Supplement to Chapter 2 - Probability and Statistics
Learning Objectives
After completing this chapter you should be able to:
1. Define and explain the terms event, mutually exclusive, sample space and Venn
diagram
2. Describe the three different approaches to assigning probabilities:
classical, relative frequency, and subjective.
3. Explain what is meant by marginal, conditional, and joint probability.
4. Contrast independent and dependent events.
5. Calculate variance, standard deviation and coefficient of variation for discrete
probability distributions and samples.
6. Compare and distinguish between discrete and continuous random variables with
examples.
7. List the assumptions and solve problems using binomial distribution.
8. State the parameters and uses of normal distribution.
9. Use the Z table to find the probability of a normally distributed random variable X
Chapter Outline
Introduction, 2
Sample space and events, 2
Concepts of Probability, 2
Probability Axioms, 3
Unions and Intersections, 3
Venn Diagrams, 3
Methods of Assigning Probabilities, 5
Classical Approach, 5
Relative Frequency Approach, 6
Subjective Approach, 6
Conditional Probability, 6
Joint Probability, 7
Marginal Probability, 7
Computing the Probability that Both Events Will Occur, 7
Random Variables, 9
Expected Value, Variance, and Standard Deviation of a Discrete
Random Variable, 9
Sample Mean and Sample Variance, 12
Coefficient of Variation, 14
Binomial Distribution, 14
Normal Distribution, 16
Introduction
The techniques and methods covered in this chapter not only will be useful in covering
the material in chapter 2, but also covering the material in the probabilistic models
section in chapters 11, 12, 13 and 14. Probabilities express the degree of certainty in
managerial decision making or any situation involving uncertainty. Assigning
probabilities to future events allows us to analyze decision options in a rational way. For
instance, a person is more likely to carry an umbrella if the probability of rain is high than
when the probability of rain is low. The ability to quantify the likelihood of the
occurrence of some future event well allows us to make sound decisions
Sample Space and Events
When we specify all possible outcomes in an experiment or a study, we are stating the
sample space. A sample space is the collection of all the experimental outcomes.
Examples of experimental outcomes may be the potential outcome of a soccer game (win,
loose or tie). For a soccer game result we just listed the sample space since our list above
included all possible outcomes. If the experiment involves rolling a fair die, the sample
space or the experimental outcomes are (1, 2, 3, 4, 5, and 6). Since all of the possible
outcomes are included, this constitutes a sample space.
An event is any subset of a sample space. It consists of one or more outcomes
with a common characteristic, such as even numbered outcomes for a single roll of die.
When a die is rolled, sample space consists of 6 outcomes i.e. S = (1, 2, 3, 4, 5, and 6).
When a local college classifies its students as S = (deposit paid, deposit unpaid). The
sample space describing the method of payment at the local grocery store might have four
outcomes: S = (cash, credit card, check, debit card). The term experiment suggests that
the outcome is uncertain prior to taking the observations. If an experiment consists of
outcomes generated by a sequential process such a flipping a coin three times and
recording whether it is heads or tails can be represented as a tree diagram which is
illustrated figure 2S-1. On each flip either a head or tail will occur. The final sequence of
heads and tails will depend on the outcome of each flip. In figure 2S-1, we can see that
There are eight possible outcomes in this experiment.
Concepts of Probability
Probabilities are stated with reference to some event. The event in question might be rain,
winning a game, and head on single flip of a coin (50% chance). All of the above
examples are situations that could be considered a random experiment. A random
experiment is a process of observation for which the outcome is uncertain.
Probability is a measure of likelihood of the occurrence of an event. Possible
values range from 0% to 100%. The closer a probability is to 100%, the more likely it is
to occur. Probabilities are often expressed as decimals. For example, if the probability of
outcome A is 75%, we would express that as P(A) = .75. There are other ways of
expressing probabilities. For instance .75 is a decimal expression of probability of A, and
3 out of 4 is a fraction expression of probability of A. Since a sample space consists of all
the possible outcomes of an experiment, the probability of a sample space is 100 %.
In doing the probability calculations, we have to consider concepts of a
compliment of an event, mutually exclusive events, and collectively exhaustive events.
The complement of an event consists of all outcomes of the event that are not part of the
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original event A. For example, the probability of complement of event A is symbolized
by P( A) , which is the probability of everything that is not in A. Mutually exclusive
events are signified by the fact that two events cannot occur jointly or simultaneously.
Therefore if two events A and B have some component in common, we can conclude
that event A and B are not mutually exclusive. Events are said to be collectively
exhaustive if the comprise the entire sample set; no other events are possible.
Probability Axioms
All probability statements must satisfy the following two probability axioms:
1. The probability of event A must be a value between zero and 1
0  P( A)  1 .
2 If Ai i = 1,2, …, j is a set of mutually exclusive and collectively exhaustive events
in the sample space S, then the sum of the probability of these events is 1.00.
J
That is:
 P( A )  1 , where j is the number of events in S.
i 1
i
Unions and Intersections
Set theory can be instrumental in explaining a few simple concepts. Let’s consider the
terms union and intersection. The union of two events A and B consists of all outcomes
belonging to either or both events A or B. In the set notation the union of A and B is
A  B. The key term in union is or. The intersection of two events A and B consists of
the outcomes that belong to event A and event B. The outcome must belong to both
events to be an intersection. In the set notation, the intersection of event A and event B is
A  B. The key term in intersection is and. In other words “both” implies “and” which
implies intersection, and “either” implies “or” which implies union.
For events that are not mutually exclusive: P( A  B)
.
0
P( A  B)  P( A)  P( B)  P( A  B)
For mutually exclusive events: P( A  B)  0
P( A  B)  P( A)  P( B)
Venn Diagrams
It can be helpful to show the sample space in a graphical form, because the graph makes
it easier to understand the relationships between events. This visualization can be
accomplished by using a Venn diagram, which depicts mutually exclusive and non
mutually exclusive events. Read Example 2S-1 and then look at the Venn diagram
shown in figure 2S-2.
Example 2S-1
The school book store orders two colors of glasses (red and green) with the school logo.
The colored glasses are packed randomly. 40% of the glasses are red. There is a 20%
3
chance that the glass will be damaged during shipment and a 8% chance that it will be a
damaged red glass. What is the probability that a glass that is randomly selected:
a) Will be red or damaged?
b) Will be green?
c) Will not be red or damaged?
d) Draw a Venn diagram for Exercise 2S-1 part a.
Solution
R= Red G = Green D = Damaged and D  Not damaged
a)
P( R  D)  P( R)  P( D)  P( R  D)
P( R  D)  (.4)  (.2)  (.08)  .52
b) P(G)= 1 - P(R)=1 - .4 = .6
c)
P  G  D   P (G )  P ( D )  P (G  D)
P  G  D   .6  .2  .12  .68
d) Since P(R) = .40, P(D) =.2 P( R  D)  .08 , giving us the following Venn diagram.
Figure 2S-1
P(R) = .32
P(R ∩ D) = .08
P(D) = .12
P(ND) = .8
P(G) = .6
Exercises
1. The personnel manager of a company has gathered information about the company.
This information is displayed in Figure 2S-2. One employee is to be selected at random.
4
Figure 2S-2
N = 500
A (125)
B (310)
N = Neither degree
A = Engineering degree
B = Business degree
a) What is the probability that the employee has an engineering degree?
b) What is the probability that the employee has a business degree?
c) What is the probability that an employee has at least one of these degrees?
d) What is the probability that the selected employee has neither degree?
Solution
a) Total = 125 + 310 +500 = 935
A/Total = 125/935 = .1337
b) B/Total = 310/935 = .3316
c) (A + B)/Total = (125 + 310)/935 = 435/935 = .4652
d) N/Total = 500 / 935 = .5348
Methods of Assigning Probabilities
There are three methods of assigning probabilities:
1. The classical approach
2. The relative frequency approach
3. The subjective approach
The Classical Approach
This approach applies when the events have equally likely outcomes. Games of chance
such as tossing a coin or rolling a die are based on the classical approach. Let’s take the
example of rolling a die. Since a die has six possible outcomes, the probability of any
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outcome on a fair die based on a classical approach is 1/6. Likewise, the probability of
head on a toss of a fair coin is ½.
The Relative Frequency Approach
The classical approach to assignment of probabilities involves situations in which
outcomes are equally likely. As you might guess the probability of outcomes are not
equally likely in many circumstances. For example, we may have surveyed 200 people
about a taste test of three brands of soft drinks: Pepsi-Cola, Coca-Cola, and RC Cola.
Suppose that 90 out of 200 preferred Pepsi-Cola. In other words P (Pepsi) = 90 / 200 =
.45 or 45%. Hence according to the relative frequency approach, we have the following
number of times A occured
definition P( A) 
total number of observations
The Subjective Approach
Probabilities computed by either the classical approach or the relative frequency
approach are called objective probabilities because these probabilities are based on
objective facts.
However, there are numerous situations that don’t lend themselves to an objective
approach. In situations in which the outcomes are not equally likely or historical data is
not readily available a subjective probability must be assigned. For example the
probability that a particular new product will be a success is a subjective
assessment of probability. Since this type of assessment is based on the personal opinion
of the decision maker, subjective probability is defined as follows: Subjective Probability
is a personal assessment of the likelihood of an event. Therefore subjective probability
involves our efforts to quantify our feeling or beliefs about something, such as the chance
of success of a new product. Disadvantages of subjective probability are:
1. Difficult to support if questioned
2. Biases can be a factor. Preconceived notions about what should happen can distort
objectivity. It is often difficult to eliminate biases because they are often
subconscious.
Conditional Probability
Conditional Probability is the likelihood that an event will occur given that some other
event is already known to have occurred. The notation used is P( A B) which is read as
the probability of A given B. B implies that B is already known to have occurred.
P( A  B)
P( B)
Example 2S-2 A company that does tax returns has determined that the probability that a
tax return will require a refund is 55%. The probability that a tax return will require a
refund and the return contains an error is 20%. A tax return has been selected and known
to require a refund Find the probability that the selected tax return has an error given
that it requires a refund.
In general, we can compute this probability by using: P( A B) 
6
Solution
P(R) = .55 (probability of refund required)
P( E  R)  .2 (probability of refund required and has an error)
We can find the conditional probability of an error given a refund.
P( R  E ) .20
P( E R) 

 .3656
P( R)
.55
Joint Probability A joint probability is the probability that two or more events occur
simultaneously or sequentially. A joint probability is indicated by the word “and”. The
word and signifies the intersection of two events. This is signified by
P( A  B) . In the case of simultaneous occurrence of two events:
P( A  B)  P( B  A)
Marginal Probability Given a sample space S and event A, the probability that A will
occurs is called a marginal probability. This is denoted by P(A).
Example 2S-3 Assume that 30 out of 50 employees at a manufacturing company are
production workers. 12 of the 30 employees are between ages 18-29, 10 of the 30
employees are between ages 30 to 49 and 8 of the 30 employees are between ages 50 to
65. What is the probability that a single selected employee is a production worker?
Solution
Let P = Production worker
A1 = The worker is between ages 18 to 29
A2 = The worker is between ages 30 to 49
A3 = The worker is between ages 50 to 65
There are two ways to do this problem. The first way does not require the knowledge of
age breakdown:
30
P( P) 
 .6
50
The second way does not require us to know the overall number of production workers
but does require us to know the number of production workers at different age levels for
all age levels.
P( P)  P( P  A1 )  P( P  A2 )  P( P  A3 ) 
12 10 8 30
 

 .6
30 30 30 50
Thus, we obtain the same answer with both methods. Note that it is permissible to add the
joint probabilities because the three events are mutually exclusive and probability axiom
2 is satisfied.
P( P) 
Computing the Probability That Two Events Will Both Occur: P( A  B)
The probability that both events will occur simultaneously is called their joint probability
and its computation depends on whether the events in question are dependent or
independent.
7
Two events are considered to be independent of each other if the occurrence of one event
is unrelated to the occurrence of the other. For instance, if two fair dice are rolled,
knowing the result of one die does not aid in determining the result of the other die.
Two events are considered to be dependent if one event occurring affects the probability
of the other occurring. For example, knowing that a potential buyer has seen a newspaper
advertisement about a product, there is a higher probability of making a sale of that
product.
The probability that two or more events will occur simultaneously or sequentially is
called a joint probability. It can be computed using the multiplication rule. A joint
probability is indicated by the word “and”, it signifies the intersection of two events. If
two events are dependent, the probability that both will occur simultaneously is equal to
the probability that the first one occurs times the conditional probability of the second
one occurring.
Example 2S-2 Let’s consider an example with two dependent events A and B. Let’s also
assume that event A occurs first. What is the formula of joint probability of event A and
event B?
Solution
P ( A  B )  P ( A) P ( B A)
or P ( A  B )  P ( B ) P ( A B)
Example 2S-3 Let’s consider an example with two independent events A and B. Let’s
also assume that event A occurs first. What is the joint probability of event A and event
B?
Solution
P( A  B)  P( A) P( B)
Let’s consider the following two examples to illustrate joint probability
Example 2S-4 Two consumers are to be selected at random from a group of five
consumers (A, B, C, D, E) to be interviewed by a product representative. What is the
probability that both C and E will be interviewed?
Solution Since event C and event E are dependent, we can employ the dependent event
formula for joint probability:
P (C  E )  P (C ) . P ( E C )
1 1 1
P (C  E )  . 
5 4 20
Example 2S-5 One-third of the consumers of a certain deodorant are women, and 70% of
consumers chose a certain deodorant because of its nice smell. Find the probability that
the next randomly selected consumer will be a woman and will have selected this product
8
because of its nice smell. Assuming these two events are independent we can employ the
multiplication rule for independent events:
Let W = Women and S = Select because of its nice smell
P(W) = .3333 P(S)=.70
Solution
P(W  S )  P(W ).P( S )  (.3333).(.70)  .2333
Random Variables
A random variable is a function that assigns numbers to the basic experimental outcomes.
Let’s consider the coin example and let X(tail) = 0 and X(head) = 1 in which case the
variable X is defined as the number of heads occurring.
Generally speaking random variables are either discrete or continuous. It is
important to distinguish between discrete and continuous random variables because
different mathematical techniques are utilized depending on which type is involved. A
random variable is said to be discrete if it can assume only a finite number of values. A
random variable is said to be continuous if it can assume any value in some interval or set
of intervals. In the continuous case, the set of possible outcomes is always infinite.
Therefore, it is not possible to list the sample space by individual values in any form. The
way to distinguish between discrete and continuous is to ask whether the values of the
random variable can be counted. The outcomes of discrete random variables can be
counted (e.g., the number of heads in 10 coin tosses number, or the number of defectives
items in batch of 300 units). The outcomes of continuous random variables are measured
rather than counted (e.g., the weight of an individual, or the length of a bolt).
Expected Value, Variance and Standard Deviation of a Discrete Random Variable
In working with discrete random variables it is helpful to know certain values that aid in
describing the distribution. The most commonly used values are those that identify the
physical center and the dispersion (the way the values are spread around the center).
Given a discrete random variable X, with probability function p(x) the expected value of
X (also called the mean and denoted E[X] ) or  (mean of random variable X) is defined
as the weighted average of the values of X may assume where weights are the
corresponding probabilities, that is:
  E[X]=  [ X .P( X )]
all X
Example 2S-6
Determine the expected number of broken tools per day for the probability distribution
given in Table 2S-1:
9
Table 2S-1 Discrete Probability Distribution
for Broken Tools
Number Broken per day P(X)
X P(X)
0
.23
0
1
.50
.50
2
.15
.30
3
.08
.24
4
.04
.16
Total
1.0
1.2
Hence the mean or expected number of broken tools per day is 1.20. This value can be
interpreted as the long run average of broken tools per day. Obviously, this value cannot
occur on any day since the average is not an integer value. However in interpreting this
value say over 50 days, the factory can expect to have (50).(1.2)= 60 broken tools. This
result does not imply that exactly 60 tools will be broken over the 50 day period, but it
does provide management with an estimate of the replacement tools that will be needed.
This figure (60 tools) can not only be used to estimate the number of replacements, it can
also be used to for other planning purposes such as estimating the cost of the
replacements, estimating the downtime due to tool breakage, etc.
It would also be useful for a decision maker to have some knowledge about the extent to
which the actual number of breakages might tend to vary from the expected value of
breakages by calculating the variance or the standard deviation.
The general formula for computing the variance is:
n
 2   [ X i  E[ X ]* P( X i )] . The variance for the tool breakage example is computed in
i 1
Table 2 S-2:
Table 2S-2 Calculation of Variance for a Discrete Random Variable
Number Broken per day P(X)
Xi –E(X)
[Xi –E(X)]2 [Xi –E(X)]2*P(X)
0
.23
-1.2
1.44
.3312
1
.50
-.2
.04
.02
2
.15
.8
.64
.096
3
.08
1.8
3.24
.2592
4
.04
2.8
7.84
.3136
Total
1.0
1.02
This computation can be performed using Excel. The associated Excel spreadsheet is
located in Exhibit2 S-1
Hence the variance of daily tool breakdowns is  2  1.02 breakdowns squared/day
Even though variance shows dispersion or variation around the mean, it is based on
squared values. The standard deviation is generally preferred measure of dispersion
because the result is in the same terms (not squared) as the original random variable (X).
It is denoted as  and is computed by taking the square root of the variance. For our
example:   1.02  1.01 breakdowns per month. A small standard deviation relative to
10
the mean can be interpreted as the values of the random variable are closely bunched to
the expected value (mean).
Exhibit 2S-1 Example Calculation of the Variance and standard deviation of a
probability distribution with Excel
Cell
Cells Copied to
B7
Formula
=SUM(B3:B7)
D3
D4:D7
=A3-C3
E3
E4:E7
=D3^2
F3
F4:F7
=B3*E3
B8
=SUM(B3:B7)
F8
=SUM(F3:F7)
F9
=SQRT(F8)
B17
=SUM(B12:B16)
C12
C17
C13:C16
=A12*B12
=SUM(C12:C16)
11
Sample Mean and Sample Variance
The mean and variance explained in the above section are more correctly referred to as
the population mean and population variance. The population mean is the expected
value of all possible values and the population variance or standard deviation is the
dispersion of all possible values around the mean. The arithmetic average of a set of
sample data is called a sample mean. For ungrouped data it is determined by the
n
following formula:
X
X
i 1
i
n
Example 2S-6 Calculate the mean GPA for the following seven students
2
3
4
5
6
Student 1
2.43
2.76
3.78
3.03
1.97
3.12
GPA
7
3.21
Solution
2.43  2.76  3.78  3.03  1.97  3.12  3.21
 2.90
7
The Sample Variance, denoted by s2, is defined as the average squared deviation from
the mean. Sample standard deviation is the square root of the sample variance. For
ungrouped data sample variance is calculated using the following formula:
X
n
s X2 
(X
i
 X )2
and the sample standard deviation is calculated as sX  sX2
n 1
Example 2S-7
For the GPA data given in example 2S-7, calculate the sample variance and the sample
standard deviation
i 1
Solution
One way to solve this problem is illustrated in Table 2S-3:
Table 2S-3 Calculation of Sample Variance
Student
GPA
Mean
1
2.43
2.90
2
2.76
2.90
3
3.78
2.90
4
3.03
2.90
5
1.97
2.90
6
3.12
2.90
7
3.21
2.90
20.3
Total
s X2 
2.0412
 .3402
7 1
GPA-Mean
-.47
-.14
.88
.13
-.93
.22
.31
(GPA-Mean)2
.2209
.0196
.7744
.0169
.8649
..0484
.0961
2.0412
s X  .3402  .5833
Another way to solve this is to use Excel.
12
Exhibit 2S-2 Example Calculation of Sample Variance and Standard Deviation with
Excel
Cell
Cells Copied to
Formula
B9
=SUM(B2:B8)
B10
=AVERAGE(B2:B8)
D2
D3:D8
=B2-C2
E2
E3:E8
=D2^2
E9
=SUM(E2:E8)
E10
=E9/(7-1)
E11
=SQRT(E10)
13
The Coefficient of Variation
Sometimes we want to measure the size of the standard deviation of population or sample
relative to the sample or population mean. This is referred to as the Coefficient of
Variation. It is defined for a population or a sample is given by:
standard deviation
Coefficient of Variation=
X 100 For instance, the coefficient of
mean
variation for Example 2S-6 is .5833 (  -standard deviation)/2.90 (  -mean) = .2011. The
higher the coefficient of variation, the more volatile the variable. Generally, in comparing
2 variables, we prefer the variable with the smaller coefficient of variation.
Binomial distribution
The simplest random variable is one that has one value. However, we would have little
interest in such a random variable. If the random variable can assume one of two possible
values, it could be used to describe an experiment that can be classified as resulting in
either “success” or “failure”. Let’s assume that the random variable assigns the value of 1
to success and value of 0 to failure with probability of p to success and probability of 1-p
to failure. This type of discrete variable is generally known as a Bernoulli random
variable. If a random experiment consists of making n independent trials from an infinite
population where the probability of success p is constant from trial to trial, the probability
of number of successes P(X) is given by the binomial distribution. The general forms of
the probability function for the family of binomial distributions given by:
C Xn p X (1  p ) n  X for X= 0,1,2,....,n 


P( X )  
0  p 1

0

elsewhere


Properties of a binomial process are listed below:
1. There are two possible outcomes for each trial. Outcomes could be yes or
no, success or failure, defective or non-defective, heads or tails and so on.
2. The probability of an outcome remains constant from trial to trial. For
example the probability of success or failure on any trial remains the same
regardless of the number of trials. If the probability of success is .30 it will
remain .30 on each trial regardless of the number of successes on previous
trials.
3. Related to number 2, outcomes of the trials are independent. In other
words, if a success occurred on a previous trial, it does not affect the
probability of success on the next trial.
4. The number of trials are discrete and integer. For example, the number of
trials can be 10 but not 10.3.
Assume a random experiment consisting of making n independent trials with X
successes. The following equation represents the number of ways to sequence X
successes among n trials.
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n!
The terms n!, X! and (n-X)! are called factorials. Factorials are
X ! n  X !
computed using the formula X! = X (X-1) (X-2) (X-3)….. (2) (1). By definition 0! is
equal to 1.
CXn 
For number of successes the formula for the mean (  ) and standard deviation ( ) are:
  np and  = np(1-p)
(2S 1)
For percentage of successes formula for the mean (  ) and standard deviation ( ) are:
p(1- p)
(2S  2)
n
The binomial distribution formula may look complicated, but it is fairly easy as
demonstrated by the following example.
  p and  
Example 2S-8
Enka Manufacturing Company (EMC) produces a product that has a known defective rate
of 5%. In a day, 10 items are randomly selected and checked to see if they are defective.
From a sample of 10 units selected, 2 units are found to be defective. What is the
probability of this occurring by chance?
Solution
We can assume that the known defective rate is constant from one day to the next and
one product to the next. The two parameters are n = 10 and probability or p = .05
The number of possibilities of selecting 2 defective items from 10 is given by the
10!
10!
combination formula C210 

 45 In other words; we can select 2
2!10  2 ! 2!8!
defective units from 10 defective units 45 different ways.
The probability of one such sequence (2 defective units and 8 non-defective units)
is: (.05)2 (.95)8  .0017 . Remember, the probability of a single unit being defective is 5%.
Since there are 45 ways of obtaining two defectives and 8 non-defectives and probability
of obtaining one such sequence is .0017: .0017* 45= .0746 (the probability of two
defective units out of 10 units when the probability of a single unit being defective is
5%). The answer to the question is: .0746.
Example 2S-9
Enka Manufacturing Company (EMC) produces a product that has a known defective rate
of 5%. In a day, 10 items are randomly selected and checked to see if they are defective.
From a sample of 10 units selected, what is the probability of finding 2 units or less to be
defective by chance?
15
Solution
Again, we can assume that the known defective rate is constant from one day to the next
and one product to the next. The two parameters are n = 10 and p = .05.
The probability of obtaining two defectives from a sample of ten by chance is:
.0017: .0017* 45= .0746
The probability of obtaining two or less defectives from a sample of ten by chance can be
found by the following: P( X  2)  P( X  1)  P( X  0) . In example 2S-8, we found
P (X=2) = .0746. P( X  1)  C110 (.05)1 (.95)9  (10)(.05)(.630249)  .315125
P( X  0)  C010 (.05)0 (.95)10  (1)(1)(.59874)  .598737
P( X  2)  P( X  1)  P( X  0)  .0746  .315125  .598737  .988462 .
Example 2S-10
Enka Manufacturing Company (EMC) produces a product that has a known defective rate
of 5%. In a day, 10 items are randomly selected and checked to see if they are defective.
In a sample of 10 units selected, what is the probability of finding 3 units or more to be
defective by chance?
Solution
Since P( X  2)  .988462 , the probability of getting 3 or more defectives from a sample
of 10 is: P( X  3) = 1-.988462 =.011538
Because our examples included very small values for n and X, we were able to work out
the problems manually. The problems involving larger values of n and X can be solved
using a calculator or using binomial tables. However, the demonstration of how to use
those tables is beyond the scope of this text.
Normal Distribution
In the last section, we introduced the concept of a binomial variable which is a discrete
random variable. Now we turn our attention to a continuous random variable, and
continuous distributions. One of the most commonly used continuous distributions is the
normal distribution. It is probably the best known and most widely used of all
probability models. Today the normal family is used at least as a good approximation to
describe situations where measurements cluster around some central value. Many
phenomena in the business world exhibit a normal distribution. Examples include scores
on a standardized college entrance examination, starting salaries of accounting college
graduates first jobs, monthly sales of a non-seasonal product, portfolio yields, and
electricity consumption. The bell shaped curve can be flatter or taller depending on the
degree to which the values of the random variable are dispersed around the center of the
distribution. The normal distribution has several attributes demonstrated in Figure 2S-1
16
1. It is symmetrical around the mean
2. It asymptotically approaches 0 at  and +.
3. It is a bell-shaped curve. Any normal variable can be converted to a standard
normal random variable Z through the use of the Z transformation. The Z
transformation is given in equation 2S-1-3.
X 
Z
(2S-3)

The Z value enables us to represent any value of X in terms of its distance from the
mean in standard deviations. If X is a normal random variable with a mean  X and a
standard deviation of  X , then the variable obtained through the Z transformation is
also normally distributed with a mean (  ) of 0 and standard deviation (  ) of 1.
Figure 2S-3 Normal Curve for Selected Values of X for the parameters
μx = 15 and σx = 1.5
Table 2S-4 demonstrates the calculation of probability for different values of X
when  X  15 and  X  1.5 .
17
Figure 2S-4 Standardized Normal Curve
1 covers 68.26% of the area
2 s cover 95.44% of the area
3 s cover 99.74% of the area
The probabilities shown in Figure 2S-3 and Figure 2S-4 can be obtained in Appendix B,
Table A. In Figure 2S-4, P(1  Z  1)  .6826
P(2  Z  2)  .9544
P(3  Z  3)  .9974
The probabilities in the Z table are indexed to two decimals (nearest hundredth) such
as, 1.00, 3.12, 1.76 and 2.56. However, the Z table is shown at the intersection of a
row and column. Values of the first decimal and the first value after the decimal are
shown on the left side (like 1.2, 2.4, or .8), while the second or last digit after the
decimal appears across the top. For example, if the value of Z=1.28, find 1.2 on the
left side of the table and .08 across the top. Where the two intersect, we find the
probability of .3997.
For another instance, in Figure 2S-1, when the value of Z is equal
X   16.5  15



 1.0  , the probability to the left of Z = 1 can be obtained
to 1  Z 

1.5


by first finding in Appendix B-Table A as the probability of P (0  Z  1) =.3413 and
then adding .5000 to obtain .8413, because 50% of the distribution is to the left of the
mean. To obtain the probability of getting a value that is to the right of Z = 1, subtract
the probability that we found to be to the left of g Z = 1 from 1,0000. Thus, it is
1.0000 - ,8413 = .1587. Note that the probability of obtaining a value that lies at
18
exactly Z = 1 is zero; the probability of any single point in a continuous distribution is
zero.
To use Appendix B-Table A, we need to know the value of Z only. As we
indicated earlier, Table A in appendix B gives the area under the standardized normal
curve from the mean to the value of Z. Since the left and right side of the distribution
is symmetric the probabilities on the right side are equal to the probabilities on the
left side since they are mirror images of each other.
Table 2S-4 Calculation of Probability
Converting X Values to Standard
Normal Z Values Where   15 and   1.5
X
Z
P(X  x )
15
0
.50
16.5
1
.8413
18
2
.9772
19.5
3
.9987
The Z, standard normal value, would have a minus sign if the value of X is less than
the mean. Table 2S-5 shows some examples of the conversion of the actual difference
between the mean and another value to the relative difference in terms of the number
of standard deviations:
Table 2S-5 Using Standard Normal Distribution Conversion from X to Z
Standard
Value of
Difference
Relative
Mean (  )
Interest
Difference
Deviation (  )
X- 
X 
Z=

30
1
32
2
+2
26
2
24
-2
-1
25
2.5
32.5
7.5
+3
15
3
10.5
-4.5
-1.5
20
3.5
20
0
0
It is also necessary to be able to work in reverse, going from Z values to actual values
of X. For instance, we might want to know what the actual value of X would be
equivalent of Z = +2. Assuming we know the mean and standard deviation and that
we are considering a normal distribution, the conversion takes the following form:
X    Z .Table 2S-6 demonstrates example calculations of this conversion:
19
Table 2S-6 Using Standard Normal Distribution Conversion from Z to X
Standard
Z (standard
Calculation
Actual Value
Mean (  )
of X
Deviation (  ) normal variable) X    Z
27
1
+3
27+3(1)
30
53
3
-1
53-1(3)
50
65
2.5
-2
65-2(2.5)
60
68.5
5
+.3
68.5 + .3(5)
70
Table 2S-7 demonstrates how to read various probabilities given the Z value and its
sign. In Table 2S-7, two values of Z are utilized twice. The first value of Z has a
positive value and the second value of Z has a negative value. Two directions (to the
right and to the left) are both being utilized. As shown in Table 2S-7, when the value
of Z is positive and direction is to the right, we subtract the probability of Z from
Table A Appendix B from .5, because we are asked the tail area. However, when the
value of Z is positive and direction is to the left, we add the probability of Z from
Table A Appendix B to .5, because we are asked the non-tail area. When the value of
Z is negative (Z = -1.25) and direction is to the right, we add the probability of Z from
Table A Appendix B to .5, because we are asked the non-tail area. However, when
the value of Z is negative and direction is to the left, we subtract the probability of Z
from Table A Appendix B to .5, because we are asked the tail area.
Table 2S-7 Finding the Probability of Greater (to the right) or
Less Than (to the left) of Certain Z Value
Z
Direction
Calculation Probability
1.5
To the right
.5-.4332
.0668
1.5
To the left
.5+.4332
.9332
-1.25
To the right
.5+.3944
.8944
-1.25
To the left
.5-.3944
.1056
Example 2S-11
Boxes of whole grain mighty wheats cereal are filled automatically and the net weight
is approximated by a normal distribution with a mean of 16 ounces and a standard
deviation of .4 ounces. What is the probability of obtaining a box with a net weight of
less than 16.7 ounces?
Solution
Any time we are attempting to find the area under the normal curve, it is advisable to
draw a picture so that we are aware of exactly what are we are evaluating. Therefore
16.7  16 .7
Z
  1.75 . From Appendix B, Table A P(0  Z  1.75)  .4599 , so
.4
.4
P( Z  1.75)  .5  .4599  .9599
Figure 2S-5 Graphical representation of P(X  16.7) for Example 2S-11
20
.9599
.0401
Example 2S-12
Boxes of whole grain “mighty wheats” cereal are filled automatically and the net
weight is approximated by a normal distribution with a mean of 16 ounces and a
standard deviation of .4 ounces. What is the probability of obtaining a box with a net
weight of less than 15.5 ounces?
Solution
Any time we are attempting to find the area under the normal curve, it is advisable to
draw a picture so that we are aware of exactly what are we are evaluating. Therefore
15.5  16 .5
Z

 1.25 . From Appendix B, Table A P(0  Z  1.25)  .3944 , so
.4
.4
P( Z  1.25)  .5  .3944  .1056
Figure 2S-6 Graphical representation of P(X  15.5) for Example 2S-12
.3944
.1056
x
z
Example 2S-13
21
Boxes of whole grain mighty wheat cereal are filled automatically and the net weight
is approximated by a normal distribution with a mean of 16 ounces and a standard
deviation of .4 ounces. What is the probability of obtaining a box with a net weight of
between 16.5 and 16.9 ounces?
Solution
Any time we are attempting to find the area under the normal curve, it is advisable to
draw a picture so that we are aware of exactly what are we are evaluating. Therefore
16.5  16 .5
Z
  1.25 . From Appendix B, Table A P(0  Z  1.25)  .3944 , so
.4
.4
16.9  16 .9
Z
  2.25 From Appendix B, Table A P(0  Z  2.25)  .4878
.4
.4
P(1.25  Z  2.25)  .4878  .3944  .0934
Figure 2S-7 Graphical representation of P( 16.5  X  16.9) for Example 2S-13
.0934
x
z
To summarize, a normal distribution is a continuous probability distribution that is
symmetrical on both sides of the mean. It resembles a bell shaped curve. The center
of a normal distribution is its mean (  ). The area under the normal curve represents a
probability and the total area under the curve is equal to 1.00.
Summary
This chapter supplement provides a summary of probability and statistics. Probability
is the measure of likelihood that a particular event will occur. Probability has a value
between 0 and 1 and it can be assigned subjectively, objectively or experimentally. In
this chapter appendix, we also covered the relationship between joint, marginal, and
conditional probabilities and we derived the multiplication rule for both statistically
dependent and independent events. We discussed the probability of mutually
exclusive and independent events. When the term “and” is used, implies joint
22
probability and to employ the multiplication rule. When the term “or” is used, employ
the addition rule. When the term “given” is used, as in” P(A) given that event B has
occurred,” that involves conditional probability.
In this chapter supplement, we introduced expected value and defined the mean to be
the expected value of a variable. As a measure of dispersion, we reviewed the concept
of variance, which is the average squared deviation from the mean and its square root,
the standard deviation. We also discussed variance and standard deviation as they
apply to sample data. In this area, we also compared the population mean, variance,
and standard deviation with sample mean, variance and standard deviation.
In this chapter supplement we also discussed random variables, and the associated
important discrete and continuous probability distributions. In practice discrete
random variables involve counting, and continuous variables involve measuring.
One of the most important discrete probability distributions- binomial distribution,
involves two outcomes, “success” or “failure” and it requires that the probability of
success remains constant from trial to trial. The most important continuous
probability distribution is the normal distribution. The conversion to the standard
normal distribution is shown and the use of Z transformation to obtain values from
the Z table.
Glossary
Addition of probabilities (mutually exclusive): P( A  B)  P( A)  P( B)
Addition of probabilities (not mutually exclusive):
P( A  B)  P( A)  P( B)  P( A  B)
Binomial distribution. When a random experiment consists of making n independent
trials from an infinite population where the probability of success p is constant from
trial to trial, the probability of number of successes P(X) is given by binomial
distribution.
Coefficient of Variation A value involving a sample or a population that measures
the variation in terms of a standard deviation relative to its mean.
P( A  B)
Conditional Probability: P( B A) 
P( A)
Continuous random variable A random variable is said to be continuous if it can
assume any value in some interval or set of intervals. It generally measures some
characteristic of an item.
Discrete random variable A random variable is said to be discrete if it can assume
only finite number of values and involve counting.
Expected Value (mean -  ) This value identifies the physical center of data.
Independence of two events: P( A B)  P( A)
P( B A)  P ( B )
23
Joint Probability: P( A  B)  P( A) P( B A)
Marginal Probability: P( A) 
P( A  B)
P( B A)
Multiplication of probabilities (dependent events): P( A  B)  P( A) P( B A)
Multiplication of probabilities (independent events): P( A  B)  P( A) P ( B )
Normal distribution Continuous bell shaped probability distribution that is
symmetrical on both sides of the mean. The area under the normal curve is measured
by determining the number of standard deviations the value of a random variable is
from the mean.
Objective probabilities Probabilities based on equally likely outcomes.
Random variable Associated numerical values with the outcomes of a random
experiment.
Relative Frequency probabilities Probabilities based on collection of data that
involves experimental results.
Sample mean The arithmetic average of a set of sample data.
Sample Standard Deviation Denoted by s, it is equal to the positive square root of
the sample variance. Its units are the same as the units of the mean.
Sample Variance Denoted by s2 is defined to be the average squared deviation of
sample data from the mean.
Standard deviation It measures distance from the mean. This value is in the same
terms as the mean.
Subjective probabilities Probabilities based on judgment.
Variance(  2 ) Measures distance from the mean. It is equal to the square of the
standard deviation.
24
Exercises
1. Of the students in a high school 30% are freshmen, 35% are sophomores, 20% are
juniors, and the rest are seniors. One of the students has won $1 million in a state
lottery. Find these probabilities:
a. The student is a senior.
b. The student is a freshman or a sophomore.
c. The student is not a freshman.
2. If three lots of casings each contain 10% defectives, what is the probability of an
inspector:
a) Not finding any defectives if he inspects one casing from each of the three lots?
b) Finding all of them to be defective?
3. A coin will be tossed four times, and we are given the following probabilities for the
number of heads that occur:
P(0)=.0625 P(3)=.2500
P(1)=.2500 P(4)=.0625
P(2)=.3750
Find the probability of the following:
a. One or two heads
b. Less than three heads
c. Four heads
d. Less than two or more than three heads
4. The probabilities of 0, 1, 2, 3, 4, 5, 6, or 7 accidents during a weekday between the
hours of 1 and 6 A.M. respectively, .08, .15, .20, .25, .18, .07, .04, and .01. Find the
probability that on any weekday morning between those hours, there will be the
following:
a. Fewer than 3 accidents
b. Three or less accidents
c. Exactly 3 accidents
d. No accidents
e. More than 7 accidents
5. A firm that manufactures glassware has a four-step inspection process. The
probability that a defective will get past any one inspection station is reported to be
approximately 20% by the company.
a. Using this 20% figure, find the probability of a defective getting through
all four inspection stations without being detected.
b. What would your answer be if a fifth station is added, with a probability of
50% of catching defectives?
6. The probability is 90% that a machine will turn out an acceptable hex nut. If
successive pieces are independent of each other, find the probability of getting the
following:
a. Two pieces in a row that are unacceptable
b. A good piece and then a bad piece, in that order
c. A good piece and a bad piece, in any order
d. Three bad pieces in a row
25
7. A firm intends to introduce three new products. The manager of marketing research
estimates that Product A has a probability of success of .40, Product B has a
probability of success of .25, and Product C has a probability of success of .20.
Assume that each product’s success is independent of the others’ success or failure.
a. What is the probability that none are successful?
b. What is the probability that at least one is successful?
8. According to a computer printout, the number of items in four warehouses,
classified by dollar value, is shown in the table below:
Warehouse
Dollar Value
1
2
3
Under $10
910
840
680
$10 to less than $100
80
55
60
$100 or more
10
5
10
4
720
75
5
a. If one item is selected at random from Warehouse 1, what is the probability its
value will be under $10?
b. What is the probability that a randomly selected item will be an item from
Warehouse 1 with a value under $10?
c. What is the probability that a randomly selected item will have a value of $100
or more?
d. What is the probability that a randomly selected item will have a value of $10
or more?
e. What is the probability that a randomly selected item will have a value of $10
or more and be from Warehouse 1 or 2?
9. Which of the following pairs of events are mutually exclusive?
Event A
Event B
a. Bid on a job
Don’t bid on a job
b. Own stocks
Own bonds
c. Draw a queen from a deck
Draw a red card from the deck
of playing cards
d. Payment is late
Payment is not late
e. Make a profit
Make a large profit
10. Identify the complement of each of these events:
a. Product does not fail during the warranty period.
b. Draw a heart from a deck of 52 playing cards.
c. Draw a red card from a deck of cards.
d. No cars were damaged in the mishap.
e. More than three complaints were received.
f. Three or more complaints were received.
26
11. Suppose that P(A)=.30, P(B)=.70, and P(A  B)=.79.
a. What is P(A  B)?
b. Compute P ( A B ).
c. Are A and B mutually exclusive? Explain.
d. Are A and B collectively exhaustive? Explain.
e. Are A and B independent? Explain.
12. Suppose that P(A)=.30, P(B)=.80, P(A  B)=.15.
a. Are A and B mutually exclusive? Explain.
b. Find P( B  ).
c. Find P (A or B).
13. Seventy percent of the managers who work for an airline company have been with the
firm for more than five years. Fifty percent of the managers earn in excess of
$40,000 a year, and 35% of these managers have been with the firm for more than
five years. Suppose one manager is selected at random from a list of managers.
Determine the following probabilities:
a. The manager has worked for the firm for less than five years.
b. The manager has a salary in excess of $40,000 per year given that the
manager has worked for the firm for more than five years.
c. The manager has not worked for the firm for more than five years and
makes $40,000 or less per year.
d. Are earning more than 40,000 and having been with the company for more
than five years are independent?
14. A hotel has two independent burglar alarm systems. One has a probability of .98 of
operating and the other, an older model, a probability of .95 of operating. What is the
probability that neither operates if (or when) a burglary occurs?
15. The manager of a parts department needs a part for a very important job. Because the
part is not on hand, it must be ordered. One vendor is known to deliver 60% of rush
orders within two days. Another is known to deliver 75% of the rush orders within
two days. Because of importance of the job, the manager decides to order the part
from both suppliers.
a. What is the probability that neither part will arrive in 2 days?
b. What is the probability that two parts will arrive in two days?
c. What is the probability that exactly one part arrives?
16. Machine breakdowns are independent of each other. Suppose there are four
machines, and their respective probabilities of breakdown are 1%, 2%, 5%, and 10%
for any given day. Find these probabilities:
a. All break down on a given day.
b. None break down.
17. Given that H=75, K=80, ( H  K )=25, and the sample space total is 150, determine
each of the following:
a. ( H  K )
b. H 
c. ( H  K ) d. K 
27
18. As part of a research study investigators examined health records of 950 residents of a
home for the elderly. They found 500 with heart disease and 350 who had high blood
pressure. Moreover, a cross-check indicated that 200 of the persons with heart
disease had high blood pressure.
a. How many of the residents did not have high blood pressure?
b. How many had at least one of the two health problems?
c. How many residents had neither problem?
d. What percentage of residents with high blood pressure had heart disease?
19. Suppose that P(A)= .4, P(B)= .6, and P(A  B)= .24
a. Find P(A  B)
b. Compute P(A/B).
c. Are A and B mutually exclusive? Explain.
d. Are A and B independent? Explain.
20. A balanced die is rolled once; find the probability of getting the following:
a. A six
b. A five or seven
c. An even-numbered face
d. A number less than four
21. A pair of balanced dice will be rolled. What is the probability that the sum of their
faces is (a) seven; (b) greater than seven?
22. The following table gives the probability distribution, p(x), of the number of
customers, X, arriving in a bank during an arbitrary 4-minute period.
X
0
1
2
3
4
5
p(x)
.14
.29
.25
.18
.10
.04
a. Use Excel to find the expected number of customers arriving in the next 4minute period.
b. Find the standard deviation of number of customers arriving in the next 4minute period.
28
23. Use Excel with the following and calculate
X
-7 -2 0
5
13
P(X)
.20 .15 .30 .25 .15
a. Expected value
b. Variance
c. Standard deviation
24. Use the following data, which represent annual salaries in $1000 of 10 junior
executives selected at random from the Custom Car Company leaf plot.
25.2
23.8
26.7 34.2
31.6
29.7
28.4
26.7
30.1
27.3
a. Determine the sample mean
b. Determine the sample variance
c. Determine the sample standard deviation
25. A sample of 25 electronic components gave time to failure (in days) of
44
101
48
49
64
74
51
59
32
56
29
62
48
60
122
45
69
52
73
61
55
39
37
57
18
Use Excel to do the following calculations:
a.
b.
c.
Calculate the average time to failure for this sample.
Calculate the variance time to failure.
Calculate the standard deviation of times to failure
26. A firm that manufactures billiard tables suspects that 2% of its output is defective in
some way. If this suspicion is correct, find the following probabilities, assuming a
sample of nine tables:
a.
There are no defective tables.
b. There is at least one defective.
27. Of the students at a community college 41% smoke cigarettes. Six students are
selected to be interviewed for their views on smoking.
a.
Find the probability that none of those selected smoke.
b.
Find the probability that all those selected smoke.
c.
Find the probability that at least two of the six smoke.
28. Eight percent of the jobs done by a certain contractor require follow-up calls. A
contractor has done 14 jobs this week. Determine the probability that the number of
follow-up calls is the following:
a.
None
b.
One
c. Two
d.
At least one
29
29. A new-car dealer has found that 80% of the cars sold are returned to the service
department to correct manufacturing defects during the first 25 days after purchase.
For the 11 cars sold during a first 5-day period, determine the following probabilities:
a. All return within 25 days for service.
b. Only one does not return.
30. Assume that 9% of the hot dogs sold at a baseball stadium are ordered without
mustard. If seven persons order hot dogs, find the following probabilities:
a. All want mustard.
b. Only one does not want mustard.
c. Two do not want mustard.
31. An oil exploration firm finds that about 5% of the test wells it drills yield a deposit of
natural gas. If it drills six wells, find the probability that at least one will yield gas.
32. A real estate firm has noted that one out of ten prospective homeowners will make an
offer on a house if they return for a second visit. In ten such cases, find the
probability that none will make an offer.
33. A recent survey indicates that only 15% of the physicians in a rural area smoke. Two
out of eight doctors selected from a list supplied by the county medical board are
found to smoke.
a.
b.
Assuming the survey was correct, what was the probability of getting this
result?
What is the probability that at least one doctor in a rural area smokes?
34. Medical research suggests that 20% of the general population suffer adverse side
effects from a new drug. Suppose a doctor prescribes the drug for four patients.
What are the following probabilities?
a. None will have side effects.
b. All will have them.
c. At least one will have side effects.
35. Traffic records show that 25% of the vehicles stopped on an interstate highway
failed to pass a safety check. If 16 vehicles are stopped, find these probabilities:
a. Two or more failing.
b. Three or more failing.
30
36. Of the students in a junior college 75% change majors at least once during their
first year, according to the college registrar. If 11 students are selected from the
freshman class, find these probabilities:
a. All have changed majors at least once.
b. At least 9 have changed.
37. From the next shipment’s batch a random sample of 10 units has less than two
defectives. If a shipment actually has 5% defectives, what is the probability that it
will pass?
38. Compute the mean and standard deviation for the number of successes for these
cases:
a. n=25, p=.5
b. n=50, p=.2
c. n=80, p=.4
39. Draw a normal curve and shade in the desired area, and then determine the
required information:
a.
b.
c.
d.
Find the area to the right of z =1.0.
Find the area to the left of z = 1.0.
Find the area to the right of z = -.34.
Find the area to the right of z = -.56 and z = -.20.
40. Find the values of z that will produce these areas:
a.
b.
c.
d.
e.
The area to the left of z is .0505.
The area to the right of z is .0228.
The area between +z and –z is .0240.
The area between 0 and z is .4772.
The area between –z and +z is .9762.
41. Find the values of z that correspond to these probabilities:
a. The area to the right of z is .5000.
b. The area to the left of z is .3520.
c. The area to the left of z is .0107
d. The area between + z and – z is .9544.
e. The area between +z and –z is .6826
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42. A normally distributed population has a mean of 40 and a standard deviation of 3.
Find actual values if X for these z values:
a.
b.
c.
d.
+.10
+2.0
+.75
-2.53
43. A normal distribution has a mean of 50 and a standard deviation of 5. What
percentage of the population of values lies in these intervals?
a.
b.
c.
d.
From 40 to 50
From 40 to 45
From 56 to 60
From 45 to 55
44. After curing for 28 days, ordinary portland cement has an average compressive
strength of 4000 pounds per square inch (psi). Suppose this compressive strength
is normally distributed, with a standard deviation of 120 psi. Find these
probabilities for a 28-day compressive strength.
a. Less than 3900
b. More than 3850
c. More than 3880
45. Suppose the mean income in a large community can be reasonably approximated
by a normal distribution with a mean of $15,000 and a standard deviation of
$3000. What percentage of the population has incomes exceeding $18,600?
46. A supplier of wrought iron claims that his product has a tensile strength that is
approximately normal, with a mean of 50,000 psi and a variance of 810,000.
Assuming his claims are true, what percentage of sample measurements would
you expect to be as follows?
a. Less than 49,750 psi
b. More than  250 psi from 50,000 psi.
47. A process for producing pipes makes pieces with a mean diameter of 2.00 inches
and a standard deviation of .01 inch. Pipe with a diameter that varies by more
than .028 from the mean is considered defective. Assume normality. What
percentage of the pipes will be defective?
48. Weights of fish caught are approximately normal, with a mean of 4.5 pounds and
a standard deviation of .5 pound.
a. What percentage of the fish weigh less than 4 pounds?
b. What percentage of the fish weigh within a pound of the average weight?
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49. The amount of beef in a 12-ounce can manufactured by Super Suds Beverages,
Inc., is known to be well approximated by a normal distribution, with a mean of
12 ounces and a standard deviation of .25 ounce.
a. What percentage of cans should have less than 11.60 ounces?
b. What percentage of cans will not vary by more than .30 ounce from the
mean?
50. A single card is drawn from a deck of 52 shuffled cards. Find the probability of
getting the following:
a. A queen
b. A face-card
c. A black card
d. A club
e. A nine of diamonds
f. A red 3 or a black 5
g. What type of method was used to assign these probabilities?
51. There are 80 beads in a bowl:
COLOR
Black
Blue
Red
Orange
Green
NUMBER
20
25
15
12
8
80
After mixing the beads one is selected. Find the probability that the one selected
is:
a.
b.
c.
d.
e.
f.
Green
Black
Blue or Red
Not Orange
Orange or Green
What type of method was used to assign these probabilities?
52. A random sample of 50 employees showed 10 with high blood pressure. What is
the probability that if another employee is selected he will have a high blood
pressure?
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53. Compute the mean and the standard deviation for the percentage of successes for
the following cases:
a. n=25, p=.5
b. n=50, p=.2
c. n=80, p=.4
54. Compute the coefficient of variation for the data in problem 22 and problem 23
and comment on which coefficient of variation is more favorable and why.
55. Compute the coefficient of variation for the data in problem 24 and 25 and
comment on which coefficient of variation is more favorable and why?
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