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Numerical Descriptions
of Data
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The arithmetic mean of a variable is computed by
determining the sum of all the values of the variable in
the data set divided by the number of observations.
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3-2
The population arithmetic mean is computed using
all the individuals in a population.
The population mean is a parameter.
The population arithmetic mean is denoted by .

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3-3
If x1, x2, …, xN are the N observations of a variable
from a population, then the population mean, µ, is
x1  x2 

N
 xN
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3-4
The sample arithmetic mean is computed using
sample data.
The sample mean is a statistic.
The sample arithmetic mean is denoted by

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x.
3-5
If x1, x2, …, xn are the n observations of a variable
from a sample, then the sample mean, x , is
x1  x2 
x
n
 xn
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3-6
EXAMPLE
Computing the population mean and sample mean
of a data set
The following data represent the travel times (in minutes) to
work for all seven employees of a start-up web
development company.
23, 36, 23, 18, 5, 26, 43
Determine the population mean of this data.
Step 1: There are N = 7 observations.
Step 2: 23 + 36 + 23 + 18 + 5 + 26 + 43 = 174
Step 3: 174 / 7 ≈ 24.85714
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3-7
EXAMPLE
Computing the population mean and sample mean
of a data set
Now suppose that from the seven times given we take a
random sample of 3 observations. Those observations
were:
5 , 36 , 26
Determine the sample mean for these observations
Step 1: There are n = 3 observations.
Step 2: 5 + 36 + 26 = 67
Step 3: 67 / 3≈ 22.33333
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3-8
The median of a variable is the value that lies in the
middle of the data when arranged in ascending order.
We use M to represent the median.
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3-9
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3-10
EXAMPLE
Computing a Median of a Data Set with an Odd
Number of Observations
The following data represent the travel times (in minutes) to
work for all seven employees of a start-up web
development company.
23, 36, 23, 18, 5, 26, 43
Determine the median of this data.
Step 1: 5, 18, 23, 23, 26, 36, 43
Step 2: There are n = 7 observations.
n 1 7 1
M = 23

4
Step 3:
2
2
5, 18, 23, 23, 26, 36, 43
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3-11
EXAMPLE
Computing a Median of a Data Set with an
Even Number of Observations
Suppose the start-up company hires a new employee. The
travel time of the new employee is 70 minutes. Determine
the median of the “new” data set.
23, 36, 23, 18, 5, 26, 43, 70
Step 1: 5, 18, 23, 23, 26, 36, 43, 70
Step 2: There are n = 8 observations.
23  26
n 1 8 1
M
 24.5 minutes

 4.5
Step 3:
2
2
2
5, 18, 23, 23, 26, 36, 43, 70

M  24.5
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3-12
EXAMPLE
Computing a Median of a Data Set with an
Even Number of Observations
The following data represent the travel times (in minutes) to
work for all seven employees of a start-up web development
company.
23, 36, 23, 18, 5, 26, 43
Suppose a new employee is hired who has a 130 minute
commute. How does this impact the value of the mean and
median?
Mean before new hire: 24.9 minutes
Median before new hire: 23 minutes
Mean after new hire: 38 minutes
Median after new hire: 24.5 minutes
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3-13
A numerical summary of data is said to be resistant if
extreme values (very large or small) relative to the data do
not affect its value substantially.
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3-14
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3-15
EXAMPLE
Describing the Shape of the Distribution
The following data represent the asking price of homes
for sale in Lincoln, NE.
79,995
99,899
105,200
128,950
130,950
131,800
149,900
151,350
154,900
189,900
203,950
217,500
111,000
120,000
121,700
132,300
134,950
135,500
159,900
163,300
165,000
260,000
284,900
299,900
125,950
126,900
138,500
147,500
174,850
180,000
309,900
349,900
Source: http://www.homeseekers.com
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Find the mean and median. Use the mean and
median to identify the shape of the distribution.
Verify your result by drawing a histogram of the
data.
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Find the mean and median. Use the mean and
median to identify the shape of the distribution.
Verify your result by drawing a histogram of the
data.
The mean asking price is $168,320 and the median
asking price is $148,700. Therefore, we would
conjecture that the distribution is skewed right.
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3-18
Asking Price of Homes in Lincoln, NE
12
10
Frequency
8
6
4
2
0
100000
150000
200000
250000
Asking Price
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300000
350000
3-19
The mode of a variable is the most frequent
observation of the variable that occurs in the
data set.
If there is no observation that occurs with the
most frequency, we say the data has no mode.
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3-20
EXAMPLE
Finding the Mode of a Data Set
The data on the next slide represent the Vice Presidents
of the United States and their state of birth. Find the
mode.
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The mode is
New York.
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Tally data to determine most
frequent observation
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The range, R, of a variable is the difference
between the largest data value and the smallest
data values. That is
Range = R = Largest Data Value – Smallest Data Value
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3-26
EXAMPLE
Finding the Range of a Set of Data
The following data represent the travel times (in minutes)
to work for all seven employees of a start-up web
development company.
23, 36, 23, 18, 5, 26, 43
Find the range.
Range = 43 – 5
= 38 minutes
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3-27
The population variance of a variable is the sum of
squared deviations about the population mean divided
by the number of observations in the population, N.
That is it is the mean of the sum of the squared
deviations about the population mean.
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The population variance is symbolically represented
by σ2 (lower case Greek sigma squared).
Note: When using the above formula, do not round until the
last computation. Use as many decimals as allowed by your
calculator in order to avoid round off errors.
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3-29
EXAMPLE
Computing a Population Variance
The following data represent the travel times (in minutes) to
work for all seven employees of a start-up web development
company.
23, 36, 23, 18, 5, 26, 43
Compute the population variance of this data. Recall that
174

 24.85714
7
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3-30
xi
μ
xi – μ
(xi – μ)2
23
36
23
18
24.85714
24.85714
24.85714
24.85714
-1.85714
11.14286
-1.85714
-6.85714
3.44898
124.1633
3.44898
47.02041
5
26
43
24.85714
24.85714
24.85714
-19.8571
1.142857
18.14286
394.3061
1.306122
329.1633
 x   
i

2
x  


i
N
2
2

902.8571
902.8571

 129.0 minutes2
7
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3-31
The Computational Formula
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3-32
EXAMPLE
Computing a Population Variance
Using the Computational Formula
The following data represent the travel times (in
minutes) to work for all seven employees of a start-up
web development company.
23, 36, 23, 18, 5, 26, 43
Compute the population variance of this data using the
computational formula.
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3-33
23, 36, 23, 18, 5, 26, 43
2
2
2
2
x

23

36

...

43
 5228
i
x
i
 23  36  ...  43  174
2 
2
x
i
x



i
N
N
2
1742
5228 
7

7
 129.0
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3-34
The sample variance is computed by determining the
sum of squared deviations about the sample mean and
then dividing this result by n – 1.
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Note: Whenever a statistic consistently overestimates or
underestimates a parameter, it is called biased. To obtain an
unbiased estimate of the population variance, we divide the
sum of the squared deviations about the mean by n - 1.
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3-36
EXAMPLE
Computing a Sample Variance
Previously, we obtained the following simple random sample for the travel time
data: 5, 36, 26.
Compute the sample variance travel time.
Travel Time, xi
Sample Mean,
Deviation about the
Mean,
Squared Deviations about the
Mean,
 x  x
2
x
xi  x
5
22.333
5 – 22.333
= -17.333
(-17.333)2 = 300.432889
36
22.333
13.667
186.786889
26
22.333
3.667
13.446889
i
 x  x
i
s
2
x  x



i
n 1
2
 500.66667
2

500.66667
3 1
 250.333 square minutes
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3-37
The population standard deviation is denoted by
It is obtained by taking the square root of the population
variance, so that
The sample standard deviation is denoted by
s
It is obtained by taking the square root of the sample variance, so
that
s  s2
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3-38
EXAMPLE Computing a Sample Standard
Deviation
Recall the sample data 5, 26, 36 results in a sample variance of
s2 

xi  x
n 1

2

500.66667
3 1
 250.333 square minutes
Use this result to determine the sample standard deviation.
s  s2 
500.666667
 15.8 minutes
3 1
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3-39
EXAMPLE
Comparing Standard Deviations
Determine the standard deviation waiting time
for Wendy’s and McDonald’s. Which is
larger? Why?
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3-40
Wait Time at Wendy’s
1.50
2.53
1.88
3.99
0.90
0.79
1.20
2.94
1.90
1.23
1.01
1.46
1.40
1.00
0.92
1.66
0.89
1.33
1.54
1.09
0.94
0.95
1.20
0.99
1.72
0.67
0.90
0.84
0.35
2.00
Wait Time at McDonald’s
3.50
0.00
1.97
0.00
3.08
0.00
0.26
0.71
0.28
2.75
0.38
0.14
2.22
0.44
0.36
0.43
0.60
4.54
1.38
3.10
1.82
2.33
0.80
0.92
2.19
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3.04
2.54
0.50
1.17
0.23
3-41
EXAMPLE
Comparing Standard Deviations
Determine the standard deviation waiting time
for Wendy’s and McDonald’s. Which is
larger? Why?
Sample standard deviation for Wendy’s:
0.738 minutes
Sample standard deviation for McDonald’s:
1.265 minutes
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3-42
Quartiles divide data sets into fourths, or four equal parts.
• The 1st quartile, denoted Q1, divides the bottom 25% the data
from the top 75%. Therefore, the 1st quartile is equivalent to the
25th percentile.
• The 2nd quartile divides the bottom 50% of the data from the top
50% of the data, so that the 2nd quartile is equivalent to the 50th
percentile, which is equivalent to the median.
• The 3rd quartile divides the bottom 75% of the data from the top
25% of the data, so that the 3rd quartile is equivalent to the 75th
percentile.
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3-44
EXAMPLE
Finding and Interpreting Quartiles
A group of Brigham Young University—Idaho students (Matthew Herring,
Nathan Spencer, Mark Walker, and Mark Steiner) collected data on the
speed of vehicles traveling through a construction zone on a state highway,
where the posted speed was 25 mph. The recorded speed of 14 randomly
selected vehicles is given below:
20, 24, 27, 28, 29, 30, 32, 33, 34, 36, 38, 39, 40, 40
Find and interpret the quartiles for speed in the construction zone.
Step 1: The data is already in ascending order.
Step 2: There are n = 14 observations, so the median, or second quartile, Q2, is the mean of the
7th and 8th observations. Therefore, M = 32.5.
Step 3: The median of the bottom half of the data is the first quartile, Q1.
20, 24, 27, 28, 29, 30, 32
The median of these seven observations is 28. Therefore, Q1 = 28. The median of the top half of
the data is the third quartile, Q3. Therefore, Q3 = 38.
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Interpretation:
• 25% of the speeds are less than or equal to the first quartile, 28 miles per
hour, and 75% of the speeds are greater than 28 miles per hour.
• 50% of the speeds are less than or equal to the second quartile, 32.5 miles
per hour, and 50% of the speeds are greater than 32.5 miles per hour.
• 75% of the speeds are less than or equal to the third quartile, 38 miles per
hour, and 25% of the speeds are greater than 38 miles per hour.
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3-47
EXAMPLE
Determining and Interpreting the
Interquartile Range
Determine and interpret the interquartile range of the speed data.
Q1 = 28
Q3 = 38
IQR  Q3  Q1
 38  28
 10
The range of the middle 50% of the speed of cars traveling through the construction
zone is 10 miles per hour.
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Suppose a 15th car travels through the construction zone at 100 miles per hour.
How does this value impact the mean, median, standard deviation, and
interquartile range?
Without 15th car
With 15th car
Mean
32.1 mph
36.7 mph
Median
32.5 mph
33 mph
Standard deviation
6.2 mph
18.5 mph
IQR
10 mph
11 mph
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3-49
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3-50
EXAMPLE
Determining and Interpreting the
Interquartile Range
Check the speed data for outliers.
Step 1: The first and third quartiles are Q1 = 28 mph and Q3 = 38 mph.
Step 2: The interquartile range is 10 mph.
Step 3: The fences are
Lower Fence = Q1 – 1.5(IQR)
Upper Fence = Q3 + 1.5(IQR)
= 28 – 1.5(10)
= 38 + 1.5(10)
= 13 mph
= 53 mph
Step 4: There are no values less than 13 mph or greater than 53 mph. Therefore,
there are no outliers.
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3-52
EXAMPLE
Obtaining the Five-Number Summary
Every six months, the United States Federal Reserve Board conducts a survey of credit
card plans in the U.S. The following data are the interest rates charged by 10 credit card
issuers randomly selected for the July 2005 survey. Determine the five-number summary
of the data.
First, we write the data is
Institution
Rate
Pulaski Bank and Trust Company
6.5%
Rainier Pacific Savings Bank
12.0%
Wells Fargo Bank NA
14.4%
Firstbank of Colorado
14.4%
Lafayette Ambassador Bank
14.3%
Infibank
13.0%
United Bank, Inc.
13.3%
First National Bank of The Mid-Cities
13.9%
Bank of Louisiana
9.9%
Bar Harbor Bank and Trust Company
14.5%
Source: http://www.federalreserve.gov/pubs/SHOP/survey.htm
ascending order:
6.5%, 9.9%, 12.0%, 13.0%,
13.3%, 13.9%, 14.3%, 14.4%,
14.4%, 14.5%
The smallest number is 6.5%.
The largest number is 14.5%.
The first quartile is 12.0%.
The second quartile is 13.6%.
The third quartile is 14.4%.
Five-number Summary:
6.5% 12.0% 13.6% 14.4% 14.5%
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3-54
EXAMPLE
Constructing a Boxplot
Every six months, the United States Federal Reserve Board conducts a survey of credit
card plans in the U.S. The following data are the interest rates charged by 10 credit card
issuers randomly selected for the July 2005 survey. Draw a boxplot of the data.
Institution
Pulaski Bank and Trust Company
Rate
6.5%
Rainier Pacific Savings Bank
12.0%
Wells Fargo Bank NA
14.4%
Firstbank of Colorado
14.4%
Lafayette Ambassador Bank
14.3%
Infibank
13.0%
United Bank, Inc.
13.3%
First National Bank of The Mid-Cities
13.9%
Bank of Louisiana
Bar Harbor Bank and Trust Company
9.9%
14.5%
Source: http://www.federalreserve.gov/pubs/SHOP/survey.htm
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3-55
Step 1: The interquartile range (IQR) is 14.4% - 12% = 2.4%. The lower and upper
fences are:
Lower Fence = Q1 – 1.5(IQR)
Upper Fence = Q3 + 1.5(IQR)
= 12 – 1.5(2.4)
= 14.4 + 1.5(2.4)
= 8.4%
= 18.0%
Step 2:
*
[
]
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3-56
The interest rate boxplot indicates that the distribution is skewed left.
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3-58
EXAMPLE
Using Chebyshev’s Theorem
Using the data from the previous example, use Chebyshev’s
Theorem to
(a) determine the percentage of patients that have serum HDL
within 3 standard deviations of the mean.
1

1

100%  88.9%

2 
 3 
(b) determine the actual percentage of patients that have serum
HDL between 34 and 80.8.
1

1  2 100%  75%
 2 
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EXAMPLE
Using the Empirical Rule
The following data represent the serum HDL
cholesterol of the 54 female patients of a family
doctor.
41
62
67
60
54
45
48
75
69
60
54
47
43
77
69
60
55
47
38
58
70
61
56
48
35
82
65
62
56
48
37
39
72
63
56
50
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44
85
74
64
57
52
44
55
74
64
58
52
44
54
74
64
59
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(a) Compute the population mean and standard
deviation.
(b) Draw a histogram to verify the data is bell-shaped.
(c) Determine the percentage of patients that have
serum HDL within 3 standard deviations of the mean
according to the Empirical Rule.
(d) Determine the percentage of patients that have
serum HDL between 34 and 69.1 according to the
Empirical Rule.
(e) Determine the actual percentage of patients that
have serum HDL between 34 and 69.1.
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3-64
EXAMPLE
Using the Empirical Rule
The following data represent the serum HDL
cholesterol of the 54 female patients of a family
doctor.
41
62
67
60
54
45
48
75
69
60
54
47
43
77
69
60
55
47
38
58
70
61
56
48
35
82
65
62
56
48
37
39
72
63
56
50
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44
85
74
64
57
52
44
55
74
64
58
52
44
54
74
64
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3-65
(a) Compute the population mean and standard
deviation.
(b) Draw a histogram to verify the data is bell-shaped.
(c) Determine the percentage of patients that have
serum HDL within 3 standard deviations of the mean
according to the Empirical Rule.
(d) Determine the percentage of patients that have
serum HDL between 34 and 69.1 according to the
Empirical Rule.
(e) Determine the actual percentage of patients that
have serum HDL between 34 and 69.1.
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(a) Using the formulas for mean and
standard deviation
  57.4 and   11.7
(b)
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22.3
34.0
45.7
57.4
69.1
80.8
92.5
(c) According to the Empirical Rule, 99.7% of the patients that have serum HDL
within 3 standard deviations of the mean.
(d) 13.5% + 34% + 34% = 81.5% of patients will have a serum HDL between 34.0
and 69.1 according to the Empirical Rule.
(e) 45 out of the 54 or 83.3% of the patients have a serum HDL between 34.0 and
69.1.
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EXAMPLE
Using Z-Scores
The mean height of males 20 years or older is 69.1 inches with a
standard deviation of 2.8 inches. The mean height of females
20 years or older is 63.7 inches with a standard deviation of 2.7
inches. Data based on information obtained from National
Health and Examination Survey. Who is relatively taller?
Kevin Garnett whose height is 83 inches
or
Candace Parker whose height is 76 inches
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3-70
83  69.1
zkg 
2.8
 4.96
76  63.7
zcp 
2.7
 4.56
Kevin Garnett’s height is 4.96 standard deviations above the mean.
Candace Parker’s height is 4.56 standard deviations above the
mean. Kevin Garnett is relatively taller.
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The kth percentile, denoted, Pk, of a set of data is a value
such that k percent of the observations are less than or equal
to the value.
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