply the number of successes divided by 5. After the second sample is
selected, your estimate of  is the number of successes in both samples
divided by 10 and so forth.

31
Binomial Distribtution n = 10 and  =
Binomial Distribution n = 10 and  =
0.20
0.50
0.400
0.300                                                                                0.400
0.300
p(x) 0.200
p(x) 0.200
0.100
0.100
0.000
0.000
0   1   2   3   4   5   6       7   8   9   10
0        1   2   3   4   5   6   7     8   9   10
Number of Successes
Number of Successes
Binomial Distribution n = 10 and  =
0.80

0.400
0.300
p(x)

0.200
0.100
0.000
0   1    2    3   4   5   6     7     8     9       10
Number of Successes

Figure 6.3 Effects of changing  when
n is fixed                                                                                                                   32
Binomial Distribution n = 5 and  = 0.20
Figure 6.4 Effects
of Changing n for a                                          0.500
0.400

fixed value of                                         p(x)
0.300
0.200
0.100
0.000
0           1        2       3        4       5
Number of Successes

Binomial Distribution n = 25 and  =                         Binomial Distribution n = 10 and  =
0.20                                                         0.20
0.450
0.400
0.400                                                        0.350
0.300
0.300
0.250
p(x) 0.200                                                p(x)   0.200
0.150
0.100                                                        0.100
0.050
0.000                                                        0.000
0   2   4   6    8 10 12 14 16 18 20 22 24                  0       1   2       3   4    5   6   7    8   9       10
Number of Statistics                                            Number of Successes                    33
Binomial Distribution for n = 10 and  = 0.50                                                                      Binomial Distribution for n = 25 and  = 0.50
0.25                                                                                                                 1.60E-01

1.40E-01
0.2
1.20E-01

1.00E-01
0.15
8.00E-02

p(x)    0.1                                                                                                            p(x) 6.00E-02
4.00E-02
0.05
2.00E-02

0.00E+00
0
0       2        4        6     8         10   12        14    16        18    20   22        24
0        1       2            3         4        5         6         7        8         9         10
Number of Successes
Number of Successes

Binomial Distribution for n = 100 and  = 0.50
Binomial Distribution for n = 50 and  = 0.50                                                      8.00E-02
1.20E-01
7.00E-02

1.00E-01                                                                                                              6.00E-02

8.00E-02                                                                                                              5.00E-02

p(x)   4.00E-02
6.00E-02
p(x)
3.00E-02
4.00E-02
2.00E-02
2.00E-02                                                                                                              1.00E-02

0.00E+00                                                                                                               0.00E+00
0   3   6       9                                                                                                     0   6
12       15   18   21   24   27       30   33   36   39   42   45   48                                    12       18   24   30    36   42   48   54    60   66   72    78   84   90   96
Number of Successes                                                                                                     Number of Successes

Figures 6.5 Binomial Distribution for large values
of n                                                                                                                                                                                                                   34

A Probability Density Function, f(x), is
a smooth curve that represents the
probability distribution of a continuous
random variable.

35
Figure 6.6. Probability
Distribution for a
Continuous Random                Probability Density Function for a Continuous Random Variable

Variable.

f(x)

Values of X

36
Figure 6.7. Probability
represented by an area
under the curve           P(x1 < X < x2)

x1   x2

37

For a Normal Random Variable, the
parameter  is the mean of the normal
random variable, X, and  is the
standard deviation.

38
Figure 6.8. Normal
Probability Curve
Normal Curve

f(x)

Values of X

39
Normal Curves with Equal Means and Different Standard
Normal Curves with Different Means and Equal Standard
Deviations                                                                                      Deviations

     
     
     
     

15   17   19   21   23   25   27   29   31   33   35   37   39   41   43   45
Values of X                                                     15   17   19   21   23   25   27   29   31   33   35   37   39   41   43   45
Values of X

Figure 6.9 Probability represented by an
area under the curve
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TRY IT NOW!
Food Expenditures
Looking at the Normal Curve

The amount of money that a person working in a large city spends each
week for lunch is a normally distributed random variable. For professional
and management personnel the random variable has a mean of \$35 and a
standard deviation of \$5. For hourly employees the mean is \$30 with a
standard deviation of \$2. Sketch the normal curves for each of the two
random variables on the same graph.
41

A Z Random Variable is normally
distributed with a mean of 0 and a
standard deviation of 1, Z ~ N(0,1).

A Standard Normal Table is a table of
probabilities for a Z random variable.

42
TRY IT NOW!
Translating from X to Z
The number of pages of a statistics textbook that a student can read
in a given hour is a normally distributed random variable with a mean of
7 pages and a standard deviation of 1.5 pages. One of the professors who
uses the book wants to know the probability that a randomly selected
student can read more than 8.5 pages of the textbook in an hour. Draw a
picture that depicts the problem to be solved and find the Z values
necessary to solve the problem.
43
Figure 6.10 Probability
given by the standard
normal table

44
Second Decimal Place
z      0.00     0.01     0.02     0.03     0.04    0.05      0.06     0.07     0.08     0.09
0.0   0.5000   0.5040   0.5080   0.5120   0.5160   0.5199   0.5239   0.5279   0.5319   0.5359
0.1   0.5398   0.5438   0.5478   0.5517   0.5557   0.5596   0.5636   0.5675   0.5714   0.5753
0.2   0.5793   0.5832   0.5871   0.5910   0.5948   0.5987   0.6026   0.6064   0.6103   0.6141
0.3   0.6179   0.6217   0.6255   0.6293   0.6331   0.6368   0.6406   0.6443   0.6480   0.6517
0.4   0.6554   0.6591   0.6628   0.6664   0.6700   0.6736   0.6772   0.6808   0.6844   0.6879
0.5   0.6915   0.6950   0.6985   0.7019   0.7054   0.7088   0.7123   0.7157   0.7190   0.7224
0.6   0.7257   0.7291   0.7324   0.7357   0.7389   0.7422   0.7454   0.7486   0.7517   0.7549
0.7   0.7580   0.7611   0.7642   0.7673   0.7704   0.7734   0.7764   0.7794   0.7823   0.7852
0.8   0.7881   0.7910   0.7939   0.7967   0.7995   0.8023   0.8051   0.8078   0.8106   0.8133
0.9   0.8159   0.8186   0.8212   0.8238   0.8264   0.8289   0.8315   0.8340   0.8365   0.8389
1.0   0.8413   0.8438   0.8461   0.8485   0.8508   0.8531   0.8554   0.8577   0.8599   0.8621
1.1   0.8643   0.8665   0.8686   0.8708   0.8729   0.8749   0.8770   0.8790   0.8810   0.8830
1.2   0.8849   0.8869   0.8888   0.8907   0.8925   0.8944   0.8962   0.8980   0.8997   0.9015

Figure 6.11 The Standard Normal Table

45
Figure 6.12 Comparison
of Upper and Lower
Probabilities

P(Z < -1)
P( Z > 1)

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Figure 6.13
Finding the area
between two Z values
0.9772

0.0228

47
TRY IT NOW!
The Standard Normal Table
Using the Table to Find Probabilities
For each of the following question, draw a picture of what you
are trying to find BEFORE you use the table to find it.
Find the probability that a Z random variable takes on a value that is less
than 2.74.
Find the probability that a Z random variable is greater than 0.85.

48
TRY IT NOW!
The Standard Normal Table
Using the Table to Find Probabilities
(con’t)
Find the probability that Z is between -1.36 and 1.87.

49
TRY IT NOW!
Solving Normal Probability Problems
The instructor who is interested in how many pages of the
statistics textbook that students can read in an hour knows that the
random variable is N(7, 1.5).

Find the probability that a student could read more than 11.5 pages
in an hour.

50
TRY IT NOW!
Solving Normal Probability Problems
(con’t)
The instructor was worried about the percentage of students who could
not finish reading a 5-page section in the given hour. What percentage
of the students is this?

51

Top p%
(p known)

X

Figure 6.14 The “inverse” normal probability
problem
52

10% or 0.1000

Figure 6.15 Bottom 10% of the normal
distribution
53

z      0.00     0.01     0.02     0.03     0.04     0.05     0.06     0.07     0.08     0.09
-1.9   0.0287   0.0281   0.0274   0.0268   0.0262   0.0256   0.0250   0.0244   0.0239   0.0233
-1.8   0.0359   0.0351   0.0344   0.0336   0.0329   0.0322   0.0314   0.0307   0.0301   0.0294
-1.7   0.0446   0.0436   0.0427   0.0418   0.0409   0.0401   0.0392   0.0384   0.0375   0.0367
-1.6   0.0548   0.0537   0.0526   0.0516   0.0505   0.0495   0.0485   0.0475   0.0465   0.0455
-1.5   0.0668   0.0655   0.0643   0.0630   0.0618   0.0606   0.0594   0.0582   0.0571   0.0559
-1.4   0.0808   0.0793   0.0778   0.0764   0.0749   0.0735   0.0721   0.0708   0.0694   0.0681
-1.3   0.0968   0.0951   0.0934   0.0918   0.0901   0.0885   0.0869   0.0853   0.0838   0.0823
-1.2   0.1151   0.1131   0.1112   0.1093   0.1075   0.1056   0.1038   0.1020   0.1003   0.0985
-1.1   0.1357   0.1335   0.1314   0.1292   0.1271   0.1251   0.1230   0.1210   0.1190   0.1170
-1.0   0.1587   0.1562   0.1539   0.1515   0.1492   0.1469   0.1446   0.1423   0.1401   0.1379
-0.9   0.1841   0.1814   0.1788   0.1762   0.1736   0.1711   0.1685   0.1660   0.1635   0.1611
-0.8   0.2119   0.2090   0.2061   0.2033   0.2005   0.1977   0.1949   0.1922   0.1894   0.1867

Figure 6.16 Normal Probability Table
Containing the Probability 0.1000
54
TRY IT NOW!
Solving the Inverse Problem

The instructor who is interested in how fast students can read
the statistics textbook would like to identify the bottom 25% of
the class, in terms of the number of pages that they can read in an hour.

Find the number of pages per hour that defines the bottom 25% of the
students.
55
Calculating Binomial Probabilities
in Excel

1. Position the cursor in an empty cell in the worksheet.
2. From the main toolbar, click on the Function Wizard icon.
The Paste Function dialog box opens. Highlight
Statistical for the function category and BINOMDIST for
the function. Click OK and the dialog box for the
BINOMDIST function opens.
3. Place the cursor in the textbox labeled Number_s. Type in
the number “25.”
56
Calculating Binomial Probabilities
in Excel (con’t)
4. Place the cursor in the textbox labeled Trials
and enter “50.”
5. Place cursor in the textbox labeled Probability_s. Type in
“0.45.”
6. The last textbox lets you indicate what kind of probability
you want. In this case we want P(X = 25), so set this value
to False.
7. Click on OK and the probability - 0.087330027 - will
appear in the cell which you started.                       57
Calculating Normal Probabilities
1. Fill in the mean,  , and the standard deviation,  ,
of the random variable in the appropriate textboxes.

2. From the section labeled Range of Interest, select the type
of probability you want. For left and right tail
probabilities, you will enter one value of X in the textbox
labeled Value. If you select Left and Right, either inside
or outside,the box will change and you will enter two
values of X.
58
Calculating Normal Probabilities

3. To solve the inverse normal problem, select Inverse Value
type in the left tail area in the box labeled Cumulative
Probability.

4. Fill in where you want the output to go and click OK to
obtain the results.

59
Generating Binomial Random Data in Excel
1. Select Binomial from the list of distributions and the
dialog box will change to allow input of the appropriate
parameters.
2. In the textbox for Number of Variables: type “1” and in
the textbox for Number of Random Numbers: type “20.”
3. In the textbox labeled p Value enter the value for  0.70 &
in the textbox for Number of Trials: enter “20.”
4. Specify where you want the output to appear and click OK.

60
Figure 6.25 The Binomial Random Variable
Dialog Box

61
Figure 6.26 Random Data from Binomial
Distribution

62
Chapter 6 Summary
In this chapter you have learned:
 Probability is more than just flipping coins,
spinning spinners, or gambling. It is important
in the study and development of statistics.
 Probability is the bridge between Descriptive
Statistics and Inferential Statistics.
 In Descriptive Statistics we use different
techniques to describe sample data.
63
Chapter 6 Summary (con’t)
 In Inferential Statistics we will make test
hypotheses about the populations from which
these samples came.
 Probability is the tool that allows us to reconcile
what happened (descriptive) with what we think
is true by determining how likely the outcomes of
the experiment we perform are.
64

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