Measures Of Central Tendency
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MEASURES OF CENTRAL TENDENCY LESSON 2
WHAT IS A MEASURE OF CENTRAL TENDENCY? Commonly referred to as an average, is a single value that represents a data set. Its purpose is to locate the center of the data set.
REMEMBER THE 3 M'S? 😃
MEAN •
The arithmetic mean, often called as the mean, is the most frequently used measure of central tendency. The mean is the only measure in which all values plays an equal role meaning to determine its values you would need to consider all the values of any given data set. The mean is appropriate to determine the central tendency of an interval or ratio data. ( x bar ) = used to represent a sample.
μ = used to denote the mean of a population.
A. Properties of Mean 1. A set of data has only one mean. 2. Mean can be applied for interval or ratio data. 3. All values in the data set are included in computing the mean 4. The mean is very useful in comparing two or more data sets. 5. Mean is affected by the extreme small or large values on the data set. 6. The mean cannot be computed for the data in a frequency distribution with an open-ended class. 7. Mean is most appropriate in symmetrical data.
B. Mean for Ungrouped Data : Mean = Sum of all values / Number of Values
Sample Mean : Where :
= sample Mean ( it is "x bar")
x = the value of any particular observations or measurement. = sum of all x's. n = total number of values in the sample.
Population Mean:
μ=
Where:
μ = population Mean ( it is read "mu" ) x = the value of any particular observations or measurement. = sum of all x's. N = total number of values in the population.
Example 1: The daily rates (in peso) of a sample of eight employees at GMS Inc. are 550, 420, 560, 500, 700, 670, 860, 480. Find the mean daily rate of employee. Solution:
=
The sample Mean daily salary of employees is 592.50.
Example 2: Find the population mean of the ages of 9 middlemanagement employees of a certain company. The ages are 53, 45, 59, 48, 54, 46, 51, 58 and 55. Solution:
μ=
μ The mean population age of middle-management employee is 52.11.
Sample Mean for Grouped Data Sample Mean: Where:
= sample Mean = frequency
X = the value of any particular observations or measurement. = sum of all products of f and x's. n = total number of values in the sample.
Population Mean:
μ Where:
μ = population f = frequency
x= the value of any particular observations or measurement. = sum of all the products of f and x's. N = total number of values in the population.
Example: SJS Travel Agency, a nationwide local travel agency, offers special rates on summer period. The owner wants additional information on the ages of those people taking travel tours. A random sample of 50 customers taking travel tours last summer revealed these ages.
18
29
42
57
61
67
37
49
53
47
24
34
45
58
63
70
39
51
54
48
28
36
46
60
66
77
40
52
56
49
19
31
44
58
62
68
38
50
54
48
27
36
46
59
64
74
39
51
55
48
Class Limits
Frequency (f)
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
•
Solution:
Step 1: Determine the midpoints on each class limit. Step 2: Multiply each class frequency (f) with the corresponding midpoint (x) to obtain the product fx. Step 3: Get the sum of product fx. Step 4: Apply the formula to obtain the value of the sample Mean.
Class Limits
Frequency (f)
Midpoints (x)
fx
18-26
3
22
66
27-35
5
31
155
36-44
9
40
360
45-53
14
49
686
54-62
11
58
638
63-71
6
67
402
72-80
2
76
152
Total
50
---
= 2,459
Thus, the mean age of the frequency distribution of people taking travel is 49.18.
MEDIAN •
Is the midpoint of the data array. When the data set is ordered
whether ascending or descending, it is called data array. Properties of Median
1. The median is unique, there is only one median for a set of data. 2. The median is found by arranging the set of data in ascending or descending form and getting the value of the middle observation. 3. Median is not affected by the extreme small or large values. 4. Median can be computed for ordinal, interval and ratio data. 5. Median is appropriate in a skewed data.
MEDIAN for UNGROUPED DATA
To determine the value of median for ungrouped we need to consider two rules: 1. If n is odd, the median is the middle ranked. 2. If n is even, then the median is the average of the two middle ranked values. Median (Rank Value) = (n+1) / 2 Note that n is the population/sample size.
Example: Find the median of the ages of 9 middle-management
employees of a certain company. The ages are 53, 45, 59, 48, 54, 46, 51, 58 and 55. Solution: Step 1: Arrange the data in order. 45, 46, 48, 51, 53, 54, 55, 58, 59 Step 2: Select the middle rank using the formula
Step 3: Identify the median in the data set. 45, 46, 48, 51, 53, 54, 55, 58, 59
HENCE, THE MEDIAN AGE IS 53 YEARS.
Example 2: The daily rates (in Peso) of a sample of eight employees at GMS Inc. are 550, 420, 560, 500, 700, 670, 860, 480. Find the median daily rate of employee. Solution: Step 1: Arrange the data in order. 420, 480, 500, 550, 560, 670, 700, 860 Step 2: Select the middle rank value using the formula
Step 3: Identify the median in the data set. 420, 480, 500, 550 560, 670, 700, 860
SINCE THE MIDDLE POINT FALLS BETWEEN 550 AND 560, WE CAN DETERMINE THE MEDIAN OF THE DATA SET BY GETTING THE AVERAGE OF THE TWO VALUES.
Therefore, the median daily rate is 555.
Median for Grouped Data Take note that the median is located in the middle value of the frequency distribution. It is the value that separates the upper half of the distribution from the lower half. It is also obvious to note that it is a measure of central tendency because it is the exact center of the scores in a distribution. Median (Ranked Value)= N/2
Where: LB = lower boundary of the median class.
N = number of sample size (or population) cf = cumulative frequency before the median class. f = frequency of the median class. i = interval.
Using the example on SJS Travel Agency, determine the median of the frequency distribution on the ages of 50 people taking travel tours. Given the table Class Limits
f
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
Step 1: Determine the median class using the formula.
Limits f cf Step 2:Class Construct a cumulative frequency column in he table. 18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
Step 3: Identify the median class by locating the nth rank in the table. Class Limits
f
cf
18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
THIS CLASS COVERS 18TH TO 31ST RANK IN THE FREQUENCY DISTRIBUTION. THE 25TH RANK BELONGS IN THIS CLASS.
Step 4: Determine the values of LB, cf, f, i, and N.
LB = 45-0.5 = 44.5
I = 27 - 18 =9 OR I = 35 - 26 =9
Class Limits
f
cf
18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
Total
50
CF
MEDIAN CLASS
FREQUENCY
Step 5: Apply the formula to compute the value of the median.
Thus, the median is 49.64, observed that the median will fall within the class boundary of the median class.
MODE •
The mode is the value in the data set that appears most frequently. Like the median and unlike the mean, extreme values in the data set do not affect the mode. A data may not contain any mode if none of the values is "most typical". Unimodal- there is one value that occurs most frequently. Bimodal- if the data has two values with the greatest frequency.
Multimodal- if the data has more than two values with the greatest frequency. No Mode - if all values in the data set have the same frequencies.
Using the example on SJS Travel Agency, determine the estimated mode of the frequency distribution on the ages of 50 people taking travel tours. Given the table Class Limits
f
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
Solving for the estimated Mode Given: LB = 44.5 Fm = 14 F m-1 = 9 F m+1 =11 i=9 Mode = 44. 5 ((14-9)/(8))x9 = 50.13
WHAT IS A MEASURE OF CENTRAL TENDENCY? Commonly referred to as an average, is a single value that represents a data set. Its purpose is to locate the center of the data set.
REMEMBER THE 3 M'S? 😃
MEAN •
The arithmetic mean, often called as the mean, is the most frequently used measure of central tendency. The mean is the only measure in which all values plays an equal role meaning to determine its values you would need to consider all the values of any given data set. The mean is appropriate to determine the central tendency of an interval or ratio data. ( x bar ) = used to represent a sample.
μ = used to denote the mean of a population.
A. Properties of Mean 1. A set of data has only one mean. 2. Mean can be applied for interval or ratio data. 3. All values in the data set are included in computing the mean 4. The mean is very useful in comparing two or more data sets. 5. Mean is affected by the extreme small or large values on the data set. 6. The mean cannot be computed for the data in a frequency distribution with an open-ended class. 7. Mean is most appropriate in symmetrical data.
B. Mean for Ungrouped Data : Mean = Sum of all values / Number of Values
Sample Mean : Where :
= sample Mean ( it is "x bar")
x = the value of any particular observations or measurement. = sum of all x's. n = total number of values in the sample.
Population Mean:
μ=
Where:
μ = population Mean ( it is read "mu" ) x = the value of any particular observations or measurement. = sum of all x's. N = total number of values in the population.
Example 1: The daily rates (in peso) of a sample of eight employees at GMS Inc. are 550, 420, 560, 500, 700, 670, 860, 480. Find the mean daily rate of employee. Solution:
=
The sample Mean daily salary of employees is 592.50.
Example 2: Find the population mean of the ages of 9 middlemanagement employees of a certain company. The ages are 53, 45, 59, 48, 54, 46, 51, 58 and 55. Solution:
μ=
μ The mean population age of middle-management employee is 52.11.
Sample Mean for Grouped Data Sample Mean: Where:
= sample Mean = frequency
X = the value of any particular observations or measurement. = sum of all products of f and x's. n = total number of values in the sample.
Population Mean:
μ Where:
μ = population f = frequency
x= the value of any particular observations or measurement. = sum of all the products of f and x's. N = total number of values in the population.
Example: SJS Travel Agency, a nationwide local travel agency, offers special rates on summer period. The owner wants additional information on the ages of those people taking travel tours. A random sample of 50 customers taking travel tours last summer revealed these ages.
18
29
42
57
61
67
37
49
53
47
24
34
45
58
63
70
39
51
54
48
28
36
46
60
66
77
40
52
56
49
19
31
44
58
62
68
38
50
54
48
27
36
46
59
64
74
39
51
55
48
Class Limits
Frequency (f)
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
•
Solution:
Step 1: Determine the midpoints on each class limit. Step 2: Multiply each class frequency (f) with the corresponding midpoint (x) to obtain the product fx. Step 3: Get the sum of product fx. Step 4: Apply the formula to obtain the value of the sample Mean.
Class Limits
Frequency (f)
Midpoints (x)
fx
18-26
3
22
66
27-35
5
31
155
36-44
9
40
360
45-53
14
49
686
54-62
11
58
638
63-71
6
67
402
72-80
2
76
152
Total
50
---
= 2,459
Thus, the mean age of the frequency distribution of people taking travel is 49.18.
MEDIAN •
Is the midpoint of the data array. When the data set is ordered
whether ascending or descending, it is called data array. Properties of Median
1. The median is unique, there is only one median for a set of data. 2. The median is found by arranging the set of data in ascending or descending form and getting the value of the middle observation. 3. Median is not affected by the extreme small or large values. 4. Median can be computed for ordinal, interval and ratio data. 5. Median is appropriate in a skewed data.
MEDIAN for UNGROUPED DATA
To determine the value of median for ungrouped we need to consider two rules: 1. If n is odd, the median is the middle ranked. 2. If n is even, then the median is the average of the two middle ranked values. Median (Rank Value) = (n+1) / 2 Note that n is the population/sample size.
Example: Find the median of the ages of 9 middle-management
employees of a certain company. The ages are 53, 45, 59, 48, 54, 46, 51, 58 and 55. Solution: Step 1: Arrange the data in order. 45, 46, 48, 51, 53, 54, 55, 58, 59 Step 2: Select the middle rank using the formula
Step 3: Identify the median in the data set. 45, 46, 48, 51, 53, 54, 55, 58, 59
HENCE, THE MEDIAN AGE IS 53 YEARS.
Example 2: The daily rates (in Peso) of a sample of eight employees at GMS Inc. are 550, 420, 560, 500, 700, 670, 860, 480. Find the median daily rate of employee. Solution: Step 1: Arrange the data in order. 420, 480, 500, 550, 560, 670, 700, 860 Step 2: Select the middle rank value using the formula
Step 3: Identify the median in the data set. 420, 480, 500, 550 560, 670, 700, 860
SINCE THE MIDDLE POINT FALLS BETWEEN 550 AND 560, WE CAN DETERMINE THE MEDIAN OF THE DATA SET BY GETTING THE AVERAGE OF THE TWO VALUES.
Therefore, the median daily rate is 555.
Median for Grouped Data Take note that the median is located in the middle value of the frequency distribution. It is the value that separates the upper half of the distribution from the lower half. It is also obvious to note that it is a measure of central tendency because it is the exact center of the scores in a distribution. Median (Ranked Value)= N/2
Where: LB = lower boundary of the median class.
N = number of sample size (or population) cf = cumulative frequency before the median class. f = frequency of the median class. i = interval.
Using the example on SJS Travel Agency, determine the median of the frequency distribution on the ages of 50 people taking travel tours. Given the table Class Limits
f
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
Step 1: Determine the median class using the formula.
Limits f cf Step 2:Class Construct a cumulative frequency column in he table. 18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
Step 3: Identify the median class by locating the nth rank in the table. Class Limits
f
cf
18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
THIS CLASS COVERS 18TH TO 31ST RANK IN THE FREQUENCY DISTRIBUTION. THE 25TH RANK BELONGS IN THIS CLASS.
Step 4: Determine the values of LB, cf, f, i, and N.
LB = 45-0.5 = 44.5
I = 27 - 18 =9 OR I = 35 - 26 =9
Class Limits
f
cf
18-26
3
3
27-35
5
8
36-44
9
17
45-53
14
31
54-62
11
42
63-71
6
48
72-80
2
50
Total
50
CF
MEDIAN CLASS
FREQUENCY
Step 5: Apply the formula to compute the value of the median.
Thus, the median is 49.64, observed that the median will fall within the class boundary of the median class.
MODE •
The mode is the value in the data set that appears most frequently. Like the median and unlike the mean, extreme values in the data set do not affect the mode. A data may not contain any mode if none of the values is "most typical". Unimodal- there is one value that occurs most frequently. Bimodal- if the data has two values with the greatest frequency.
Multimodal- if the data has more than two values with the greatest frequency. No Mode - if all values in the data set have the same frequencies.
Using the example on SJS Travel Agency, determine the estimated mode of the frequency distribution on the ages of 50 people taking travel tours. Given the table Class Limits
f
18-26
3
27-35
5
36-44
9
45-53
14
54-62
11
63-71
6
72-80
2
Solving for the estimated Mode Given: LB = 44.5 Fm = 14 F m-1 = 9 F m+1 =11 i=9 Mode = 44. 5 ((14-9)/(8))x9 = 50.13
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