Download Data Mean

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Measures of Central Tendency
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 1
Definitions
• An average is a number that is representative
of a group of data.
• The arithmetic mean, or simply the mean is
symbolized by x , when it is a sample of a
population or by the Greek letter mu, , when
it is the entire population.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 2
Mean
• The mean, is the sum of the data divided by
the number of pieces of data. The formula for
calculating the mean is
Sx
x
n
where Sx represents the sum of all the data and n represents
the number of pieces of data.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 3
Example-find the mean
• Find the mean amount of money parents
spent on new school supplies and clothes if 5
parents randomly surveyed replied as follows:
$327 $465 $672 $150 $230
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 4
Solution
327  465  672  150  230 1844
x

 368.8
5
5
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 5
Example
• Malcolm’s average on 5 exams is 81.
Determine the sum of all his exam scores.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 6
Example
An average of 60 on 7 exams is needed to pass a
course. On her first 6 exams, Sheryl received grades of:
51, 72, 80, 62, 57, and 69.
• What grade must she receive on her last exam to
pass the course?
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 7
Example - continued
• An average of 70 is needed to get a C in the course.
Is it possible for Sheryl to get a C? If so, what grade
must she receive on the 7th exam to get a C in the
course?
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 8
Example - continued
• If her lowest grade of the exams already taken is dropped:
• What grade must she receive on her last exam to pass the
course?
• What grade must she receive on her last exam to get a C
the course?
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 9
Median
• The median is the value in the middle of a set
of ranked data.
• Example: Determine the median of
$327 $465 $672 $150 $230.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 10
Solution
Rank the data from smallest to largest.
$150 $230 $327 $465 $672
middle value
(median)
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 11
Example: Median (even data)
• Determine the median of the following set of
data: 8, 15, 9, 3, 4, 7, 11, 12, 6, 4.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 12
Solution
Rank the data:
3 4 4 6 7 8 9 11 12 15
There are 10 pieces of data so the median will
lie halfway between the two middle pieces the 7
and 8.
The median is (7 + 8)/2 = 7.5
3 4 4 6 7 8 9 11 12 15
(median) middle value
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 13
Mode
• The mode is the piece of data that occurs
most frequently.
• Example: Determine the mode of the data set:
3, 4, 4, 6, 7, 8, 9, 11, 12, 15.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 14
Solution
• The mode is 4 since it occurs twice and the
other values only occur once.
3, 4, 4, 6, 7, 8, 9, 11, 12, 15.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 15
Midrange
• The midrange is the value halfway between
the lowest (L) and highest (H) values in a set of
data.
lowest value + highest value
Midrange 
2
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 16
Example
• Find the midrange of the data set $327, $465,
$672, $150, $230.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 17
Example
• Find the midrange of the data set $327, $465,
$672, $150, $230.
150 + 672 822
Midrange 

 411
2
2
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 18
Example
• The weights of eight Labrador retrievers
rounded to the nearest pound are 85, 92, 88,
75, 94, 88, 84, and 101. Determine the
a)
b)
c)
d)
mean
median
mode
midrange
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 19
Example--dog weights 85, 92, 88, 75,
94, 88, 84, 101
a. Mean
85  92  88  75  94  88  84  101 707
x

 88.375
8
8
b. Median-rank the data
75, 84, 85, 88, 88, 92, 94, 101
The median is 88.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 20
Example--dog weights 85, 92, 88, 75,
94, 88, 84, 101
c. Mode-the number that occurs most
frequently. The mode is 88.
d. Midrange = (L + H)/2
= (75 + 101)/2 = 88
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 21
Measures of Position
• Measures of position are often used to make
comparisons.
• Two measures of position are percentiles and
quartiles.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 22
To Find the Quartiles of a Set of Data
1. Order the data from smallest to largest.
2. Find the median, or 2nd quartile (Q2), of the
set of data.
3. The first quartile, Q1, is the median of the
lower half of the data; that is, Q1, is the
median of the data less than Q2.
4. The third quartile, Q3, is the median of the
upper half of the data; that is, Q3 is the
median of the data greater than Q2.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 23
Example: Quartiles
• The weekly grocery bills for 23 families are as follows.
Determine Q1, Q2, and Q3.
170
330
225
75
95
210
80
225
160
172
270
170
215
130
190
270
240
310
74
280
270
50
81
• Order the data:
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 24
Example: Quartiles continued
• Order the data:
50 75 74 80 81 95 130
160 170 170 172 190 210 215
225 225 240 270 270 270 280
310 330
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 25
Example: Quartiles continued
Q2 is the median of the entire data set which
is 190.
Q1 is the median of the numbers from 50 to
172 which is 95.
Q3 is the median of the numbers from 210 to
330 which is 270.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 26
Example – Interpreting Statistics
The following statistics represent weekly salaries at the Midtown
Construction Company:
Mean
Median
Mode
$550
$540
$530
1st quartile
$510
3rd quartile
$575
83rd percentile $615
• What is the most common salary?
• What percent of employees’ salaries surpassed $540?
• What percent of employees’ salaries surpassed $575?
• What percent of employees’ salaries were less than $510?
• What percent of employees’ salaries surpassed $615?
• If the company has 100 employees, what is the total weekly
salary of all employees?
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 27
Measures of Dispersion
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 28
Measures of Dispersion
• Measures of dispersion are used to indicate
the spread of the data.
• The range is the difference between the
highest and lowest values; it indicates the
total spread of the data.
Range = highest value – lowest value
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 29
Example: Range
• Nine different employees were selected and
the amount of their salary was recorded. Find
the range of the salaries.
$24,000 $32,000 $26,500
$56,000 $48,000 $27,000
$28,500 $34,500 $56,750
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 30
Solution
• Range = $56,750  $24,000 = $32,750
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 31
Standard Deviation
• The standard deviation measures how much
the data differ from the mean. It is symbolized
with s when it is calculated for a sample, and
with  (Greek letter sigma) when it is
calculated for a population.
s
Copyright © 2009 Pearson Education, Inc.

S xx

2
n 1
Slide 13 - 32
To Find the Standard Deviation of a Set
of Data
1. Find the mean of the set of data.
2. Make a chart having three columns:
Data
Data  Mean
(Data  Mean)2
3. List the data vertically under the column
marked Data.
4. Subtract the mean from each piece of data
and place the difference in the Data  Mean
column.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 33
To Find the Standard Deviation of a Set
of Data continued
5. Square the values obtained in the Data 
Mean column and record these values in the
(Data  Mean)2 column.
6. Determine the sum of the values in the
(Data  Mean)2 column.
7. Divide the sum obtained in step 6 by n  1,
where n is the number of pieces of data.
8. Determine the square root of the number
obtained in step 7. This number is the
standard deviation of the set of data.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 34
Example
• Find the standard deviation of the following
prices of selected washing machines:
$280, $217, $665, $684, $939, $299
Find the mean.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 35
Example
• Find the standard deviation of the following
prices of selected washing machines:
$280, $217, $665, $684, $939, $299
Find the mean.
280  217  665  684  939  299 3084
x

 514
6
6
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 36
Example continued, mean = 514
Data
217
280
299
665
684
939
Data  Mean
Copyright © 2009 Pearson Education, Inc.
(Data  Mean)2
Slide 13 - 37
Example continued, mean = 514
Data
217
280
299
665
684
939
Data  Mean
297
234
215
151
170
425
0
Copyright © 2009 Pearson Education, Inc.
(Data  Mean)2
(297)2 = 88,209
54,756
46,225
22,801
28,900
180,625
421,516
Slide 13 - 38
Example continued, mean = 514
• Find the standard deviation:
s

S xx

2
n 1
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 39
Example continued, mean = 514
s

S xx

2
n 1
421,516

 84303.2  290.35
5
• The standard deviation is $290.35.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 40
Related documents