Download 14-2 Measures of Central Tendency

Advanced Mathematical Concepts
Chapter 14
Lesson 14-2
Example 1
Find the mean of the set {21, 18, 17, 23, 32, 15, 21}.
sum of the values in the set of data
number of values in the set
21 + 18 + 17 + 23 + 32 + 15 + 21
X =
7
147
X = 7 or 21
X =
The mean of the set of data is 21.
Example 2
REAL ESTATE The data below shows the selling price of homes that recently sold in a particular
community.
124,500
191,000
142,000
164,000
183,000
394,000
165,000
148,000
a. Find the mean of the data.
b. Find the median of the data.
c. Find the mode of the data.
d. State which measures of central tendency seem most representative of the data. Explain.
a. Since there are 8 homes, n = 8.
8
1 X = 1(124,500 + 142,000 + 183,000 + 165,000 + 191, 000 +
1
8
8
i 1
164,000 + 394,000 + 148,000) or 188,937.50
The mean selling price is $188,937.50.
b. To find the median, order the data.
124,500 142,000 148,000
164,000
165,000 183,000
191,000 394,000
Since there are an even number of data, the median is the mean of the two middle numbers, 164,000
and 165,000. Therefore, the median selling price is $164,500.
c. Since all elements in the set of data have the same frequency, there is no mode.
d. Notice that the mean is affected by the extreme value 394,000 and does not accurately represent the
data. The median is a more representative measure of central tendency.
Advanced Mathematical Concepts
Chapter 14
Example 3
COLLEGE The data below represent the ages of students in a Probability and Statistics course at a
local university.
22 25
20 41
20 31
20 21 33
32 27 19
29 40 23
29 18 19 21
18 20 19 22
21 19 20 21
22
24
25
a. Make a stem-and-leaf plot of the students’ ages.
b. Find the mean of the data.
c. Find the median of the data.
d. Find the mode of the data.
e. What is a good representative for the average age of a student in the class?
a. Since the ages range from 18 to 41, we will use the tens place for the stems. List the stems and draw a
vertical line to the right of the stems. Then, list the leaves, which in this case will be the ones digit. It
is often helpful to list the leaves as you come to them and then rewrite the plot with the leaves in
order from the least to the greatest.
stem
leaf
1
8 8 9 9 9 9
2
0 0 0 0 0 1 1 1 1 2 2 2 3 4 5 5 7 9 9
3
1 2 3
4
0 1
1 | 8 = 18
b. Enter the data in the L1 list of a
graphing calculator. Use the statistics
mode of the calculator to find the mean.
The mean is 24.03.
c. Since the median is the middle value, it
is the mean of the 15th and 16th leaves.
The median is 21.5.
d. The stem-and-leaf plot shows the modes by repeated digits for a particular stem. There are five 0’s
with a stem of 2. The mode is 20.
e. The value of the mean is slightly distorted by the few higher ages in the class. The median, 21.5, is a
better representation of the average age of a student in the class.
Advanced Mathematical Concepts
Chapter 14
Example 4
EDUCATION Estimate the mean of the scores of 100 students on a statistics test given the
following frequency distribution.
Class Limits
90 – 100
80 – 90
70 – 80
60 – 70
50 – 60
40 – 50
Class Marks (X)
95
85
75
65
55
45
Frequency (f)
21
34
17
13
9
6
6
 f1  100
i 1
7770
X = 100 or 77.7
fX
1995
2890
1275
845
495
270
6
 f
i
 X i   7770
i 1
The mean is approximately 78.
Example 5
EDUCATION Estimate the median of the data in the frequency distribution in Example 4.
Since there are 100 scores in the distribution, 50 scores are below the median and 50 scores are above the
median. From the chart above, find the least cumulative frequency that is greater than or equal to 50. So,
the median class is 80 – 90.
You can use a proportion to find the value of the median by finding the ratios of the differences in the
cumulative frequencies and the upper limits of the classes.
79 – 45 = 34
90 – 80 = 10
50 – 45 = 5
Md – 80 = x
34 10
5 = x
x  1.470588235
Use a calculator.
Md – 80 = x
Md – 80  1.5
Md  81.5
The median of the data is approximately 81.5.