business statistics
DESCRIPTION
Business statisticsTRANSCRIPT
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CHAPTER 1
INTRODUCTION TO STATISTICAL ANALYSIS
Reading
Newbold 1.1, 1.3, parts of 1.2.
Anderson, Sweeney, and Williams Chapter 1
Wonnacott and Wonnacott Chapter 1
James T Mc Clave, P. George Benson Chapter 1
Introductory Comments
This Chapter sets the framework for the book. Read it carefully, because the ideas
introduced are a basis to this subject and research Methodology.
1. Random Sampling, Deductive and Inductive Statistics.
Random Sampling
Only in exceptional circumstances is it possible to consider every member of the
population. In most cases only a sample of the population can be considered and
the results contained from this sample must be generalized to apply to the
population.
In order that these generalizations should be accurate the sample must be random,
that is, every possible sample has an equal chance of selection and the choice of a
member of the sample must not be influenced by previous selection; this is simple
random sampling.
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Example 1
Suppose that a population consists of six measurements, 1, 2, 3, 4, 5, and 7. List
all possible different samples of two measurements that could be selected from
the population. Give the probability associated with each sample in a random
sample of 2n measurement selected from the populations.
Solution
All possible samples are listed below
Sample Measurements 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1,2
1,3
1,4
1,5
1,7
2,3
2,4
2,5
2,7
3,4
3,5
3,7
4,5
4,7
5,7
Now let us suppose that I draw a single sample of n = 2 measurement from the 15
possible sample of two measurements. The sample selected is called a random sample if
every sample had an equal probability (1/15) being selected.
It is rather unlikely that we would ever achieve a truly random sample, because the
probabilities of selection will not always be exactly equal. But we do the best we can.
One of the simplest and most reliable ways to select a random sample of n measurements
from a population is to use a table of random numbers (See Appendix B). Random
number tables are constructed in such a way that, no matter where you start in the tables
no matter what direction you move, the digits occur randomly and with equal probability.
Thus if we wished to choose a random sample of n = measurements from a population
containing 100 measurements, we could label the measurements in the population from
0 to 99 (or 1 to 100). Then referring to Appendix Vii and choosing a random starting
point, the next 10 two-digit numbers going across the page would indicate the labels of
the particular measurements to be included in the random sample. Similarly, by moving
up or down the page, we would also obtain a random sample.
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Example 2
A small community consists of 850 families. We wish to obtain a random sample of 20
families to ascertain public acceptance of a wage and price freeze. Refer to Appendix B
to determine which families should be sampled.
Solution
Assuming that a list of all families in the community is available such as a telephone
directory), we could label the families from 0 to 849 (or equivalently, from 1 to 850).
Then referring to the Appendix, we choose a starting point. Suppose we have decided to
start at line 1, column 4. Going down the page we will choose the first 20 three-digit
numbers between 000 and 849 from Table B, we have
511 791 099 671 152
584 045 783 301 568
754 750 059 498 701
258 266 105 469 160
These 20 members identify the 20 families that are to be included in our example/
Deductive and Inductive Statistics.
The reasoning that is used in statistics hinges on understanding two types of logic,
namely deductive and inductive logic. The type of logic that reasons from the particular
(sample) to the general (Population) is known as inductive logic, while the type that
reasons from the general to the particular is known as deductive logic.
Learning Objectives
After working through this chapter, you should be able to:
Explain what random sampling is
Explain the difference between a population and a sample
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CHAPTER 2
METHODS OF ORGANISING AND PRESENTING DATA
Reading
Newbold Chapter 2
James T Mc Clave and P George Benson Chapter 2
Tailoka Frank P Chapter 3
Introductory Comments
This Chapter contains themes to do with the understanding of data. We find graphical
representations from the data, which allow one to easily see its most important
characteristics. Most of the graphical representations are very tedious to construct
without the use of a computer. However, one understands much more if one tries a few
with pencil and a paper.
Graphical Representations Of Data
Types of business data; methods of representation of qualitative data, cumulative
frequency distribution.
Types of business data. Although the number of business phenomena that can be
measured is almost limitless, business data can generally be classified as one of two
types: quantitative or qualitative.
Quantitative data are observations that are measured on a numerical scale. Examples of
quantitative business data are:
i. The monthly unemployment percentage ii. Last years sales for selected firms. iii. The number of women executives in an industry.
Qualitative data is one that is not measurable, in the sense that height is measured, or
countable, as people entering a store. Many characteristics can be classified only in one
of asset of category. Examples of qualitative business data are:
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i) The political party affiliations of fifty randomly selected business executives.
Each executive would have one and only one political party affiliation.
ii) The brand of petrol last purchased by seventy four randomly selected car owners.
Again, each measurement would fall into one and only one category.
Notice that each of the examples has nonnumerical or qualitative measurements.
Graphical methods for describing qualitative data.
(a) The Bar Graph
For example, suppose a womans clothing store located in the downtown area of a large city wants to open a branch in the suburbs. To obtain some information
about the geographical distribution of its present customers, the Store manager
conducts a survey in which each customer is asked to identify her place of
residence with regard to the citys four quadrants. Northwest (NW), North east (NE), Southwest (SW), or Southeast (SE). Out of town customers are excluded
from the survey. The response of n = 30 randomly selected resident customers
might appear as in Table 1.1 (note that the symbol n is used here and throughout
this course to represent the sample size i.e. the number of measurements in a
sample). You can see that each of the thirty measurements fall in one and only
one of the four possible categories representing the four quadrants of the city.
Table 1.1. Customer resident Survey: n = 30
Customer Resident Customer Residence Customer Residence
1
2
3
4
5
6
7
8
9
10
NW
SE
SE
NW
SW
NW
NE
SW
NW
SE
11
12
13
14
15
16
17
18
19
20
NW
SE
SW
NW
SW
NE
NE
NW
NW
SW
21
22
23
24
25
26
27
28
29
30
NE
NW
SW
SE
SW
NW
NW
SE
NE
SW
A natural and useful technique for summarizing qualitative data is to tabulate the
frequency or relative frequency of each category.
Definition:
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The frequency for a category is the total number of measurements that fall in the
category. The frequency for a particular category, say category i will be denoted by the
symbol if .
The relative frequency for a category is the frequency of that category divided by the
total number of measurements; that is, the relative frequency for category I is
Relative frequency = n
f i
Where n = total number of measurements in the sample
if = frequency for the i category.
The frequency for a category is the total number of measurements in that category,
whereas the relative frequency for a category is the proportion of measurements in the
category. Table 1.2 shows the frequency and relative frequency for the customer
residences listed in Table 1.1. Note that the sum of the frequencies should always equal
the total number of measurements in the sample and the sum of the relative frequencies
should always equal 1 (except for rounding errors) as in Table 1.2.
Category Frequency Relative Frequency
NE
NW
SE
SW
5
11
6
8
5/30 = .167
11/30 = .367
6/30 = .200
8/30 = .267
Total 30 1
A common means of graphically presenting the frequencies or relative frequencies for
qualitative data is the bar chart. For this type of chart, the frequencies (or relative
frequencie) are represented by bars-one bar for each category.
The height of the bar for a given category is proportional to the category frequency (or
relative frequency). Usually the bars are placed in a vertical position with the base of the
bar on the horizontal axis of the graph. The order of the bars on the horizontal axis is
unimportant. Both a frequency bar chart and a relative frequency bar chart for the
customers residence are shown in Figure 1.1.
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10
Relative
5 Frequency
Frequency
0
NE NW SE SW
Residential quadrant
a) A frequency bar chart.
.50
.25
0
NE NW SE SW
Residential Quadrant
b) A Relative Frequency bar chart.
Figure 1.1
b) The Pie Chart
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The second method of describing qualitative data sets is the pie chart. This is
often used in newspaper and magazine articles to depict budgets and other
economic information. A complete circle (the pie) represents the total number of
measurements. This is partitioned into a number of slices with one slice for each
category. For example, since a complete circle spans 360o, if the relative
frequency for a category is .30, the slice assigned to that category is 30% of 360
or (.30) (36) = 108o.
108o
Figure 1.2 The portion of a pie char corresponding to a relative frequency of .3.
Graphical Methods for Describing Quantitative Data.
The Frequency Histogram and Polygon.
The histogram (often called a frequency distribution) is the most popular graphical
technique for depicting quantitative data. To introduce the histogram we will use thirty
companies selected randomly from the 1980 Financial Magazine (the top 500 companies
in sales for calendar year 1979). The variable X we will be interested in is the earnings
per share (E/S) for these thirty companies. The earnings per share is computed by
dividing the years net profit by the total number of share of common stock outstanding. This figure is of interest to the economic community because it reflects the economic
health of the company.
The earnings per share figures for the thirty companies are shown (to the nearest ngwee)
in Table 1.3.
Company E/S Company E/S` Company E/S
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1
2
3
4
5
6
7
8
9
10
1.85
3.42
9.11
1.96
6.48
5.72
1.72
.8.56
0.72
6.28
11
12
13
14
15
16
17
18
19
20
2.80
3.46
8.32
4.62
3.27
1.35
3.28
3.75
5.23
2.92
21
22
23
24
25
26
27
28
29
30
2.75
6.58
3.54
4.65
0.75
2.01
5.36
4.40
6.49
1.12
How to construct a Histogram
1. Arrange the data in increasing order, from smallest to largest measurement.
2. Divide the interval from the smallest to the largest measurement into between five
and twenty equal sub-intervals, making sure that:
a) Each measurement falls into one and only one measurement class.
b) No measurement falls on a measurement class boundary.
Use a small number of measurement classes if you have a small amount of
data; use a larger number of classes for large amount of data.
3. Compute the frequency (or relative frequency) of measurements in each
measurement class.
4. Using a vertical axis of about three-fourths the length of the horizontal axis, plot
each frequency (or relative frequency) as a rectangle over the corresponding
measurement class.
Using a number of measurements, n = 30, is not large, we will use six classes to
span the distance between the smallest measurements, 0.72, and the largest
measurement, 9.11. This distance divided by 6 is equal to
Largest measurement smallest measurement = 9.11 0.72 Number of intervals 6
1.4
By locating the lower boundary of the first class interval at 0.715 (slightly below the
smallest measurement) and adding 1.4, we find the upper boundary to be 2.115. Adding
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1.4 again, we find the upper boundary of the second class to be 3.515. Continuing this
process, we obtain the six class intervals shown in the table below. Note that each
boundary falls on a 0.005 value (one significant digit more than the measurement), which
guarantees that no measurement will fall on a class boundary.
The next step is to find the class frequency and calculate the class relative frequencies
Class Measurement
Class
Class
Frequency
Class relative
Frequency
1
2
3
4
5
6
0.715 2.115 2.115 3.515 3.515 4.915 4.915 6.315 6.315 7.715 7.715 9.115
8
7
5
4
3
3
8/30 = .267
7/30 = .233
5/30 = .167
4/30 = .133
3/30 = .100
3/30 = .100
Total 30 1.00
Table 1.4
Definition
The class frequency for a given class, say class i, is equal to the total number of
measurements that fall in that class. The class frequency for class I is denoted by the
symbol if .
Definition
The class relative frequency for a given class, say class i, is equal to the class frequency
divided by the total number n of measurements, i.e.
Relative frequency for class i = n
fi
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8
6
4
2
0
0.517 2.115 3.515 4.915 6.315 7.715 9.115
Earnings per share
a) Frequency Histogram.
.3
.2
.1
0.715 2.115 3.515 4.915 6.315 7.715 9.115
Earnings per share
(b) Relative Frequency histogram
Cumulative Frequency Distribution
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It is often useful to know the number or the proportion of the total number of
measurements that are less than or equal to those contained in a particular class. These
quantities are called the class cumulative frequency and the class cumulative relative
frequency respectively.
For example, if the classes are numbered from the smallest to the largest values of x, 1, 2,
3, 4, . . . , then the cumulative frequency for the third class would equal the sum of the
class frequencies corresponding to classes 1, 2, and 3.
Cumulative frequency for class 3213 fff
Similarly, cumulative relative frequency for class n
fff 3213
where n is the total
number of measurements in the sample.
Cumulative frequencies and cumulative relative frequencies for earning per share data.
Class No. Measurement
class
Class
Frequency
Cumulative
frequency
Class Relative
Frequency
Class
Cumulative
Relative
Frequency
1
2
3
4
5
6
0.715 - 2.115
2.115 3.515
3.155 4.915
4.915 6.315
6.315 7.715
7.715 9.115
8
7
5
4
3
3
30
8
(8 + 7) = 15
(15 + 5) = 20
(20 + 4) = 24
(24 + 3) = 27
(27 + 3) = 30
8/30 = .267
7/30 = .233
5/30= .167
4/30 = .133
3/30 = .100
3/100 = .100
8/30 =.267
15/30 = .500
20/30 = .667
24/30 = .800
27/30 = .900
30/30 = 1.00
Cumulative relative frequency Distribution for earnings per share data.
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1.0
Cumulative
Relative .8
Frequency
.6
.4
.2
0.715 2.115 3.115 4.915 6.315 7.715 9.115
Earnings per share
Learning Objective
After working through this Chapter you should be able to:
Draw a pie chart, bar chart and also construct frequency tables, relative frequencies, and histogram.
Interpret the diagrams. You will understand the importance of captions, axis labels and graduation of axes.
CHAPTER 3
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DESCRIPTIVE MEASURES
Reading
Newbold Chapter 2
Wonnacott and Wonnacolt Chapter 2
Tailoka Frank P. Chapter 4
James T McClave , Lawrence Lapin L and P George Benson Chapter 3
Introductory Comments
This Chapter contains themes which allow one to easily se the most important
characteristics of data. The idea is to find simple numbers like the mean, variance which
will summarize those characteristics.
3. Numerical Description of Data.
The Mode; A measure of Central tendency.
Definition.
The mode is the measure that occurs with the greatest frequency in the data set.
Because if emphasizes data concentration, the mode has application in marketing
as well as in description of large data sets collected by state and federal agencies.
Unless the data set is rather large, the mode may not be very meaningful. For
example, consider the earning per share measurements for the thirty financial
companies we used in the previous chapter. If you were to re-examine these data,
you would find that none of the thirty measurements is duplicated in this sample.
This, strictly speaking, all thirty measurements are mode for this sample.
Obviously, this information is of no practical use for data description. We can
calculate a more meaningful mode by constructing a relative frequency histogram
for the data. The interval containing the most measurements is called the modal
class and the mode is taken to be the midpoint of this class interval.
The modal class, the one corresponding to the interval 0.715 2.115 lies to the left side of the distribution. The mode is the midpoint of this interval; that is
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Mode = 415.12
115.2715.0
In the sense that the mode measures data concentration, it provides a measure of central
tendency of the data.
The Arithmetic mean
A measurement of Central Tendency
The most popular and best understood measure of central Tendency for a quantitative
data set is the arithmetic (or simply the mean):
Definition
The mean of a set of quantitative data is equal to the sum of the measurements divided by
the number of measurement contained in the data set. The mean of a sample is denoted
by x (read x bar) and represent the formula for this calculation as follows:-
Example 1
Calculate the mean of the following five simple measures,. 5, 3, 8, 5,6.
Solution
Using the definition of the sample mean and demand shorthand notation we find
.4.55
27
5
65835
5
5
11
ixx
The mean of this sample is 5.4
The sample mean will play an important role in accomplishing our objective of making
inferences about populations based on sample information. For this reason it is important
to use a different symbol when we want to discuss the mean of a population of
measurement s i.e. the mean of the entire set of measurements in which we are interested.
We use the Greek letter (mu) for the population mean
The Median: Another measure of Central Tendency
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The median of a data set is the number such that half the measurements fall below the
median and half fall above. The median is of most value in describing large data sets. If
the data set is characterized by a relative frequency histogram, the median is the point on
the x-axis such that half the area under the histogram lies above the median and half lies
below. For a small, or even a large but finite, number of measurements, there may be
many numbers that t satisfy the property indicated in the figure on the next page. For this
reason, we will arbitrarily calculate the media of a data.
Calculating a median
1. If the number of n of measurements in a data set is odd, the median is the middle
number when the measurements are arranged in ascending (or descending) order.
2.. If the number of n of measurements is even, the median is the mean of the two
middle measurements when the measurements are arranged in ascending (or
descending) order.
Example 2
Consider the following sample of n = 7 measurements.
5, 7, 4, 5, 20, 6, 2
a) Calculate the median of this sample
b) Eliminate the last measurement (the 2) and calculate the median of the remaining n = 6 measurements.
Solution
a) The seven measurements in the sample are first arranged in ascending order
2, 4, 5, 5, 6, 7, 20
Since the number of measurements is odd, the median is the middle measure.
Thus, the median of this sample is 5.
b) After removing the 2 from the set of measurements, we arrange the sample
measurements in ascending order as follows:
4, 5, 5, 6, 7, 20
Now the number of measurements is even, and so we average the middle two
measurements. The median is (5+6)/2 = 5.5.
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Comparing the mean and the median
1. If the median is less than the mean, the data set is skewed to the right.
Relative
Frequency
Median Mean
Rightward Skewness measurement units
deviationdards
medianmean
deviationdards
ModeMeanSkewness
tan
)(3
tan
2. The median will equal the mean when the data set is symmetric.
Median Mean
Measurement unit
Symmetry
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3. If the median is greater than the mean, the data set is skewed to the left.
Mean Median
The range: A measure of variability
Measures of Variation
Definition:
The range of a data. Set is equal to the largest measurement minus the smallest measure.
When dealing with grouped data, there are two procedures which are not adopted for
determining the range.
1. Range = class mark of highest class class mark of lowest class. 2. Range = upper class boundary of highest class lower class boundary of lowest
class.
Variance and Standard Deviation
The Sample Variance for a sample of n measurements is equal to the squared distances
from the mean divided by (n-1). In symbols using 2S to represent the simple variances,
1
)(1
2
2
n
xx
S
n
i
i
The second step in finding a meaningful measure of data variability is to calculate the
standard deviation of the data set.
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The sample standard deviation , s, is defined as the positive square root of the sample
variance, 2S thus,
1
)(1
2
2
n
xx
SS
n
i
i
The corresponding quantity, the population standard deviation, measure the variability of
the measurements in the population and is denoted by (sigma). The population
variances will therefore be denoted by 2 .
Example 3
Calculate the standard deviation of the following sample. 2, 3, 3, 3, 4.
Solution
For this set of data, .3x Then
71.05.04
2
15
)34()33()23()32( 2222
S
Shortcut formular for simple variance
1
1
)()(
2
1
1
1
2
2
2
n
n
x
x
n
n
tmeasuremensampleofsumtmeasuremensampleofsquareofsum
S
n
in
i
i
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Example 4
Use the shortcut formula to compute the variances of these two samples of five measures
each.
Sample 1: 1, 2, 3, 4, 5 Sample 2:2, 3, 3, 3, 4
Solution
We first work with sample 1. The quantities needed are:
n
i
x1
1 = 1 + 2 + 3 + 4 + 5 = 15, and
552516941
543215
1
222222
1
i
x
5.24
10
4
4555
4
5
)15(55
15
5
2
1
25
12
12
n
i
i
ix
x
S
Similarly, for sample 2 we get
5
1i
ix = 2 + 3 + 3 + 3 + 4 = 15
Add 47169994433325
1
222222
1 i
x
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Then the variance for sample 2 is
5.04
2
4
4547
4
5
)15(47
15
5
2
1
25
12
12
n
i
i
ix
x
S
Example 5
The earnings per share measurements for thirty companies selected randomly from 1980
Financial/Daily mail are listed here. Calculate the sample variance 2S and the standard
deviation, S, from these measurements.
1.85
3.42
9.11
1.96
6.48
5.72
1.72
8.56
0.72
6.28
2.80
3.46
8.32
4.62
3.27
1.35
3.28
3.75
5.23
2.92
2.75
6.58
3.54
4.65
0.75
2.01
5.36
4.40
6.49
1.12
Solution
The calculation of the sample variance , 2S , would be very tedious for this example if we
tried to use the formula,
130
)(30
1
2
2
i
i xx
S
because it would be necessary to compute all thirty squared distances from the mean.
However, for the shortcut formula we need only compute:
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4331.5
29
30
)47.122(5239.657
130
30
5239.57.6)12.1(...)42.3()85.1(
47.12212.1...42.385.1
230
1
230
1
1
2
2
30
1
2222
30
1
i
i
i
i
i
i
i
x
x
S
x
andx
Notice that we retained four decimal places in the calculation of 2S to reduce rounding
errors, even though the original data were accurate to only two decimal places.
The standard deviation is
33.24331.52 SS
Interpreting the Standard Deviation
If we are comparing the variability of two samples selected from a population, the sample
with the larger standard deviation is the more variable of the two. Thus, we know how to
interpret the standard deviation on a relative or comparative basis, but we have not
explained how it provides a measure of variability for a single sample.
One way to interpret the standard deviation as a measure of variability of a data set would
be to answer questions each as the following. How many measurements are within 1
standard deviation of the mean? How many measurements are within 2 standard
deviation of the mean? For a specific data set, we can answer the questions by counting
the number of measurements in each of the intervals. However, if we are interested on
obtaining a general answer to these questions, the problem is more difficult. There are
two guidelines to help answer the questions of how many measurements fall within 1, 2,
and 3 standard deviations of the mean. The first set, which applied to any sample, is
derived from a theorem proved by the Russian Mathematician Chebyshev. The second
set, the Empirical Rule is based on empirical evidence that has accumulated over time
and applies to samples that posses mould shaped frequency distributions those that are
approximately symmetric, with a clustering of measurement about the mid point of the
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distribution (the mean, median and mode should all be about the same) and that laid off
as we move away from the center of the histogram.
Aids to the Interpretation of a Standard deviation.
1. A rule (from Chebyshevs theorem) that applied to any sample of measure regardless of the shape of the frequency distribution.
a. It is possible that none of the measurements will fall within 1 standard
deviation of the means ).( SxtoSx
b. At least of the measurement will fall within 2 standard deviations of the
mean ).22( SxtoSx
c. At least 8/9 of the measurements will fall within 3 standard deviations of
the mean ).33( SxtoSx
2. A rule of thumb, called the empirical rule, that applies to samples with frequency
distributions that are mould-shaped:
a) Approximately 68% of the measurements will fall within 1 standard
deviation of the mean ).( SxtoSx
b) Approximately 95% of the measurements will fall within 2 standard
deviations of the mean ).22( SxtoSx
c) Essentially all the measurements will fall within 3 standard deviations of
the mean ).33( SxtoSx
Example 6
Refer to the data for earnings per share for thirty companies selected randomly from the
1980 Financial/Daily Mail. .33.2,08.4 Sx Calculate the fraction of the thirty
measurements that lie within the intervals ,3,2, SxandSxSx and compare the
results with those of the Chebyshev and Empirical rule.
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Solution
), SxSx )41.6,75.1()33.208.4,33.208.4(
A check of the measurements show that 19 of the 30 measurements i.e., approximately
63% are within 1 standard deviation of the mean.
)74.8,58.0()66.408.4,66.408.4()2,2( SxSx
Contains 29 measurements, or approximately 97% of the n = 30 measurements. Finally
the 3 standard deviation interval around x
).07.11,91.2()99.608.4,99.608.4()3,3( SxSx
contains all the measurements. These 1, 2 and 3 standard deviations percentages (63, 97,
and 100) agree fairly well with the approximations of 68%, 95% and 100%, given by the
Empirical Rule for mould-shape distributions.
Example 7
The aid for interpreting the value of a standard deviation can be put to an immediate
practical use as a check on the calculation of the standard deviation. Suppose you have a
data set for which the smallest measurement is 20 and the largest is 80. You have
calculated the standard deviation of the data set to be S = 190.
How can you use the Chebyshev or empirical rule to provide a rough check on your
calculated value of S?
Solution
The larger the number of measurements in a data set, the greater will be the tendency for
very large or very small measurements (extreme values) to appear in the data set. But
from the Rules, you know that most of the measurements (approximately 95% if the
distribution is mould-shaped) will be within 2 standard deviations of the mean, and
regardless of how many measurements are in the data set, almost all of them will fall 3
standard deviations of the mean. Consequently we would expect the range to be between
4 and 6 standard deviations i.e. between 4s and 6s.
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Range largest measurement smallest measurement = 80 20 = 20.
Sx 2 x Sx 2
Range 4S
The relation between the range and the Standard deviation.
Then if we let the range equal 6S, we obtain
Range = 6S
60 = 6S
S = 10
Or, if we let the range equal 4S, we obtain a larger (and more conservative) value for S,
namely
Range = 4S
60 = 6S
S = 15
Now you can see that it does not make much difference whether you let the range equal
4S (which is more realistic for most data set) or 6S (which is reasonable for large data
sets). It is clear than your calculated value, S = 190, is too large, and you should check
your calculations.
-
26
Calculating a mean and standard Deviation from Grouped data
If your data have been grouped in classes of equal width and arranged in a frequency
table, you can use the following formulas to calculate x , S2, and S
ix Midpoint of the ith class
if = Frequency of the ith class
K = Number of classes
2
2
1
12
12
1
1
SS
n
n
fx
fx
S
n
fx
x
K
i
K
i
ii
i
K
i
ii
Example 8
Compute the mean and standard deviation for the earnings per share data using the
grouping shown in the frequency Table 1.4.
Solution
The six class interval, midpoints, and frequencies are shown in the accompanying table.
Table 1.4 Earnings per share
Class Class Midpoint Class frequency
if
0.715 2.115
2.115 3.515
3.515 4.915
4.915 6.315
6.315 7.015
7.715 9.115
1.415
2.815
4.215
5.615
7.015
8.415
8
7
5
4
3
3
30 ifn
-
27
1
03.430
85.120
30/)3)(415.8(...)5)(215.4()7)(815.2()8)(415.1(
2
1
12
12
1
n
n
fx
fx
S
n
fx
x
K
i
K
i
ii
i
K
i
ii
We found
K
i
ii fx1
= 120.85 when we calculated x, therefore
.35.25060.5
5060.5
29
82408.48649875.646
130
30/)85.120())3()415.8(...)7()815.2()8()415.1(( 32222
S
S
You will notice that values of ,,2Sx and S from the formulas for grouped data usually do
not agree with these obtained for the raw data ( 03.4x and S = 2.311). This is because
we have substituted the value of the class mid point for each value of x in a class
interval. Only when every value of a x in each class is equal to its respective class
midpoint will the formulas for grouped and for ungrouped data give exactly the same
answers for ,,2Sx and S. Otherwise, the formulas for grouped data will give only the
approximations to these numerical descriptive measures.
Measures of Relative Standing
Descriptive measures of the relationship of a measurement to the rest of the data are
called measure of relative standing.
One measure of relative standing of a particular measurement is its percentile ranking.
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28
Definition
Let nxxx ,...,, 21 be a set of n measurements arranged in increasing (or decreasing)
order. The pth percentile is a number x such that p% of the measurements fall below the
pth percentile and (100 p)% fall above it.
For example: if oil company A report that its yearly sales are in the 90th
percentile of all
companies in the industry, the implication is that 90% of all oil companies have yearly
sales less that As, and only 10% have yearly sales exceeding company As.
Relative
Frequency
.90
.10
Company As sales. Yearly sales.
Another measure of relative standing in popular use is the Z-score. The Z-score makes
use of the mean and standard deviation of the data set in order to specify the location of a
measurement.
Definition
The sample Z-score for a measurement x is
S
xxZ
The population Z-Score for a measurement x is
xZ
The Z-score represents the distance between a given measurement x and the mean
expressed in standard units.
-
29
Example 9
Suppose 200 steel workers are selected, and the annual income of each is determined.
The mean and standard deviation are 000,2,000,14 KSKx
Suppose Chipos annual income is K12, 000 what is his sample Z-score?
K8,000 K12,000 K14,000 K20,000
Sx 3 x x Sx 3
Annual income of steel workers.
Solution
Chipos annual income lies below the mean income of the 200 steel workers.
We compute 0.12000
1400012000
S
xxZ
Which tells us that Chipos annual income is 1.0 standard deviation below the sample mean, in short, his sample Z-score is 1.0.
Example 10
Suppose a female bank executive believes that her salary is low as a result of sex
discrimination. To try to substantiate her belief, she collects information on the salaries
of her counterparts in the banking business. She finds that their salaries have a mean of
K17, 000 and a standard deviation of K1, 000. Her salary is K13, 500. Does this
information support her claim of sex discrimination?
Solution
The analysis might proceed as follows: First, we calculate the Z-score for the womans salary with respect to those of her male counterparts. Thus
5.31000
1700013500
Z
-
30
The implication is that the womans salary is 3.5 standard deviations below the mean of the male distribution. Furthermore, if a check of the male salary data shows that the
frequency distribution is mould-shaped, we can infer that very few salaries in this
distribution should have a Z-score less than 3, as shown in the figure.
Relative
Frequency
Z-Score = -3.5
13.500 17,000 Salary (K)
Male Salary Distribution
Therefore, a Z-score of 3.5 represents either a measurement from a distribution different from the male salary distribution or a very unusual (highly improbable) measurement for
the male salary distribution.
Well, which of the two situations do you think prevails? Do you think the womans salary is simply an usually low one in the distribution of salaries, or do you think her
claim of salary discrimination is justified? Most people would probably conclude that
her salary does not come from the male salary distribution.
However, the careful investigator should require more information before inferring sex
discrimination as the case. We would want to know more about the data collection
technique the woman used, and more about her competence at her job. Also perhaps
other factors like the length of employment should be considered in the analysis.
-
31
Learning Objectives
After working through this Chapter you should be able to
Calculate the arithmetic mean, standard deviation, variance, median, and quartiles for grouped or ungrouped data.
Explain the use of all the above quartiles.
-
32
Sample Examination Questions
1. (a) Briefly state, with reasons, the type of chart which would best convey the information for each of the following:
(i) Students at the University classified by programme of study.
(ii) Members of a professional association classified by age.
(iii) Numbers of cars taxed for 2002, 2003 and 2004 in areas A, B and C of a city.
(b) The weekly cost (K) of rented accommodation was recorded for 100
students living in an area.
Amount in Thousand of
Kwachas
Frequency
0 4 3 5 9 17
10 14 24 15 19 31 20 24 19 25 - 29 6
(i) Draw a histogram.
(ii) Give the median and the interquartile range.
(iii) Calculate the mean, mode, and standard deviation.
(iv) What conclusions can you draw from the data?
-
33
2. The data below are per capita per week numbers of cigarettes sold for 38 states in a country.
19.20 26.82 19.24 27.18 25.96 30.14
29.27 21.10 28.91 29.92 29.64 21.94
22.58 29.92 26.91 43.40 30.18 23.86
28.56 24.75 24.32 24.78 22.17
20.96 27.38 24.44 26.89 41.46
21.08 23.57 15.80 32.10 24.44
29.04 31.34 29.60 23.12 17.08
(a) Plot the data using an approximate graphical method.
(b) Give the mean, the median and the mode.
(c) Assuming this is a normal distribution, and given a standard deviation of these figures of 4.387, what proportion of the states would expect to have
more than 20 cigarettes smoked per capita per week?
(d) How does this compare with the actual situation as shown in the table above?
3. (a) Briefly state, with reasons, the type of chart which would best convey in each of the following:
(i) A countrys total import of cigarettes by source.
(ii) Students in higher education classified by age.
(iii) Number of students registered for secondary school in year 2001, 2002 and 2003 for areas X, Y, and Z of a country.
(b) The weekly cost (K000) of rented accommodation was recorded for 40 students living in an area.
35 56 33 30 31 55 29 27
21 32 43 33 29 27 30 29
26 26 27 26 35 32 28 27
31 27 33 24 27 28 33 49
22 19 46 36 26 38 36 55
-
34
(i) Summarize the data in a frequency distribution table.
(ii) Calculate the mean and the standard deviation from your frequency table.
(iii) Plot a histogram for these data. What is the value of the median?
(iv) What conclusions can you draw from these data?
4. (a) Given below is a sample of 25 observations, calculate:
(i) The range (ii) The arithmetic mean
(iii) The median (iv) The lower quartile
(v) The upper quartile (vi) The quartile deviation
(vii) The mean deviation (viii) The standard deviation
5 18 29 42 50 61
8 20 33 43 54 63
10 21 35 46 56 67
11 25 39 48 58 69
14
(b) Explain the term measure of dispersion and state briefly the advantage and disadvantage of using the following measures of dispersion:
(i) Range
(ii) Mean deviation
(iii) Standard deviation
-
35
5. A machine produces the following number of rejects in each successive period of five minutes.
(a) Construct a frequency distribution from these data, using seven class
intervals of equal width.
(b) Using the frequency distribution, calculate:
(i) the mean (ii) the standard deviation
(c) Briefly explain the meaning of your calculated measures.
20 55 58 40 15 28 21 29 30 17
84 58 7 40 41 67 28 19 26 26
16 25 55 43 22 66 32 29 11 21
26 42 57 73 27 66 7 23 17 35
27 42 13 28 24 37 34 27 24 12
-
36
CHAPTER 4
PROBABILITY
Reading
Newbold Chapter 3
Tailoka Frank P Chapter 8
Wonnacott and Wonnacolt Chapter 3
Introductory Comments
Probability is more abstract than other parts of this subject, and solving the problems may
be difficult. The concepts are very important for statistics because it is the rules of
probability that allow one to reason about uncertainty. Independence and conditional
probability are important to understand clearly for the purpose of statistical investigation.
4. Elementary Probability
Counting Techniques. Introduction of the probability concept. The event and the
event relationships. Probability trees, conditional probability and statistical
independence.
Counting techniques: In calculating probabilities, it is essential to be able to work
out n(s) and n(E) as straight-forwardly as possible. Permutations and
combinations are very helpful here. We begin with the following basic principle.
Fundamental principle of counting. If two operations A, B are carried out, and
there are M different ways of carrying out A and k different ways of carrying out
B, then the combined A and B may be carried out in M x K different ways.
Example 1
Suppose a license plate contains two distinct letters followed by three digits with
the first digit not zero. How many different license places can be printed?
-
37
The first letter can be printed in 26 different ways, the second letter in 25 different ways
(since the letter printed first cannot be chosen for a second letter, the first digit in 9 ways
and each of the other two digits in 10 ways. Hence
26.25.9.10.10 = 585,000
Different plates can be printed.
Example 2.
A toy manufacturer makes a wooden toy in two parts, the top part may be coloured red,
white or blue and the bottom part brown, orange, yellow or green. How many differently
coloured toys can be produced?
A red top part may be combined with a bottom part of any of the four possible colours.
Similarly, either a white or a blue top part may be combined with each of the four
different coloured parts. Hence the number of different coloured toys is
1243
Permutations; An arrangement of a set of n objects in a given order is called a
permutation of the objects (taken all at a time). An arrangement of any nr of these objects in a given order is called an r-permutation or a permutation of the objects taken r at a time.
Example 3
Consider the set of letters a, b, c and d. Then
i) bdca, dcba and acdb are permutations of the 4 letters (taken all at a time).
ii) bad, adb and bca are permutations of 4 letters taken 3 at a time.
iii) ad, ca, da and bd are permutations of the 4 letters taken 2 at a time.
-
38
Example 4
The telephone switchboard in the company requires two operators whose chairs
(positions) are side by side. When the telephone operators go to lunch, two of the four
Secretaries take their places. If we make a distinction between the two operatorss positions, in how may ways can the four secretaries fill them?
We can answer this question by determining the number of possible permutations of 4
things taken 2 at a time. There are 4 secretaries, A, B, C and D, to fill the first position.
Once this position has been filled, there are only 3 secretaries to fill the second positions.
The figure below
Ways to fill Ways to fill second Counting the number of
First position position permutations
B 1
A C 2
D 3
A 4
B C 5
D 6
A 7
C B 8
D 9
A 10
D B 11
-
39
C 12
The tree diagram on the page illustrates that there are 4.3 = 12 possible permutations of
four things taken two at a time. Suppose that n is the number of distinct objects from
which an ordered arrangement is to be derived, and r is the number of objects in the
arrangement. The number of possible ordered arrangements is the number of
permutations of things taken r at a time. This is written symbolically as ),( rnP in
general, or rn P .
)1()1(..).2)(1(),( rnnnnrnP
We multiply the right-hand side of equation (1) by
)!/()!( rnrn
This is equivalent to multiplying by 1, we obtain
)!1(
!
)!(
)!)(1(...)2)(1(
)!(
)!1()1(..).2)(1(),(
n
n
rn
rnrnnnn
rn
nrnnnnrnP
Example 5
i) In a stock room, 5 adjacent bins are available for storing 5 different items. The
stock of each item can be stored satisfactorily in any bin. In how many ways can
we assign the 5 items to the 5 bins?
We get the answer by evaluating P(5, 5) which is
1201.2.3.4.5)!55(
!5)5,5(
P
ii) Suppose that there are 6 different parts to be stocked, but only 4 bins are
available.
To find the number of possible arrangements, we need to determine the number of
permutations of 6 things taken 4 at a time, which is
-
40
360!2
1.2.3.4.5.6
)!46(
!6)4,6(
P
Example 6
How many permutation are there of 3 objects, say, a, b and c?
There are 63.2.1!3)!33(
!3)3,3(
P such permutations.
These are abc, acb, bac, bca, cab, cba.
Permutation with repetitions:
The number of permutations of n objects of which 1n are alike, 2n are alike of another
kind . . . . rn are alike of a further kind, is given by
rnnnnwhere
nnn
n
...
!..!.!
!
21
21
Example 7
Find the number of permutation of the word ACCOUNTANTS
Total number of letters in ACCOUNTANTS is 11 out of which there are two Cs, two Ns, and two ts. So the required number of permutation s
.2494800!2!2!2!2
!11
Combinations
A combination is an arrangement of objects without regard to order.
-
41
Example 8
The combinations of the letters a, b, c, d taken 3 at a time are
{a, b, c}, {a, b, d}, (a, c, d}, (b, c, d} or simply
abc, abd, acd, bcd, . Observe that the following combinations are equal.
abc, acb, bac, bca, cab, cba.
That is, each denotes the same set a, b, c
The number of combinations of n objectives taken r at a time will be denoted by
r
nCorrnC ),( .
Example 9
We determine the number of combinations of the four letters, a, b, c, d taken 3 at a time.
Note that each combination consisting of three letters determine 3! = 6 permutations of
the letters in the combination.
Combinations Permutations
abc abc, acb, bac, bca, cab cba
abd abd, adb, bad, bda, dab, dba
acd acd, adc, cad, cda, dac, dca
bcd bcd, bdc, cbd, cbd, dbc, dcb
Thus the number of combinations multiplied by 3! Equals the number of permutations
-
42
)!(!
!),(
.4)3,4(;6!3242.3.4)3,4(
!3
)3,4(
)3,4()3,4(!3).3,4(
rnr
nrnCThus
abovenotedashenceCandPNow
P
orCPC
Example 10
A perfume manufacturer who makes 10 fragrances wants to prepare a gift package
containing 6 fragrances. How many combinations of fragrances are available?
The answer is
2101.2.3.4!.6
!6.7.8.9.10
)610(!6
!10)6,10(
C
Tree Diagrams
A tree diagram is a device used to enumerate all the possible outcomes of a sequence of
experiments where each experiment can occur in a finite number of ways. The
construction of tree diagrams is illustrated in the following examples.
Example 11
Find the product A x B x C where
A = {1, 2}, B{a, b, c} and C = {3, 4}. The tree diagram follows:
3 (1, a, 3)
a
4 (1, a, 4)
1 b 3 (1, b, 3)
4 (1, b, 4)
c
3 (1, c, 3)
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43
0
4 (1, c, 4)
a 3 (2, a, 3)
5 (2, a, 4)
b 3 (2, b, 3)
2
4 (2, b, 4)
3 (2, c, 3)
c
4 (2, c, 4)
Observe that the tree is constructed from left to right, and that the number of branches at
each prints corresponds to the number of possible outcomes of the next experiment.
Example 12
Mumba and Ened are to play a tennis tournament. The first person to win two games in a
row or who wins a total of three games wins the tournament. The following diagram
shows the possible outcomes of the tournament.
M M
M
M
M E E
E E
0
M
E M
M M
E
E
-
44
E E
Observe that there are 10 end points which correspond to the 10 possible outcomes of the
tournament.
MM, MEMM, MEMEM, MEMEE, MEE, EMM, EMEMM, EMEME, EMEE, EE
The path from the beginning of the tree to the end point indicates who won which game
in the individual tournament.
Basic Of Probability
Given a sample spaces S, we need to assign to each event that can be obtained from S a
number, called the probability of the event. This number will indicate the relative
likelihood of the various events.
For events that are equally likely, the probability of the event can be found from the
following basic probability principle. Then the probability that event E occurs, written P
(E), is
P(E) = m (1)
n
This same result can also be given in terms of the cardinal number of a set. Where n (E)
represents the number of elements in a finite set E. With the same assumptions given
above,
P(E) = n(E) . (2)
n(S)
-
45
Example 1
Suppose a fair coin is tossed twice. The sample space is S = (HH), (HT), (TH), (TT). Set S contains 4 outcomes, all of which are equally likely. (This makes n = 4 in the
formula (1) above.) Find the probability of the following outcomes.
a) E = (HT), (TH)
Event E contains two elements, so
P (E) = 2 = 1
4 2
By this result, a head or tail will show up 1/2 of the time when a fair coin is tossed
twice.
b) Two heads
Let event F = (HH) be the event two heads are observed when a fair coin is
tossed twice. Event F contains one element, so
P (F) =
c) Three heads
A fair coin tossed twice can never show three heads. If G is the event, then G =
, and P (G) =4
0 = 0.
The event is impossible.
-
46
Example 2
If a single paying card is drawn at random from an ordinary 52-card bridge deck,
find the probability of each of the following events.
a) An ace is drawn
There are four aces on the deck, out of 52 cards, so
P(ace) =13
1
52
4
b) A face card is drawn
Since there are 12 face cards
P (face card) =13
3
52
12
c) A spade is drawn
The deck contains 13 spaces, so
P (spade) = 4
1
54
13
d) A spade or heart is drawn
Besides the 13 spades, the deck contains 13 hearts, so
P (spade or heart) =2
1
52
26
-
47
Example 3
The Manager of a department store has decided to make a study on the size of purchases
made by people coming into the store. To begin he chooses a day that seems fairly
typical and gathers the following data. (Purchases have been rounded to the nearest
Kwacha) with sales tax ignored.
Amount of purchase Number of customer Probability (relative
frequency)
K0 and under
160 0.280
K2250 and under
K11250
84 0.147
K11250 and under
K13500
50 0.088
K13500 and under
K20250
136 0.239
K20250 and under
K22500
77 0.135
K22500 and over 63 0.111
570 1.000
Probability Distributions.
In Example 3 the outcomes were various purchase amounts, and a probability was
assigned to each outcome. By this process, a probability distribution can be set up; that is
to each possible outcome of an experiment, a number, called the probability of that
outcome, is assigned.
-
48
Example 4
Set up a probability distribution for the number of heads observed when a fair coin is
tossed twice.
_______________________________________
Number of heads Probability _______________________________________
0 1
4
1 2
4
2 1
4
_________
Total 1
_______________________________________
The probability distribution that was set up suggests the following properties of
probability.
Let S = S1, S2, S3, , Sn be the sample space obtained from the union of n distinct
simple events S1 , (S2 , S3 ,, Sn with associated probabilities P1, P2, P3, ,
Pn. Then
1. 0 P1 1, 0 P2 1, , 0 Pn 1
(All probabilities are between 0 and 1 inclusive);
2. P1 + P2 + P3 + + Pn = 1;
(The sum of all probabilities for a sample space is 1.);
3. P (S) = 1
4. P() = 0
-
49
Addition Principle
Suppose nSSSE 21, , where nSSS ,, 21 are distinct simple events then
P (E) = P( S1 ) + P( S2 ) + ... + P ( Sn )
Example 5
Refer to the previous Example and find the probability that a customer spends at least
K11, 250 but less than K20250.
This event is union of two simple events spending K11, 250 to K20, 250. The probability
of spending at least K11, 250 but less than K20, 250 can thus be found by the addition
principle. Let this event A, then
P (A ) = P(Spending K11250 K13500) + P(spending K13500 -K20250)
Addition for Mutually Exclusive Events .
For mutually exclusive events E and F
P (EUF) = P(E) + P(F)
Example 6
Use the probability distribution of Example 5 to find the probability that we get at least
one head on tossing a fair twice.
Event E At least one head is the union of three mutually exclusive events, two heads, one head one tail and one tail one head.
P(E) = P(2 heads) + 2P(one head one tail)
= 4
3
4
2
4
1
Complement: P(E ') = 1 - P(E' ) and P(E) = 1 - P(E)
-
50
In a particular experiment, P(E) 8
3 . Find P(E')
P(E') = 1 - P(E) = 8
5
8
31 .
Example 7
In example 3 above, find the probability that a customer spends less than K22500. Let E
to be the event a customer spends less than K22500.
P(E) = 0.281 + 0.147 + 0.088 + 0.2394 + 0.135 = 0.889
Alternatively E' is the event that a customer spends K22500 and over from the table.
P(E') = 0.111, and 1-P( E ) = P(E) = 1 - 0.111 = 0.889
Odds
The Odds in favor of an event E is defined as the ratio of P(E) to P(E') , or P(E)
P(E')
Example 8
Suppose the weather forecaster says that the probability of rain tomorrow is 5
2 . Find
the odds in favor of rain tomorrow.
Let E be the event rain tomorrow. Then E is the event no rain tomorrow. Since
P(E) 5
2
We have P( E ) =5
3. By the definition of odds, odds in favor of rain = 2/5 written 2 to
3 or 3:2 3/5 .
-
51
In general, if the odds favoring event E are m to n, then
P(E) =nm
m
and P( E ) =
nm
m
Example 9
The odds that a particular bid will be the low bid are 8 to 13. Find the probability that the
bid will be the low bid.
Solution
Odds of 8 to 13 show 8 favorable chances out of 8 + 13 = 21 chances altogether.
P (bid will be low bid) = 21
8
138
8
There is a 21
13chance that the bid will not be the low bid
Extended Addition Principle
For any two events, E and F form a sample space S,
P(EUF) = P(E) + P(F) - (E F)
-
52
Example 10.
If a single card is drawn from an ordinary deck, find the probability that it will be red or a
face card.
Let R and F represent the events red and face card respectively. Then
P(R) =52
26, P(F) =
52
12, and P (R F) =
52
6
(There are six red face cards in a deck) By the extended addition principle,
P(R F) = P(R) + P(F) - P(R F)
= 26 + 12 - 6 = 32 = 8
52 52 52 52 13
Example 11
Suppose two fair dice care rolled. Find each of the following probabilities.
a) The first die show a 2 or the sum is 6
A B
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
(1,4) (2,4) (3,4) 4,4) (5,4) (6,4)
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
-
53
P(A) =36
6 , P(B) =
36
5 , P(An B) =
36
1
By the extended addition principle
P(AB) = P(A) + P(B) P(A B)
= 18
5
36
10
36
1
36
5
36
6
b) The sum is 5 or the second die is 4.
P(sum is 5) = 36
4 , P(second die is 4) =
36
6
P(sum is 5 and second die is 4) =36
1
= 9 = 1
36 4
Often we are interested in how certain events are related to the occurrence of
other events. In particular, we may be interested in the probability of the
occurrence of an event given that another related event has occurred. Such
probabilities are referred to as Conditional Probabilities.
The conditional Probability of event E given event F, written P(EF), is
P(EF) = P(E F), P(F) 0
P(F)
-
54
Example 11
The Training Manager for a large stockbrokerage firm has noticed that
some of the of firms brokers use the firms research advice, while other brokers tend to go with their own feelings of which stocks will go up. To
see if the research department is better than just the feelings of the brokers,
the manager conducted a survey of 100 brokers, with results as shown in
the following table.
Picked stocks
That went up
Didnt pick stocks
That went up
Total
Used research 30 15 45
Didnt use research 30 25 55
Totals 60 40 100
Letting A represent the event picked stocks that went up, and letting B represent the event used research, we can find the following probabilities.
P(A) = 100
60 = 0.6 P(A') =
100
40 = 0.4
P(B) =100
45 = 0.45 P(B') =
100
55 = 0.55
Suppose we want to find the probability that a broker using research will pick stocks that
go up. From the table above, of the 45 brokers who use research, 30 picked stocks that
went up, with
P(broker who uses research picks stocks that go up)
= 30 = 0.667.
45
This is a different number than the probability that a broker picks stocks that go up, 0.6,
since we have additional information (the broker uses research) which reduced the
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55
sample space. In other words, we found the probability that a broker picks stocks that go
up, A, given the additional information that the broker uses research, B. This is called the
conditional probability of event A, given that event B has occurred, written P(A/B). In
the example above,
P(AB) = P(A B)
P(B)
= 30 = 0.667.
45
Product Rule: For any events E and F
P(EF) = P(F). P(E/F)
Example 12.
A class is5
2 women and
5
3men . Of the women, 25% are business majors. Find the
probability that a student chosen at random is a woman business major.
Solution
Let B and W represent the events business major and woman, respectively. We want
to find P(B W) . By the product rule,
P(B W) = P(W). P(BW)
Using the given information, P(W) =5
2 = 0.4 and P(BW) = 0.25.
Thus P(B W) = 0.4(0.25) = 0.10
Example 13
Suppose an investment firm is interested in the following events:
A = Common stock in XYZ Corporation gains 10% next year
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56
B = Gross National Product gains 10% next year
The firm has assigned the following probabilities on the basis of available information.
P(AB) = 0.8, P(B) = 0.3
That is, the Investment Company believes the probability is 0.8 that the XYZ common
stock will gain 10% in the next year assuming that the GNP gains 10% in the same time
period. In addition, the company believes the probability is only 0.3 that the GNP will
gain 10% in the next year. Use the formula for calculating the probability of an
intersection to calculate the probability that XYZ common stock and the GNP gain 10%
in the next year.
Solution.
We want to calculate P(AB). The formula is
P(AB) = P(B) P(AB) = (0.3) (0.8) = 0.24
Thus, the probability, according to this investment firm, is 0.24 that both XYZ common
stock and the GNP will gain 10% in the next year.
In the previous section we showed that the probability of an event A may be substantially
altered by the assumption that the event B has occurred. However, this will not always
be the case. In some instances the assumption that event B has occurred will not alter the
probability of event A at all. When this is true, we call events A and B independent.
Events A and B are independent if the assumptions that B has occurred does
not alter the probability that A has occurred, i.e
P(AB) = P(A)
When events A and B are independent it will also be true that
P(BA) = P(B)
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57
Events that are not independent are said to be dependent.
Example 14
The probability that interest rates will rise has been assessed as 0.8. If they do rise, the
probability that the stock market index will drop is estimated to be 0.9. If the interest
rates do not rise, the probability that the stock market index will still drop is estimated as
0.4. What is the probability that the stock market index will drop?
Solution
P(A) = P(Interest rates rise) = 0.8.
P(B) = P(Stock market index drops) = ?
Then, the probability of A , the complement of A, interest rates do not rise is P( A ) =
1 0.8 = 0.2.
P(BA) = P(stock market index dropsinterest rates rise) = 0.9
P(B A) = P(stock market index dropsinterest rates do not rise) = 0.4.
By the multiplication rule
P(B and A) = P(A) P(BA) = 0.8 x 0.9 = 0.72 and
P(B and A ) = P( A ) P(B A ) = 0.2 x 0.4 = 0.08 = 0.80
Example 15
Suppose we toss a fair die, let B be the event observe a number less or equal to 4 and A to
be the event an even number is observed. Are event A and B independent?
P(B) = ,3
2
6
4 since B = { 1, 2, 3, 4}
P(A) = 2
1
6
3 since A = 2, 4,
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58
P(A B) = 3
1
6
2 where A B = 2, 4
Now given A has occurred
P(BA) = P(AU B) = 1/3 = 2 = P(B)
P(A) 3
Similarly P(AB) )(3
2
2/1
3/1
)(
)(BP
AP
BAP
)(2
1
2/1
3/1
)(
)()( AP
BP
BAPBAP
Therefore the events A and B are independent.
If events A and B are independent, the probability of intersection of A and B equals the
product of the probabilities of A and B, i.e,
P(A B) = P(A) P(B).
In the toss experiment
P(AB) = P(A). P(B) = 3
1
3
2.
2
1
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59
Bayes Theorem
A posteriori Probabilities
Suppose three machines, A, B, and C, produce similar engine components. Machine A
produces 45 percent of the total components, machine B produces 30 percent, and
Machine C, 25 percent. For the usual production schedule, 6 percent of the components
produced by machine A do not meet established specifications; for machine B of machine
C, the corresponding figures are 4 percent and 3 percent. One component is selected at
random from the total output and is found to be defective. What is the probability that
the component selected was produced by machine A?
The answer to this question is found by calculating the probability after the outcomes of
the experiment have been observed. Such probabilities are called a posteriori
probabilities as opposed to a prior probabilities probabilities that give the likelihood
that an event will occur.
D is the event that a defective component is produced by machine A, machine B or
machine C.
A
DA
B
D
DB
C
DC
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60
The three mutually exclusive events A, B and C form a partition of the sample spaces.
Apart from being mutually exclusive, their union is precisely S.
The event D may be expressed as:
1. )()()( DCDBDAD
2. The event that a component is defective and is produced by machine A is given
by
.DA
Thus, a posterior probability that a defective component selected was produced by
machine a is given by )(
)()/(
Dn
DAnDAP
)()()(
)((
)(
)()/(
DCPDBPDAP
DAP
DP
DAPDAP
(1)
Next, using the product rule, we may express
)/()()(
),/()()(
)/()()(
CDPCPDCP
andBDPBPDBP
ADPAPDAP
so that (1) may be expressed in the form
)/()()/()()/()(
)/()()/(
CDPCPBDPBPADPAP
ADPAPDAP
(2)
which is a special case of a result known as Bayes Theorem.
Observe that the expression on the right of (2) involves the probabilities P(A), P(B), P(C)
and the conditional probabilities P(D/A),P(D/B), and P(D/C), all of which may be
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61
calculated in the usual fashion. Infact, by displaying these quantities on a tree diagram,
we obtain Figure 1.0. We may compute the required probability by substituting the
relevant quantities into (2), or we may make use of the following device.
P(A/D) = Product of probabilities along the limb through A
Sum of products of the probabilities along each limb terminating at D
Step 1 Step 2 Probability of
outcome
Machine Condition
45.0)( AP 06.0)/( ADPA
)/().()( ADPAPADP
D = 0.027
30.0)( BP 94.0)/( ADP )/().( ADPADPD = 0.423
)/().()(04.0)/( BDPBPBDPDBDPB = 0.012
25.0)( CP
)/().()(96.0)/( BDPBPBDPDBDP
=0.288
)/().()(03.0)/( CDPCPCDPDCDPC = 0.0075
)./(.).()(97.0)/( CDPCPCDPDCDP
=0.2425
In either case, we obtain
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62
581.00465.0
027.0
0075.0012.0027.0
027.0
)03.0)(25.0()04.0)(3.0()06.0)(45.0(
)06.0)(45.0()/(
DAP
Before looking at any further examples, let us state the general form of Bayes Theorem.
Let nAAA ,...,, 21 be a partition of a sample space S and let E be an event of the
experiment such that .0)( EP Then the posterior probability )1)(/( niEAP i is
given by
)3()/().(...)()/()/().(
)/()()/(
2211
11
nn
iAEPAPAPAEPAEPAP
AEPAPEAP
Problems
1) In a certain city, 40 percent of the people consider themselves movement for
multiparty democracy (MMD), 35 percent consider themselves to be United Party
for Nation Development (UPND) and 25 percent consider themselves to be
independents (1). During a particular election, 45 percent of the MMDs voted, 40
percent of the UPND voted and 60 percent of the independents voted. Suppose a
person is randomly selected:
a) Find the probability that the person voted. b) If the person voted, find the probability that the voter is
i) MMD ii) UPND iii) Independent.
2) Three girls Chanda, Mumba and Chileshe, pack okra in a factory. From the batch
allotted to them Chanda packs 55%, Mumba, 30% and Chileshe 15%. The
probability that Chanda breaks some okra in a packet is 0.7, and the respective
probabilities for Mumba and Chileshe are 0.2 and 0.1. What is the probability
that a packet with broken okra found by the Checker was packed by
a) Chanda?
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63
b) Mumba? c) Chileshe?
3) A publisher sends advertising material for an accounting text to 80% of all
Professors teaching the appropriate Accounting Courses. Thirty percent of the
Professors who received this material adopted the books, as did 10% of the
professors who did not receive the material. What is the probability that a
Professor who adopts the book has received the advertising material?
Solutions
MMD UPND Independent
40.0)( MMDP 35.0)( UPNDP 25.0)( IP
45.0)/( MMDVP 40.0)/( UPNDVP 60.0)/( IVP
a) )/()()/().()/().()( IVPIPUPNDVPUPNDPMMDVPMMDPVP
= .40(.45) + .35(.40) + .25(.60)
= 0.18 + 0.14 + 0.15 = 0.47
b) i) )(
)/()/(
VP
VMPVMP
383.047.0
18.0
)/().()/().()/().(
)/().(
IVPIPUVPUPMVPMP
MVPMP
ii) )(
)()/(
VP
VUPVUP
298.047.0
14.0
)(
)/().(
VP
UVPUP
iii) 319.047.0
15.0)/( VIP
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64
2. Chanda, Mumba Chileshe
(D) (M) (H)
46.0015.006.0385.0
)1.0(15.)2.0(30.)7.0(55.
)/().()/().()/().()(
1.0)/(,2.0)/(,7.0)/(
15.)(30.)(,55.)(
HBPHPMBPMPDBPDPBP
HBPMBPDBP
HPMPDP
a) 837.046.0
385.0
)(
)/().()/(
BP
DBPDPBDP
b) 1304.046.0
06.0
)(
)/().()/(
BP
MBPMPBMP
c) 0326.046.0
015.0
)(
)/().()/(
BP
HBPHPBHP
3. Let R be the event the Professor received material. A be the even the Professor a
adopted the book
P(R).P(A/R)
P(A/R) = 0.30
AP( /R) = 0.10
P(R) = 0.8
P(A/ R ) = 0.10
P( R ) = 0.2
P( A / R ) = 0.90
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65
.923.0
26.0
24.0
02.024.0
24.0
)10.0(2.0)30.0(8.0
)30.0(8.0
)/(.)()/().(
)/().(
)(
)()/(
RAPRPRAPRP
RAPRP
AP
ARPARP
Learning Objectives
After working through this Chapter, you should be able to
List the rules of probability.
Explain conditional probability, independent events, mutually exclusive events.
Apply the Bayes Theorem to find conditional probabilities
Define combinations, permutation and be able to apply such results to problems.
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66
CHAPTER 5
PROBABILITY DISTRIBUTION
Reading
Newbold Chapters 4 (not 4.4) and only 5.5 in Chapter 5
Wonnacott and Wonnacott Chapter 4
Tailoka Frank P Chapter 9
Introductory Comments
This Chapter introduces the three useful standard distributions for two counts (Discrete
Probability distribution) and one for (Continuous probability Distribution). These are so
often used that everyone should be familiar with them. We need to know the mean, the
variance and how to find simple probabilities.
5.0 Discrete Random Variables
A random variable maybe defined roughly as a variable that takes on different
numerical values because of chance. Random variables are classified as either
discrete or continuous. A discrete random variable is one that can take on only a
finite or countable number of distinct values. For example, the number of people
entering a shop is finite the values are 0, 1, 2, etc., the outcomes on 1 roll of a fair
die are limited to 1, 2, 3, 4, 5 and 6.
A random variable is said to be continuous in a given range if the variable can
assume any value in any given interval. A continuous variable can be be
measured with any degree of accuracy by using smaller and smaller units of
measurements. Examples of continuous variables include weight, length,
velocity, distance, time, and temperature. While discrete variables can be
counted, continuous variable can be measured with some degree of accuracy.
A probability distribution of a discrete random variable x whose value at x is
)(xf possess the following properties.
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67
1. 0)( xf for all real values of x
2. x
xf 1)(
Property 1: simply states that probabilities are greater than or equal to zero. The
second property states that the sum of the probabilities in a probability
distribution is equal to 1. The notation
x
xf )( means sum of the values f for all the values that x takes on. We will
ordinarily use the term probability distribution to refer to both discrete and
continuous variables; other terms are sometimes used to refer to probability
distributions (also called probability functions).
Probability distributions of discrete random variables are often referred to as
probability mass functions or simply mass functions because the probabilities are
massed at distinct points, for example along the x axis.
Probability distributions of continuous random variables are referred to as
probability density functions or density functions.
5.1 Cumulative Distribution Functions
Given a random variable x , the values of the cumulative distribution function at
x , denoted )(xF , is the probability that x takes on values less than or equal to
x . Hence
)1()()()( xxpxf
In the case of a discrete random variable, it is clear that
cx
xfcf
)2()()(
The symbol cx
xf )(
Means sum of the values of x for all values of x less than or equal to c.
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68
Example 1
Shoprite is interested in diversifying its product line into the soft goods market.
Mr Phiri, Vice president in charge of mergers and acquisitions, is negotiating the
acquisition of quick-save, a discount shop. The determine the price Shoprite
would have to pay per share for quick save, she sets up the probability distribution
for the stock price shown in the table below.
Probability distribution and cumulative distribution for the price of Quick
save common stock.
Price of Quicksave
Common stock x Probability
xf Cumulative Probability
xF K74 250
76 500
78 750
81 000
83 250
0.08
0.15
0.53
0.20
0.04
0.08
0.23
0.76
0.96
1.00
The probability that the price would be K78 750 or less is
23.0)50076()50076(
76.053.015.008.0)78750()75078(
KFKxP
KFKxP
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69
A graph of the cumulative distribution function is a step function that is the values
change in discrete steps at the indicated integral values of the random variable x.
)(xF
1.00
0.80
0.60
0.40
0.20
0.00
K74 250 76 500 78 750 81 000 83 250 x
Price of stock
Graph of cumulative distribution of the price of Quicksave common
stocks.
5.2 Probability Distribution of Discrete Random Variables
We will discuss the binomial and Poisson probability distribution of discrete
random variables.
xAll
xxPxE )()(
The variance of discrete random variable x is
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70
xAll
xpxxE )()()( 222
In general, if g(x) is any function of the discrete random variable x, then
xAll
xXPxgxgE )()()]([
For example
)()5()5(
)()(
)(20)20(
22
xXPxXE
xXPxxE
xXxPxE
Example 2
The random variable X has the following distribution for .4,3,2,1x
X 1 2 3 4
)( xXP 0.02 0.35 0.53 0.10
Calculate:
)25()
8)(6)
)()
)35()
)()
2
2
xEe
xEd
XEc
xEb
xEa
Solution
a) )()( xXxPxE
71.2
40.059.170.002.0
)10.0(4)53.0(3)35.0(2)02.0(1
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71
b) 3)(5)35( xExE
55.10
355.13
3)71.2(5
3)]10.0(4)53.0(3)35.0(2)02.0(1[5
3)(5
xXxP
c) )()( 22 xXPXXE
79.7
6.177.44.102.0
)10.0(4)53.0(3)35.0(2)02.0(1 2222
d) 8)(68)(6 xXxPxE = 6(2.71) + 8 = 16.26 + 8
= 24.26
e) 2)(5)25( 22 xExE
95.40
2)79.7(5
2)(5
2)(5
2
2
xXPxxE
In general, the following results hold when X is a discrete random variable.
1) aaE )( where a is any constant.
2) ),()( XaEaxE where a is any constant
3) ,)()( bxaEbaXE where a and b are any constants.
4) 21221 )([)]([)]()([ fandfwherexfExfExfxfE are functions of X.
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72
Variance, Var (x)
As for the variance, the following results are useful.
1) 0)( aVar where a is any constant
2) )var()( 2 xaaxVar where a is any constant
3) )var()( 2 xabaxVar where a and b are any constants.
Example 3
For the data in Example 2, calculate the following:
)23()
)4()
)var(25)35()
xVarc
xVarb
xxVara
Solution
a) )var(25)35( xxVar
We will need to find )()()(22 xExExVar
.71.2
)()(
xXxPXE
1475.11)35var(
)4459.0(25
)var(25)35(
4459.0
)71.2(79.7
)()()(
79.7
)()(
2
22
22
xTherefore
xxVar
xEXExVar
xXPXXE
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73
0131.4)4459.0(9
)var(9)23()
1344.7)4459.0(16
)var(16)4()
xxVarc
xxVarb
Example 4
A risky investment involves paying K300 000 that will return K2, 700,000 (for a net
profit of K2, 400,000) with probability 0.3 or K0 .00 (for a net loss of K300 000) with
probability 0.7. What is your expected net profit from this investment?
Solution
x )(xP
2,400,000 0.3
-300,000 0.7
(Note that a loss is treated as a negative profit.)
Then 000,510000,210000,720)7.0)(000,300()3.0(000,400,2)()( xxPxE Your expected net profit on an investment of this kind is K510, 000. If you were to make
a very large number of investments, some would result in a net profit of K7200, 000, and
others would result in a net loss of K300, 000. However, in the long run, your Average
net profit per investment would be K510, 000.
5.3 The Binomial Distribution
The Binomial distribution, in which there are two possible outcomes on
each experimental trial, is undoubtedly the most widely applied
probability distribution of a discrete random variable. It has been used to
describe a large variety of processes in business and the social sciences as
well as other areas. The Bernoulli process after James Bernoulli (1654 1705) gives rise to the Binomial distribution.
The Bernoulli process has the following characteristics.
a) On each trial, there are two mutually exclusive possible outcomes, which are referred to as success and failure. In somewhat different language sample space of possible outcomes on each experimental trial is S =
(failure, success).
b) The probability of a success will be denoted by P , P remains constant
from trial to trial. The probability of a failure will be denoted by q , q is
always equal to P1 .
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74
c) The trials are independent. That is, the outcomes on any given trial or
sequence of trials does not affect the outcomes on subsequent trials.
Suppose we toss a coin 3 times, then we may treat each toss as one Bernoulli trial.
The possible outcomes on any particular trial are a head and a tail. Assume that
the appearance of a head is a success. For example, we may choose to refer to the
appearance for a defective item in a production process as a success, if a series of
births is treated as a Bernoulli process, the appearance of female 9male0 may be
classified as a success.
Consider the experiment of tossing a fair coin three times, then the sequence of
outcome is
HTH, HHH, HHT, THH, TTT, THT, TTH, HTT
Since the probability of a success and failure on a given trial are respectively, P
and , the probability of the outcome for instance qppqpHTH 2}{ where p is
the probability of observing a head and q is the probabilit