Measures of Central Tendency
I. Mode
The mode is the data item that appears most frequently. It is possible for a
distribution to have no mode or more than one mode. A distribution with two modes
is called a bimodal distribution. For grouped data, it is customary to identify the
interval having the greatest frequency as the modal interval and to call its midpoint
the crude mode. Since the mode indicates maximum frequency, or maximum
popularity, the value of the mode provides important information to producers,
designers, and storekeepers who their products at specific markets.
II. Median
The median is the data item that is greater than or equal to half of the other terms and
equal to or less than half of them. If there is an odd number of data items, then the
median is the middle one; if there is an even number of data items, the median is the
average of the two middle ones. Half the area under a histogram lies to the left of the
median, and half to the right. Since the median is not affected by the size of extreme
data items, it is useful with distributions in which a relatively small number of items
are extreme in one direction or the other.
Example 1: Given the following histogram, estimate the median.
%/cig
(3)
3
(2)
2
(1)
1
(.5)
0
0
10
20
40
Number of cigarettes
First, determine the area of each block:
0  x  10 : (10 cig)(1%/cig) = 10%
10  x  20 : (10 cig)(3%/cig) = 30%
20  x  40 : (20 cig)(2%/cig) = 40%
40  x  80 : (40 cig)(0.5%/cig) = 20%
1
80
%/cig
3
(2)
2
30% 10%
1
10%
30%
20%
0
0
10
20 M
40
Number of cigarettes
80
Thus, the median lies in the interval 20  x  40 . (M-20)(2)=10  M = 25.
III. Mean
The mean, or (arithmetic) average, is the value that is obtained by adding the terms
and then dividing their sum by the number of terms. This can be symbolized in the
X
formula X 
, where X is the mean, X is the sum of the data items, and N
N
is the number of data items. In any distribution, the sum of the differences between
each data item and the mean equals zero. Conversely, if the sum of the differences
between each data item and some number equals zero, that number is the mean. This
implies that the histogram of a distribution will balance when supported at the mean.
Since the mean is sensitive to the size of extreme data items, it should not be used
with distributions in which a relatively small number of items are extreme in one
direction or the other.
Example 2: Given the histogram in example 1, approximate the mean.
The mean of a histogram can be approximated by calculating a “weighted”
average. First, use the midpoint of each class interval as the one value for all the
data items in that interval. Next, use the area of each block as the “weight” for
that block; convert the area percent to a decimal. Finally, sum up the products of
the midpoint value and the “weight”. Thus, X = (5)(.1) + (15)(.3) + (30)(.4) +
(60)(.2) = .5 + 4.5 + 12 + 12 = 29 cigarettes.
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Practice Sheet – Measures of Central Tendency
I. The following are 25 final averages in a math class:
46
49
53
60
61
64
66
66
67
71
72
74
75
76
79
79
79
80
83
88
89
91
94
95
98
(1) What is the mode?
(2) What is the median?
(3) What is the mean (average)?
II. Given the following data: 5, 7, 11, 12, 13, 18
(1)
(2)
(3)
(4)
(5)
What is the average?
Add 8 to each data item. What is the new average?
Subtract 6 from each data item. What is the new average?
Multiply each data item by 7. What is the new average?
Divide each data item by 5. What is the new average?
III. In the histogram given below, the class intervals include the right endpoint, not the
left:
%/$1000
1.25
1.00
0.75
0.50
0.25
0
0
20
40
80
100
Income (in $1000)
120
(1) What is the modal interval? What is the crude mode?
(2) What is the median?
(3) What is the mean?
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IV.
Class A
N = 20
X  70
(1)
(2)
(3)
(4)
Find
Find
Find
Find
Class B
N = 30
X  80
X for class A.
X for class B.
X for the combined classes.
X for the combined classes.
V. (1) A student made 65, 78, and 85 on three tests. What must she make on the fourth
test in order to have an average of 80?
(2) A student had an average of 75 on four tests. If he makes 95 on the fifth test,
what is his average for the five test?
(3) In a certain English class, quizzes make up 15% of the final average, major tests
make up 35%, papers make up 20%, and the final exam makes up 30%. If a
student has an average of 90 on quizzes, 80 on major tests, 75 on papers, and 85
on the final exam, what is his final average?
(4) Given the following data: 5, 8, 9, 13, 19, 22, 23, 29. What is the median?
(5) Nicole averaged 19.8 points/game for the first 10 games. In the 11th game, she
scored 33 points. What is her average for the 11 games?
In problems 6 and 7, use the formula: Batting Average = Hits/Times at Bat.
(6) John had a batting average of .300 for 400 times at bat and then had an average
of .400 for the next 100 times at bat. What was his final batting average for the
500 times at bat?
(7) During 2001 baseball season, Tom had 40 hits in 200 times at bat and Andy
had 100 hits in 400 times at bat. During the 2002 season, Tom had 93 hits in
300 times at bat and Andy had 32 hits in 100 times at bat. What was Tom’s
batting average for 2001, for 2002, and for the two years combined? What was
Andy’s batting average for 2001, for 2002, and for the two years combined?
In problems 8 and 9, use the formula: Average Speed = Distance/Time.
(8) Jane averaged 50 mph for 3 hours and then averaged 65 mph for the next 2 hours.
What was her average speed of the entire 5 hours?
(9) Jean averaged 60 mph for 120 miles and then averaged 40 mph for the next 80
miles. What was her average speed for the entire 200 miles?
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Solution Key for Measures of Central Tendency
I. (1) 79
(2) .5(25) = 12.5  round to next higher integer 13  13th number is median = 75
(3) 74.2
II. (1)
(2)
(3)
(4)
(5)
11
19
5
77
2.2
III. (1) 40  x  80 , 60
(2) 56
(3) 56
IV. (1)
(2)
(3)
(4)
1400
2400
3800
76
V. (1) 92
(2) 79
(3) 82
(4) 16 
13  19
, the average of the two middle numbers
2
(5) 21
(6) .320 
160hits
500atbats
(7)
Tom
Andy
2001
.200
.250
2002
.310
.320
Combined
.266
.264 !!!!
(8) 56 mph
(9) 50 mph
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