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Lecture 3+4
Measure of Central
Tendency
Average
Average (Central Tendency )
Quantities
Observation
the Values
the center of
The average means
Average is a number which represents a whole data.
TYPES OF AVERAGE
Mean
Mode
Median
Midrange
Mean (Arithmetic Mean)
The mean is the sum of the values, divided by the total number of values.
๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ =
๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
าง
๐‘ฅ =
๐‘ฅ1+๐‘ฅ2+๐‘ฅ3+๐‘ฅ4+โ‹ฏ+๐‘ฅ๐‘›
๐‘›
าง
๐‘ฅ =
ฯƒ ๐‘ฅ๐‘–
๐‘›
Example 1 : Find Mean of 5 , 7 , 8 ,10,5
Solution : As ๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ =
๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
So าง
๐‘ฅ =
5+7+8+10+5
5
=
35
5
=7 Answer
Example 2 : The number of calls that a local police department
responded to for a sample of 9 months is shown. Find the
mean. 475, 447, 440, 761, 993, 1052, 783, 671, 621
Solution: ๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ =
๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘ 
So าง
๐‘ฅ =
475+ 447+ 440+761+993+ 1052+ 783+ 671+ 621
9
าง
๐‘ฅ =
6243
9
เดค
x โ‰ˆ 693.7 (Approximately)
Mean of Grouped Data
Mean for
grouped Data าง
๐‘ฅ =
ฯƒ ๐‘“๐‘–๐‘ฅ๐‘–
๐‘›
Where
าง
๐‘ฅ is mean
๐‘“๐‘– is frequency of class
๐‘ฅ๐‘– is mid point of class
๐‘› is cumulative frequencies
Example: Find Mean of these data represent the net worth
(inMillions) of 46 national cooperations.
Class Limits Frequencies
10-20 3
21-31 8
32-42 15
43-53 7
54-64 10
65-75 3
The Weighted Mean
Sometimes you must find the mean of a data set in which not
all values are equally represented.
When there number values of a quantity , we use weighed
mean technique to calculate.
The weighted mean of a variable X is calculated by
multiplying each value by its corresponding weight and
dividing the sum of the products by the sum of the weights.
Example
The Median
The median is the midpoint of the data array.
Before finding this point, the data must be arranged in
ascending or increasing order.
When the data set is ordered, it is called a data array.
Example: The number of tornadoes that have occurred in
the United States over an 7-year period follows. Find the
median. 684, 764, 656, 702, 856, 1133 , 1132
Solution
Step 1 : Arranging the data values in ascending order.
684, 764, 656, 702, 856, 1133 , 1132
Step 2: Data in ascending order
656, 684, 702, 764, 856, 1132, 1133
Step 3: the middle value in the data is 764 .
Hence Median is 764
Example: The number of tornadoes that have occurred in
the United States over an 8-year period follows. Find the
median. 684, 764, 656, 702, 856, 1133 , 1132,1303
Solution
Step 1 : Arranging the data values in ascending order.
684, 764, 656, 702, 856, 1133 , 1132, 1303
Step 2: Data in ascending order
656, 684, 702, 764, 856, 1132, 1133,1303
Step 3: the middle values in the data are 764 and 856
Hence Median is
764+856
2
=
1620
2
= 810 Answer
MEDIAN OF GROUPED DATA
C is cumulative frequency preceding median class
โ„Ž is width of median class
๐‘“ is frequency of median class
๐‘› is sum of frequencies of all classes
Where ๐‘™ is lower class boundary
Median of
Grouped Data
= ๐‘™ +
โ„Ž
๐‘“
๐‘›
2
โˆ’ ๐ถ
Steps for finding the Median
Step 1:
๐‘›
2
half of total observations
Step 2 : Median Class
Step 3 : terms of median class ๐‘“ ๐‘Ž๐‘›๐‘‘ โ„Ž
Step 4 : Cumulative Frequencies (C) before median
class
Example: Find Median of these data represent the net worth
(inMillions) of 46 national cooperations.
Class Limits Frequencies
10-20 3
21-31 8
32-42 15
43-53 7
54-64 10
65-75 3
Class Limits Frequencies Cummulative Frequency C
10-20 3 3
21-31 8 11
32-42 15 26
43-53 7 33
54-64 10 43
65-75 3 46
เท ๐‘“ = 46
Solution :
32-42 is median class because C is greater than
๐’
๐Ÿ
here.
So Frquency of median class is ๐‘“ = 15
Width of of median class is โ„Ž = 11 ๐‘Ž๐‘›๐‘‘ ๐‘™ = 31.5
Median for grouped data is given by
= ๐‘™ +
โ„Ž
๐‘“
๐‘›
2
โˆ’ ๐ถ
= 31.5 +
11
15
46
2
โˆ’ 11
= 31.5 + 8.8
= 40.3 Answer
Mode
The value that occurs most often in a data set is called the mode.
Example : Find the mode of the temperature โ„ƒ of week of June in Karachi, Sindh
40 , 45 , 44 , 44,43,42 , 44
Solution : Here most frequent number is 44
so 44 is mode
There can be more than one Mode(s) or
no mode at all
Example 1: 4,7,7,8,9,3,7 here 7 is mode (Unimodal data)
Example1: 5,6,5,6,5,6 here 5 and 6 are modes (Bimodal data )
Example 2: 3,6,4,3,6,4 here 3,4 are 6 are modes (Multimodal data)
MODE OF GROUPED DATA
โ„Ž is width of median class
๐‘“2 is frequency succeding modal class
๐‘“๐‘š is frequency of modal class
๐‘“1 is frequency preceding modal class
Where ๐‘™ is lower class boundary of modal class
Mode of Grouped
Data = ๐‘™ +
(๐‘“๐‘šโˆ’๐‘“1)
(๐‘“๐‘šโˆ’๐‘“1)+(๐‘“๐‘šโˆ’๐‘“2)
. โ„Ž
Example: Find Mode of these data represent the net worth
(inMillions) of 46 national cooperations.
Class Limits Frequencies
10-20 3
21-31 8
32-42 15
43-53 7
54-64 10
65-75 3
Solution: Table Modal Class is 32-42 with frequency ๐’‡๐’Ž = ๐Ÿ๐Ÿ“
Class Limits Frequencies
10-20 3
21-31 8
32-42 15
43-53 7
54-64 10
65-75 3
๐’‡๐Ÿ = ๐Ÿ– , ๐’‡๐Ÿ = ๐Ÿ• , ๐’‰ = ๐Ÿ๐Ÿ ๐’‚๐’๐’… ๐’ = ๐Ÿ‘๐Ÿ. ๐Ÿ“
As Mode = ๐‘™ +
(๐‘“๐‘šโˆ’๐‘“1)
(๐‘“๐‘šโˆ’๐‘“1)+(๐‘“๐‘šโˆ’๐‘“2)
. โ„Ž
= 31.5 +
(15โˆ’8)
(15โˆ’8)+(15โˆ’7)
. 11
= 31.5 + 5.133
= 36.633
Midrange
The midrange is a rough estimate of the middle
The midrange is defined as the sum of the smallest
and largest values in the data set, divided by 2.
๐‘€๐‘–๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘”๐‘’ =
๐‘ ๐‘š๐‘Ž๐‘™๐‘™๐‘’๐‘ ๐‘ก ๐‘ฃ๐‘Ž๐‘™๐‘ข + ๐ฟ๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘  ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’
2
Example : The number of bank failures for a recent
five-year period is shown.
Find the midrange. 3, 30, 148, 157, 71
Solution : As ๐‘€๐‘–๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘”๐‘’ =
๐‘ ๐‘š๐‘Ž๐‘™๐‘™๐‘’๐‘ ๐‘ก ๐‘ฃ๐‘Ž๐‘™๐‘ข +๐ฟ๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘  ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’
2
Smallest Value =3 and Largest Value = 157
so ๐‘€๐‘… =
3+157
2
=
160
2
MR = 80 Answer
Lecture 3 & 4 Measure of Central Tendency.pdf
Lecture 3 & 4 Measure of Central Tendency.pdf
Lecture 3 & 4 Measure of Central Tendency.pdf

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Lecture 3 & 4 Measure of Central Tendency.pdf

  • 1. Lecture 3+4 Measure of Central Tendency Average
  • 2. Average (Central Tendency ) Quantities Observation the Values the center of The average means Average is a number which represents a whole data.
  • 4. Mean (Arithmetic Mean) The mean is the sum of the values, divided by the total number of values. ๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ = ๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘  ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘  าง ๐‘ฅ = ๐‘ฅ1+๐‘ฅ2+๐‘ฅ3+๐‘ฅ4+โ‹ฏ+๐‘ฅ๐‘› ๐‘› าง ๐‘ฅ = ฯƒ ๐‘ฅ๐‘– ๐‘›
  • 5. Example 1 : Find Mean of 5 , 7 , 8 ,10,5 Solution : As ๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ = ๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘  ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘  So าง ๐‘ฅ = 5+7+8+10+5 5 = 35 5 =7 Answer
  • 6. Example 2 : The number of calls that a local police department responded to for a sample of 9 months is shown. Find the mean. 475, 447, 440, 761, 993, 1052, 783, 671, 621 Solution: ๐ด๐‘ฃ๐‘’๐‘Ÿ๐‘Ž๐‘”๐‘’ = ๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘  ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‰๐‘Ž๐‘™๐‘ข๐‘’๐‘  So าง ๐‘ฅ = 475+ 447+ 440+761+993+ 1052+ 783+ 671+ 621 9 าง ๐‘ฅ = 6243 9 เดค x โ‰ˆ 693.7 (Approximately)
  • 7. Mean of Grouped Data Mean for grouped Data าง ๐‘ฅ = ฯƒ ๐‘“๐‘–๐‘ฅ๐‘– ๐‘› Where าง ๐‘ฅ is mean ๐‘“๐‘– is frequency of class ๐‘ฅ๐‘– is mid point of class ๐‘› is cumulative frequencies
  • 8.
  • 9. Example: Find Mean of these data represent the net worth (inMillions) of 46 national cooperations. Class Limits Frequencies 10-20 3 21-31 8 32-42 15 43-53 7 54-64 10 65-75 3
  • 10. The Weighted Mean Sometimes you must find the mean of a data set in which not all values are equally represented. When there number values of a quantity , we use weighed mean technique to calculate. The weighted mean of a variable X is calculated by multiplying each value by its corresponding weight and dividing the sum of the products by the sum of the weights.
  • 12. The Median The median is the midpoint of the data array. Before finding this point, the data must be arranged in ascending or increasing order. When the data set is ordered, it is called a data array.
  • 13. Example: The number of tornadoes that have occurred in the United States over an 7-year period follows. Find the median. 684, 764, 656, 702, 856, 1133 , 1132 Solution Step 1 : Arranging the data values in ascending order. 684, 764, 656, 702, 856, 1133 , 1132 Step 2: Data in ascending order 656, 684, 702, 764, 856, 1132, 1133 Step 3: the middle value in the data is 764 . Hence Median is 764
  • 14. Example: The number of tornadoes that have occurred in the United States over an 8-year period follows. Find the median. 684, 764, 656, 702, 856, 1133 , 1132,1303 Solution Step 1 : Arranging the data values in ascending order. 684, 764, 656, 702, 856, 1133 , 1132, 1303 Step 2: Data in ascending order 656, 684, 702, 764, 856, 1132, 1133,1303 Step 3: the middle values in the data are 764 and 856 Hence Median is 764+856 2 = 1620 2 = 810 Answer
  • 15. MEDIAN OF GROUPED DATA C is cumulative frequency preceding median class โ„Ž is width of median class ๐‘“ is frequency of median class ๐‘› is sum of frequencies of all classes Where ๐‘™ is lower class boundary Median of Grouped Data = ๐‘™ + โ„Ž ๐‘“ ๐‘› 2 โˆ’ ๐ถ
  • 16. Steps for finding the Median Step 1: ๐‘› 2 half of total observations Step 2 : Median Class Step 3 : terms of median class ๐‘“ ๐‘Ž๐‘›๐‘‘ โ„Ž Step 4 : Cumulative Frequencies (C) before median class
  • 17. Example: Find Median of these data represent the net worth (inMillions) of 46 national cooperations. Class Limits Frequencies 10-20 3 21-31 8 32-42 15 43-53 7 54-64 10 65-75 3
  • 18. Class Limits Frequencies Cummulative Frequency C 10-20 3 3 21-31 8 11 32-42 15 26 43-53 7 33 54-64 10 43 65-75 3 46 เท ๐‘“ = 46 Solution : 32-42 is median class because C is greater than ๐’ ๐Ÿ here. So Frquency of median class is ๐‘“ = 15 Width of of median class is โ„Ž = 11 ๐‘Ž๐‘›๐‘‘ ๐‘™ = 31.5
  • 19. Median for grouped data is given by = ๐‘™ + โ„Ž ๐‘“ ๐‘› 2 โˆ’ ๐ถ = 31.5 + 11 15 46 2 โˆ’ 11 = 31.5 + 8.8 = 40.3 Answer
  • 20. Mode The value that occurs most often in a data set is called the mode. Example : Find the mode of the temperature โ„ƒ of week of June in Karachi, Sindh 40 , 45 , 44 , 44,43,42 , 44 Solution : Here most frequent number is 44 so 44 is mode
  • 21. There can be more than one Mode(s) or no mode at all Example 1: 4,7,7,8,9,3,7 here 7 is mode (Unimodal data) Example1: 5,6,5,6,5,6 here 5 and 6 are modes (Bimodal data ) Example 2: 3,6,4,3,6,4 here 3,4 are 6 are modes (Multimodal data)
  • 22. MODE OF GROUPED DATA โ„Ž is width of median class ๐‘“2 is frequency succeding modal class ๐‘“๐‘š is frequency of modal class ๐‘“1 is frequency preceding modal class Where ๐‘™ is lower class boundary of modal class Mode of Grouped Data = ๐‘™ + (๐‘“๐‘šโˆ’๐‘“1) (๐‘“๐‘šโˆ’๐‘“1)+(๐‘“๐‘šโˆ’๐‘“2) . โ„Ž
  • 23. Example: Find Mode of these data represent the net worth (inMillions) of 46 national cooperations. Class Limits Frequencies 10-20 3 21-31 8 32-42 15 43-53 7 54-64 10 65-75 3
  • 24. Solution: Table Modal Class is 32-42 with frequency ๐’‡๐’Ž = ๐Ÿ๐Ÿ“ Class Limits Frequencies 10-20 3 21-31 8 32-42 15 43-53 7 54-64 10 65-75 3 ๐’‡๐Ÿ = ๐Ÿ– , ๐’‡๐Ÿ = ๐Ÿ• , ๐’‰ = ๐Ÿ๐Ÿ ๐’‚๐’๐’… ๐’ = ๐Ÿ‘๐Ÿ. ๐Ÿ“
  • 25. As Mode = ๐‘™ + (๐‘“๐‘šโˆ’๐‘“1) (๐‘“๐‘šโˆ’๐‘“1)+(๐‘“๐‘šโˆ’๐‘“2) . โ„Ž = 31.5 + (15โˆ’8) (15โˆ’8)+(15โˆ’7) . 11 = 31.5 + 5.133 = 36.633
  • 26. Midrange The midrange is a rough estimate of the middle The midrange is defined as the sum of the smallest and largest values in the data set, divided by 2. ๐‘€๐‘–๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘”๐‘’ = ๐‘ ๐‘š๐‘Ž๐‘™๐‘™๐‘’๐‘ ๐‘ก ๐‘ฃ๐‘Ž๐‘™๐‘ข + ๐ฟ๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘  ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’ 2
  • 27. Example : The number of bank failures for a recent five-year period is shown. Find the midrange. 3, 30, 148, 157, 71 Solution : As ๐‘€๐‘–๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘”๐‘’ = ๐‘ ๐‘š๐‘Ž๐‘™๐‘™๐‘’๐‘ ๐‘ก ๐‘ฃ๐‘Ž๐‘™๐‘ข +๐ฟ๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘  ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’ 2 Smallest Value =3 and Largest Value = 157 so ๐‘€๐‘… = 3+157 2 = 160 2 MR = 80 Answer