Download ch06 - Gordon State College

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Slide 6-1
Chapter 6
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Measures of Center in a
Distribution
6-A
The mean is what we most commonly call the average
value. It is defined as follows:
mean =
sum of all values
total number of values
The median is the middle value in the sorted data set (or
halfway between the two middle values if the number of
values is even).
The mode is the most common value
(or group of values) in a distribution.
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Slide 6-3
6-A
Middle Value for a Median
6.72
3.46
3.46
3.60
3.60
6.44
6.44
6.72
26.70
26.70
(sorted list)
(odd number of values)
exact middle
median is 6.44
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Slide 6-4
6-A
No Middle Value for a Median
6.72
3.46
3.46
3.60
3.60
6.44
6.44
6.72
(sorted list)
(even number of values)
3.60 + 6.44
2
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median is 5.02
Slide 6-5
6-A
Mode Examples
a. 5 5 5 3 1 5 1 4 3 5

Mode is 5
b. 1 2 2 2 3 4 5 6 6 6 7 9

Bimodal (2 and 6)
c. 1 2 3 6 7 8 9 10

No Mode
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Slide 6-6
Symmetric and Skewed
Distributions
Mode
=
Mean
=
6-A
Median
SYMMETRIC
Mean
Mode
Median
SKEWED LEFT
(negatively)
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Mean
Mode
Median
SKEWED RIGHT
(positively)
Slide 6-7
6-B
Why Variation Matters
Big Bank (three line wait times):
4.1
5.2
5.6
6.2
6.7
7.2
7.7
7.7
8.5
9.3
11.0
Best Bank (one line wait times):
6.6
6.7
6.7
6.9
7.1
7.2
7.3
7.4
7.7
7.8
7.8
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Slide 6-8
Five Number Summaries &
Box Plots
Big Bank
low value (min) =
lower quartile =
median =
upper quartile =
high value (max) =
4.1
5.6
7.2
8.5
11.0
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Best Bank
low value (min) =
lower quartile =
median =
upper quartile =
high value (max) =
6-B
6.6
6.7
7.2
7.7
7.8
Slide 6-9
6-B
Standard Deviation
Let A = {2, 8, 9, 12, 19} with a mean of 10. Use the data
set A above to find the sample standard deviation.
x (data value)
2
8
9
12
19
x – mean
(deviation)
2 – 10 = -8
8 – 10 = -2
9 – 10 = -1
12 – 10 = 2
19 – 10 = 9
Total
standard deviation =
(deviation)2
(-8)2 = 64
(-2)2 = 4
(-1)2 = 1
(2)2 = 4
(9)2 = 81
154
154 = 6.2
5 1
2
standard deviaton = sum of (deviations from the mean)
total number of data values  1
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Slide 6-10
Range Rule of Thumb for
Standard Deviation
6-B
standard deviation  range
4
We can estimate standard deviation by taking an approximate
range,
(usual high – usual low), and dividing by 4. The reason for the 4 is
due to the idea that the usual high is approximately 2 standard
deviations above the mean and the usual low is approximately 2
standard deviations below.
Going in the opposite direction, on the other hand, if we know the
standard deviation we can estimate:
low value  mean – 2 x standard deviation
high value  mean + 2 x standard deviation
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Slide 6-11
Outliers for Modified Boxplots
For purposes of constructing modified
boxplots, we can consider outliers to be
data values meeting specific criteria.
In modified boxplots, a data value is an
outlier if it is . . .
Data > Q3 + 1.5  IQR
or
Data < Q1 - 1.5  IQR
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Inc.
Modified Boxplots - Example
Pulse rates of females listed in Data Set 1 in
Appendix B.
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Education, Inc.
Percentiles
are measures of location. There are 99
percentiles denoted P1, P2, . . . P99,
which divide a set of data into 100
groups with about 1% of the values in
each group.
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Education, Inc.
Finding the Percentile
of a Data Value
Percentile of value x =
number of values less than x
total number of values
Do problems 17-28 on page 125
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Education, Inc.
• 100
The 68-95-99.7 Rule for
a Normal Distribution
6-C
(mean)
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Slide 6-16
The 68-95-99.7 Rule for
a Normal Distribution
6-C
(just over two thirds)
68% within
1 standard deviation of the mean
-
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(mean)
+
Slide 6-17
The 68-95-99.7 Rule for
a Normal Distribution
6-C
(most)
95% within
2 standard deviations of the mean
68% within
1 standard deviation of the mean
-2
-
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(mean)
+
+2
Slide 6-18
The 68-95-99.7 Rule for
a Normal Distribution
6-C
99.7% within 3 standard deviations of the mean
95% within
2 standard deviations of the mean
68% within
1 standard deviation of the mean
(virtually all)
-3
-2
-
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(mean)
+
+2
+3
Slide 6-19
6-C
Also known as the Empirical Rule
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Slide 6-20
6-C
Z-Score Formula
standard score = z = data value  mean
standard deviation
Example:
If the nationwide ACT mean were 21 with a standard
deviation of 4.7, find the z-score for a 30.
What does this mean?
z = 30  21 = 1.91
4.7
This means that an ACT score of 30 would be about 1.91
standard deviations above the mean of 21.
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Slide 6-21