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The term "measures of central tendency"
refers to
finding the mean, median and mode.
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Mean: |
Average.
The sum of a set of data divided by the number of data.
(Do not round your answer
unless directed to do so.) |
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Median: |
The
middle value, or the mean of the
middle two values, when the data is
arranged in numerical order. Think of a
"median" being in the middle of a highway.
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Mode: |
The value ( number) that appears the
most.
It is possible to have more than one mode, and it is possible to have no mode. If
there is no mode-write "no mode", do
not write zero (0) . |
Consider this set of test score
values:
Normal listing of scores. |
Scores with the lowest score replaced with outlier. |
The two sets of scores above are
identical except for the first score. The set on the left
shows the actual scores. The set on the right shows what would
happen if one of the scores was WAY out of range in regard to the
other scores. Such a term is called an
outlier.
With the outlier, the mean
changed.
With the outlier, the median
did NOT change.
How do I know which measure of central
tendency to use?
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MEAN
Use the mean to describe the middle of a set of data that does not have an outlier.
Advantages:
• Most popular measure in fields such as business,
engineering and computer science.
• It is unique - there is only one answer.
• Useful when comparing sets of data.
Disadvantages:
•
Affected by extreme
values (outliers)
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MEDIAN
Use the median to describe the middle of a set of data that does have an outlier.
Advantages:
• Extreme values (outliers) do not affect the median as
strongly as they do the mean.
• Useful when comparing sets of data.
• It is unique - there is only one answer.
Disadvantages:
•
Not as popular as
mean. |
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MODE
Use the mode when the data is non-numeric or when
asked to choose the most popular item.
Advantages:
•
Extreme values (outliers) do not affect the mode.
Disadvantages:
• Not as popular as
mean and median.
• Not necessarily unique - may be more than one answer
• When no values repeat in the data set, the mode is every
value and is useless.
• When there is more than one mode, it is difficult to
interpret and/or compare. |
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What will happen to the
measures of central tendency if we add the same amount to
all data values, or multiply each data value by the same
amount? |
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Data |
Mean |
Mode |
Median |
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Original Data Set: |
6,
7, 8, 10, 12, 14, 14, 15, 16, 20 |
12.2 |
14 |
13 |
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Add 3 to each data value |
9,
10, 11, 13, 15, 17, 17, 18, 19, 23 |
15.2 |
17 |
16 |
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Multiply 2 times
each data value |
12, 14, 16, 20,
24, 28, 28, 30, 32, 40 |
24.4 |
28 |
26 |
When added: Since all values are shifted
the same amount, the measures of central tendency all
shifted by the same amount. If you add 3 to each data
value, you will add 3 to the mean, mode and median.
When multiplied: Since all values are
affected by the same multiplicative values, the measures of
central tendency will feel the same affect. If you
multiply each data value by 2, you will multiply the mean,
mode and median by 2. |
Example #1
Find the mean, median and mode for the following data: 5, 15, 10,
15, 5, 10, 10, 20, 25, 15.
(You
will need to organize
the data.)
5, 5, 10, 10, 10, 15,
15, 15, 20, 25
Mean:
Median: 5, 5, 10, 10,
10, 15, 15, 15, 20, 25
Listing the data in order is the easiest way to find the median.
The numbers
10 and 15 both fall in the middle.
Average these two numbers to get the median.
10 + 15 = 12.5
2
Mode: Two numbers appear
most often:
10 and 15.
There are three 10's and three
15's.
In this example there are two answers for the mode. |
Example #2
For what value of x will 8 and
x have the same mean (average) as 27 and 5?
| First, find the mean of 27 and
5:
27 + 5 = 16
2 |
Now, find the x value, knowing
that the average of x and 8 must be 16:
x + 8 = 16
2
32 = x + 8 cross multiply
-8
- 8
24 = x and solve |
Example #3 :
On his first 5 biology tests, Bob received the following scores:
72, 86, 92, 63, and 77. What test score must Bob earn on his sixth test so that his average (mean score) for all six tests will be 80?
Show how you arrived at your answer.
Possible solution:
Set up an equation
to represent the situation. Remember to use all 6 test
scores:
72 + 86 + 92 + 63 + 77 + x
= 80
6
cross multiply and solve:
(80)(6) = 390 + x
480 = 390 + x
- 390 -390
90 = x
Bob must get a 90 on the sixth test.
Example #4
The mean (average) weight of three dogs is 38 pounds. One of the dogs, Sparky, weighs 46 pounds. The other two dogs, Eddie and Sandy, have the same weight. Find Eddie's weight.
Let x = Eddie's weight ( they weigh the same, so
they are both represented by "x".)
Let x = Sandy's weight
Average: sum of the data divided by the number of data.
x + x + 46 = 38
cross multiply and solve
3(dogs)
(38)(3) = 2x + 46
114 = 2x + 46
-46 -46
68 = 2x
2 2
34 = x Eddie weighs 34 pounds.
You can always check your work with a calculator!!
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See how to use
your
TI-83+/TI-84+ graphing calculator with
mean, mode, median.
Click calculator. |
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See how to use
your
TI-83+/TI-84+ graphing calculator with
mean, mode, median and grouped data.
Click calculator. |
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