CHAPTER SIX
MEASURES OF CENTRA
L TENDENCY
6.
1Introduction
A type o
f
descriptive s
tatistics kno
wn as
measure
s of central tendency i
s sometimes referred to as meas
urements of location or central
value. It entails identifying a
single measure that characteri
zes a collection of scores. Th
ese metrics are used by the
researcher to categorize a g
roup of data that each have
a single value or number. It comes in handy a lot when describing overall performance in the educational industry as a researcher. The outcome produced by the computation of a locational measure captures or stands for the usual or average score acquired by a set of participants. That is, a whole collection of scores may be represented by a measure value.
The most popular descriptive statistics of this kind, mean, median, and mode, will be covered in this chapter. Each measure of central tendency is a rough estimate for a distinct scale.
6.2Objectives
By the end of this chapter, you should be able to:
describe the three most commonly used measures
compute the three most commonly measures of location.
explain major characteristics of mean, median, and mode and their uses in research studies.
Compute the mean using the assumed mean
Calculate the mean using the coding factor
Compute mean using coding factor
6.3Mean
The mean, which is sometimes referred to as the arithmetic average, is derived by summing the scores for each topic individually and dividing the result by the number of subjects. Typically, the sign for mean is used x̄ (read x ‘bar’). The formula is given as:
Mean = x̄ = \\frac{\\sum \\mathbf{X}}{\\mathbf{n}}
Where x̄ = the arithmetic mean
∑ = Greek letter sigma meaning '‘sum of’'
X = scores
n = number of sample size. This is for ungrouped data.
For the grouped data, the formula for calculating the mean is given as:
x̄ = \\frac{\\sum \\mathbf{f}\\mathbf{X}}{\\sum \\mathbf{f}}
Where fx = the products of frequencies and individual scores
∑f = the summation of frequencies.
The Mean is sub-divided into: Assumed Mean, Geomettric Mean, Quadratic Mean and Harmonic Mean.
Assumed Mean (AM): this method is carryout by deducting each value of a variable from an assumed mean which could be the class mark of any group or class interval. Assumed mean can be employed for both ungrouped and group Data.
\\overline{\\boldsymbol{x}}=\\mathrm{A}+\\frac{\\sum fd}{N}
Where A is the assumed mean
d is the deviation from assumed mean (\\mathrm{A}) for each of the value of variable
ie \\overline{\\boldsymbol{x}}-\\mathbf{A}
\\boldsymbol{N} is the sample size \\sum f
Geometric Mean (GM): The geometric mean of n positive value is the root of their product for a set of observations:
x_{1}, x_{2,}x_{3},----,x_{u},geometric mean is calculated as:
\\mathbf{G}\\mathbf{M}=\\sqrt[]{\\left(x_{1}\\right)\\left(x_{2}\\right)\\left(x_{3}\\right)\\left(x_{u}\\right)}
Quadratic Mean (QM): it is the square root of the mean of the sqaures of the given values, it is calculated as:
\\mathbf{Q}\\mathbf{M}= \\sqrt[]{\\frac{{\\mathrm{x}_{1}}^{2}+{\\mathrm{x}_{2}}^{2}+{\\mathrm{x}_{3}}^{2}+---,{\\mathrm{x}_{\\mathrm{n}}}^{2}}{\\mathrm{N}}}
Harmonic Mean: it is the recent reciprocal of the arithmetic mean of the rciprocal of the observation and it is calculated as:
\\mathbf{H}\\mathbf{M}=\\frac{\\mathrm{N}}{\\Sigma _{i}n=1}=\\frac{N}{\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1}{x_{3}}+---+\\frac{1}{x_{n}}}
Major Properties of Mean
It is the arithmetic mean of the measurement in the data set.
It is only one value for a data set.
The extreme measurement influenced its value and trimming can assist to decrease the degree of outlier influence.
In determining the mean of the complete data set, means of subjects can be combined.
It is only used for quantitative data.
6.3.1Mean of Ungrouped Data
Considering the following scores in Geometry Achievement Test
12, 0, 13, 6, 2, 3, 9, 8, 14, 6
To calculate the mean, the simple formula x̄=\\frac{\\sum \\mathbf{X}}{\\mathbf{n}} can be used when the number of scores is small.
This is shown below:
Mean = x̄ = \\frac{12+0+13+6+2+3+9+8+16+6}{10} = 7.5
In this case, the mean value of students which stand at 7.5 typically represent the overall performance of the group. This simple formula can only be applied when the scores are fewer and not arranged in a frequency table.
However, if the scores are arranged in a frequency table, the second formula provided can be used that is x̄ = \\frac{\\sum \\mathbf{f}\\mathbf{x}}{\\sum \\mathbf{f}}. To apply this formula, you have the following steps to be followed:
Step 1:Create a frequency distribution table as indicated in table 6.1. That is first column for the scores while the second column for the frequencies.
Step 2:Develop a third column for the product of each score with the corresponding frequency provided as (Fx)
Step 3:Sum-up the frequencies of the scores in column 2, (that is ∑f)
Step 4:Add-up the product of each score with its frequency, that is ∑fx
Step 5:Substitute in the formular that the results obtained in step 3 and step 4
Thus, the procedures can be illustrated using the example below.
Suppose the scores below are obtained for 30 students in geometry performance test
5, 8, 10, 8, 1, 7, 7, 10, 8, 8, 7, 10, 7, 7, 7, 5, 4, 3, 2, 4, 3,2, 4, 7, 3, 9, 2, 6, 5, 6, 5, 9, 6, 9, 6, 6, 9, 5, 3, 2, 2.
Table 6.1: Scores in Geometry Performance Test
Column 1: Score X | Column 2: Frequency | Column 3: FX |
10 | 3 | 30 |
9 | 5 | 45 |
8 | 4 | 32 |
7 | 7 | 49 |
6 | 6 | 36 |
5 | 5 | 25 |
4 | 2 | 08 |
3 | 3 | 09 |
2 | 4 | 08 |
1 | 1 | 01 |
\\sum \\mathbf{F}=30 \\sum \\mathbf{F}\\mathbf{X}=243 | ||
Mean = x̄ = \\frac{\\sum \\mathbf{f}\\mathbf{X}}{\\sum \\mathbf{f}}=\\frac{243}{30}=8.1
The Researcher sometimes obtained large scores and may decide to group them before the calculation of mean. That remind you about the discussion in chapter Two. The following procedures will guide you in calculating the mean of grouped data.
6.3.2Mean of Grouped Data
A Researcher who is interested in grouping data for calculating the mean of a frequency distribution should follow these steps:
Step 1:Group the data in class interval (cl)
Step 2:Record the frequency (f)
Step 3:Calculate the class mark of each class (x)
Step 4:Find the products of class mark and frequency (fx).
Step 5:Sum up the products resulted in fx
Step 6:Divide \\frac{\\sum \\mathbf{f}\\mathbf{x}}{\\sum \\mathbf{f}}
The data provided in table 6.3 gives the details on how to compute the mean for group data.
Table 6.2: Scores in Geometry Performance Test
Class Interval | Frequency | Class Mark | FX |
61—63 | 4 | 92.5 | 370 |
64—66 | 17 | 97 | 1649 |
67—69 | 36 | 101.5 | 3654 |
70—72 | 26 | 106 | 2756 |
73—74 | 7 | 110.5 | 773.5 |
\\sum \\mathrm{f}=90 \\sum \\mathrm{f}\\mathrm{x}=9202.5 | |||
By the formula = \\frac{\\sum \\mathbf{f}\\mathbf{x}}{\\sum \\mathbf{f}}=\\frac{9202.5}{90}=102.25
6.3.3Mean Scores in Research
A Researcher in Mathematics Education who is interested to investigate on factors affecting the teaching and learning of Mathematics in Junior Secondary Schools. The Researcher developed a questionnaire on a 4-point Likert scale to seek for the opinions of Mathematics Teachers over a number of factors.
Strongly Agree (SA) = 4, Agree (A) = 3, Disagree (D) = 2, Strongly Disagree (SD) = 1.
Table 6.3 shows the response of 30 Teachers to the research questionnaire administered by the Researcher.
Table 6.3: Responses of Teachers to the Questionnaire
S/N | Factors | SA (4) | A (3) | D (2) | SD (1) |
Attitudes of Mathematics teachers towards teaching of Mathematics | 7 | 15 | 4 | 4 | |
Understanding of Mathematics language by students | 10 | 5 | 5 | 10 | |
Inadequate of lecture method used by all teachers | 15 | 5 | 5 | 5 |
To find out whether attitude of Mathematics Teacher is a contributing factor, the Researcher can adopt this process as follows:
Table 6.4: Responses on Attitude of Mathematics Teachers
Score (x) | F | |
1 | 4 | |
2 | 4 | |
3 | 15 | |
4 | 7 | |
Total | 30 | |
Mean score of the response on the factor; attitude of Mathematics Teacher could be calculated by following these steps:
Step 1:Drawing a frequency distribution Table from Table 6.4
Step 2:Find the product of each score with its frequency.
Step 3:Sum-up the products.
Step 4:Divide the result obtained in Step 3 by the sum of scores.
Adopting the steps above, then the computation of the mean for data in Table 6.5 is shown below
Table 6.5: Mean for Data in Table 6.5
X | F | FX |
1 | 4 | 4 |
2 | 4 | 8 |
3 | 15 | 45 |
4 | 7 | 28 |
n = 30 85 | ||
x̄ = \\frac{\\sum \\mathbf{f}\\mathbf{x}}{\\mathbf{n}}=\\frac{85}{30}=2.8
Furthermore, the Researcher can continue to calculate the mean scores for the other factors by the following the steps explained above. Mean scores of the other factors are showed in Table 6.6
Table 6.6: Mean Scores of Factors that Affect Teaching and Learning of Mathematics
S/N | Factors | Mean Scores |
Attitudes of Mathematics teachers towards teaching of Mathematics | 2.8 | |
Understanding of Mathematics language by students | 2.5 | |
Inadequate of lecture method used by all teachers | 3 |
In order to answer the research question asked that what factors affect the teaching and learning of mathematics in J. S.S based on the perception of mathematics teacher. We need to draw benchmark for comparing the mean score, since the mean on this scale is given as :
\\frac{1+2+3+4}{4}=2.5
Decision can be taken by the Researcher:
Retain as factor if the x̄ ≥ 2.5
Reject as a factor if the x̄ < 2.5
By this benchmark, the Researcher will conclude that all the three factors affect teaching and learning of Mathematics in Junior Secondary Schools.
6.4Median
This is the score in the middle of the distribution, dividing the collection of scores into two equal-sized groups. The scores must be put in either ascending or descending order to get the median.
However, when the scores are ordered in order of their size, the median for an even number of points is the average of the middle score. The median serves as the middle score when there are an odd number of points.
Major Properties of Median
It is another measure of central, which the score lies. The major properties of median are:
It is only one median for any data set
The medians cannot add to find the median of the complete data set
Extreme measurement does not affect the median score
Median score in group data is stable, even when the data organized into different classes.
6.4.1Median for Ungrouped Data
Determine the median of the sets of scores in a test
0, 7, 5, 13, 8, 9, 6, 1, 2
7, 4, 5, 3, 1, 2, 6, 7, 4, 8
Solution
0, 1, 5, 6, 7, 8, 9, 13, 2
The middle number may be calculated as 6. 6 is the median.
1, 2, 3, 4, 4, 5, 6, 7, 7, 8
Here we have 4 & 5 as middle number then the median is \\frac{4+5}{2}=\\frac{9}{2}=4.5
To find the median of ungrouped data it is easy to use the following steps:
Step 1:Rearrange the score in ascending or descending order
Step 2:Counting equal number of scores from each end then the middle score in the median, but in a case number of scores in even then two middle scores area picked.
Step 3:Calculate the average of these middle scores, the result is the median as
illustrated in above example ‘b’.
When the researcher obtained large number of scores this will be described in 6.4.2
6.4.2Median for Grouped Data
To find the median of grouped data, it is slightly more difficult when compared with that of ungrouped dat. Since the real values of the scores are unknown, you can only say the median occur in particular class interval and you cannot locate it within the interval.
To calculate the median in this case, you should use the formula:
\\mathrm{M}\\mathrm{e}\\mathrm{d}\\mathrm{i}\\mathrm{a}\\mathrm{n}=\\mathrm{L}+\\frac{\\left(\\mathrm{N}/2-\\mathrm{F}\\right)\\mathrm{C}}{\\mathrm{f}}
Where L = lower class boundary of the class interval consisting of the middle value.
F = Total number of scores below L
f = Frequency of the median class
C = Class interval size
N = Number of scores
In order to find the median with large number of scores, the data in table 6.8 can be used for illustration.
Table 6.7: Cumulative Frequency Distribution of Score
Class Interval | F | Cf |
29—31 | 8 | 28 |
26—28 | 5 | 20 |
23—25 | 3 | 15 |
20—22 | 2 | 12 |
17—19 | 4 | 10 |
14—16 | 6 | 6 |
Total (N) = 28 | ||
Applying the formula to obtain the median
Median = L + \\frac{\\left(\\mathrm{N}/2-\\mathrm{F}\\right)\\mathrm{C}}{\\mathrm{f}}
Where L = 22.5
N = 28
F = 12
f = 3
C = 3
Substituting in the total formula, then we have:
Median = 22.5 + \\frac{\\left(\\mathrm{N}/2-\\mathrm{F}\\right)3}{3}
= 22.5 + \\frac{2\\times 3}{3}
= 22.5 + 2
= 24.5
The median as one measure of location is only useful when the distribution of scores is skewed. At that particular case the median serves very useful. However, the median has its own limitation because it is largely considered the mid-point of the distribution, where it lies without concerning for other scores. It is not useful for calculation of higher statistics but better the mode.
To obtain the median of grouped data the following procedure can be adopted.
Step 1: Prepare a cumulative frequency distribution table.
Step 2: Determine half of the distribution as the case may be.
Step 3: Locate the cumulative frequency where the median lies.
Step 4: Find the lower boundary of the score that contains the median.
Step 5: Apply the formula: Median = L + \\frac{\\left(\\mathbf{N}/2-\\mathbf{F}\\right)\\mathbf{C}}{2}
6.5Mode
The mode is the highest frequency that occur in a distribution or the most observed score in a set of data. The calculation of mode in data is easy and simple for example, consider this distribution of scores in a test:
Score (X) | 20 | 15 | 12 | 10 | 9 |
Frequency (F) | 5 | 7 | 2 | 3 | 4 |
The score that is most typical is 15 because it has the highest frequency, which is 7. It is important to recognize that a distribution of scores may have more than one mode. When distribution of scores indicates one mode is known as unimodal, likewise when distribution have two modes is said to be bimodal. In case, it is more than two modes is multi-modal in nature. The following referred as major properties of mode.
Major Properties of Mode
It is the highest frequency in a distribution.
Sometimes the mode can be more than one.
Extreme measurement does influence the mode.
It cannot be combined to find the mode of the complete data set.
It can be applied for both quantitative and qualitative data.
It is important to note the mode as a measure of central tendency has limitations. One of the limitations is that mode is not necessarily the representation of other values in the distribution. Another short coming of the mode it cannot be used in the calculation of other statistics. Thus, with these limitations, it is better for a researcher to try other computation of higher statistics.
6.5.1Mode for Ungrouped Data
Let’s see how to identify modes in the following cases.
Example 6.1
In the distribution below, identify the mode:
10, 20, 11, 35, 20, 15, 23, 25, 20, 12, 19, 20.
By observation, you will discover that 20 is repeated 4 times. The mode of score is 20.
Example 6.2
Determine the mode in the frequency table below:
X | 20 | 19 | 18 | 17 | 16 | 15 | 14 | 13 |
F | 7 | 3 | 5 | 6 | 7 | 6 | 2 | 3 |
Again, by observation, you will discover that 20 and 16 has the highest frequency of the value 7 in the distribution. Therefore, the distribution has two modes in nature, the modes are 20 and 16.
6.5.2 Modes for Grouped Data
In order to calculate mode of grouped data, the formula below can be applied
X =\\mathrm{L}+\\frac{\\left(\\mathrm{\\DifferentialD }^{1}\\right)^{\\mathrm{c}}}{\\mathrm{d}^{1}+\\mathrm{d}^{2}}
Where L = the exact lower limit of the modal class
d1 = difference between frequency of modal class and frequency of the
class before the modal class.
d2 = difference between frequency of modal class and frequency of the
class after the modal class.
c = the class size.
Example 6.3
Calculate the mode in the frequency distribution table provide below:
Table 6.9: Scores in Distribution
Class Interval | Frequency |
75—79 | 2 |
70—74 | 2 |
65—69 | 7 |
60—61 | 9 |
55—59 | 11 |
50—54 | 6 |
By inspection, the modal class is 55—59
Applying the formula \\mathrm{L}+\\frac{\\left(\\mathrm{\\DifferentialD }^{1}\\right)^{\\mathrm{c}}}{\\mathrm{d}^{1}+\\mathrm{d}^{2}}
Where L = 54.5, C = 5, d1 = 11—6 = 5, d2 = 11—9 = 2
X = 54.5 + \\frac{\\left(5\\right)^{5}}{5+2}=54.5+\\frac{\\left(5\\right)^{5}}{7}
= 54.5 + 446.43
= 500.93
6.6 Assumed Mean (A.M)
Computation of mean can be carryout using the assumed mean. If the data collected by researcher consist of large items. It is a short-cut method for calculating the mean. This method involves the use of an arbitrary value from the list of items considered for the purpose of calculating the arithmetic mean is known as Assumed Mean (A.M)
A number is very close to the middle score is usually suggested to avoid ambiguity. The deviation (i.e the assumed mean) subtracted from each of the score denoted by the letter ‘u’ while the mean of all the deviations recorded is known as mean deviation.
The actual arithmetic mean of the distribution is sum of the assumed mean and deviation
i.e. Actual Mean = Assumed Mean + Mean Deviation
the formula is given as:
\\overline{x}=\\mathrm{A}+\\frac{\\Sigma \\mathrm{f}\\alpha }{\\Sigma \\mathrm{f}}
Where A = Assured Mean
\\alpha = Deviation from the Assumed Mean
f = frequency
\\overline{x} = Arithmetic Mean
Example 6:4
A frequency table showing scores of nine students in achievement test
26,16,22,43,12,19,14,35,13
Using an assumed mean of 14 to calculate the mean score of the distribution.
Solution
Arrange the scores in ascending order
Table 5.10: Data for Test Scores
Scores (x) | Deviations: d = x—AM |
12 | 12—14 = –2 |
13 | 13—14 = –1 |
14 | 14—14 = 0 |
16 | 16—14 = 2 |
19 | 19—14 = 5 |
22 | 22—14 = 8 |
26 | 26—14 = 12 |
35 | 35—14 = 21 |
43 | 43—14 = 29 |
\\Sigma d=74 |
Assumed Deviation =\\frac{\\Sigma d}{n}=\\frac{79}{9}=8.2
Actual Mean = Assumed Mean + Assumed Deviation
=14+8.2
=22.2
Example 6.5
The table below indicates the masses of a number of students in a class. Calculate the Actual Mean using the Assumed Mean as 62.
Table 6.11: Masses of Students
Masses (kg) | 48 | 51 | 54 | 62 | 65 | 68 | 72 |
Frequency | 7 | 4 | 3 | 5 | 1 | 4 | 6 |
Solution
Let prepare the require table for the distribution
Table 6.12 Data obtained as Masses of Students
Masses (kg) | Dev. (\\propto )=-\\mathbf{A}\\mathbf{M} |
\\boldsymbol{F}\\boldsymbol{r}\\boldsymbol{e}\\boldsymbol{q}\\boldsymbol{u}\\boldsymbol{e}\\boldsymbol{n}\\boldsymbol{c}\\boldsymbol{y}(\\boldsymbol{f}) |
\\boldsymbol{f}\\boldsymbol{d} |
48 | 48—62 = - 14 | 7 | –91 |
51 | 51—62 = –11 | 4 | –41 |
54 | 54—62 = –8 | 3 | –24 |
62 | 62—62 = 0 | 5 | 0 |
65 | 65—62 = 3 | 1 | 3 |
68 | 68—62 = 6 | 4 | 24 |
72 | 72—62 = 10 | 6 | 60 |
A.M = 62 |
\\sum \\boldsymbol{f}=30 |
\\sum \\boldsymbol{f}\\boldsymbol{d}=\\hbox{--}72 |
Assumed =\\frac{\\sum fd}{\\sum f}=\\frac{\\hbox{--}72}{30}=\\hbox{--}2.4
Actual Mean =A.M+A.D=62\\hbox{--}2.4=59.6
Coded Factor
When the distribution data in a grouped data is large, coded factor is used with the addition of the assumed Mean. The coded is the class size (class width) and denoted by V. Itis obtained through the class width to divide each of the deviation.
The table below shows the distribution weight of students in a class.
Table 6:13 Weight (kg) of Students
Weight (kg) | No. of students |
10—19 | 6 |
20—29 | 13 |
30—39 | 10 |
40—49 | 6 |
50—59 | 11 |
60—69 | 1 |
70—79 | 3 |
Using an assured Mean of 39.5, calculate the Mean weight of the distribution
Solution
A frequency table showing the different columns and their values is constructed for the distribution as required.
Table 6.14: Distribution Data of Students’ Weight.
Weight | Class Mark | Frequency |
\\boldsymbol{d}=\\boldsymbol{x}\\hbox{--}\\boldsymbol{A}.\\boldsymbol{M} |
\\boldsymbol{u}=\\sfrac{\\boldsymbol{d}}{\\boldsymbol{c}} |
\\boldsymbol{f}\\boldsymbol{d} |
\\boldsymbol{f}\\boldsymbol{u} |
10—19 | 14.5 | 6 | –25 | –2.5 | –150 | –15 |
20—29 | 24.5 | 13 | –15 | –1.5 | –195 | –19.5 |
30—39 | 34.5 | 10 | –5 | –0.5 | –50 | –5 |
40—49 | 44.5 | 6 | 5 | 0.5 | 30 | 3 |
50—59 | 54.5 | 11 | 15 | 1.5 | 165 | 16.5 |
60.69 | 64.5 | 1 | 25 | 2.5 | 25 | 2.5 |
70—79 | 74.5 | 3 | 35 | 3.5 | 105 | 10.5 |
\\sum \\boldsymbol{f}=50 |
\\sum \\boldsymbol{f}=\\hbox{--}70 |
\\sum \\boldsymbol{f}\\boldsymbol{u}=\\hbox{--}7 |
Note that the class size (c) = 10
Also,
\\mathrm{A}=39.5
\\sum f=50
\\sum fd=\\hbox{--}70
\\sum fu=\\hbox{--}7
Then compute actual mean using Assumed Mean:
Therefore, \\overline{x}=A+\\frac{\\sum fu}{\\sum f}
=39.5+\\frac{\\hbox{--}70}{50}
=39.5\\hbox{--}1.4
=38.1kg
Alternatively using the Assumed mean and coded factor
\\overline{x}=A+\\left(\\frac{\\sum fu}{\\sum f})^{c}
=39.5+\\left(\\frac{\\hbox{--}70}{50})^{10}
=39.5+\\left(0.14\\right)10
=38.1kg
Mid Range
It is another form of calculating average that is sought from set of Data. Midrange is the addition of the highest score divided by two. For example, the following set of scores in achievement test are:
22,24,25,28,29,32,47and60.
Highest score = 60
Lowest score = 22
Midrange = \\frac{60+22}{2}=41
6.7Coding Method for Computing Mean
When researcher has large values of variable and computation of the mean used become difficult, and tedious, the coding method can be used. This is done by converting the x-values into simpler values for the computation, and then later convert it back again.
This is carryout subtracting or adding from each original, if that is possible, then divide or multiply these new values which are easily manageable. Then find the mean of the x-values \\overline{x} and by employing a suitable decoding formula to obtain\\overline{x}. An example will be given to show the method in detail.
To apply the coding method to find mean score, the following steps can be adopted.
Step I:Choose a convenient value of x (i.e., near the middle of the range)
Step II:Subtract these values chosen from every other value of x and recorded it in column 2.
Step III:Then convert the values into an even simpler from by dividing the values in column 2 by chosen value. Record the result in column 3.
Step IV:Add all class frequencies, write summation in column 4
Step V:find the product of values in each frequency. Record the result in column 5.
Step VI:Using the column 4 and 5 to find the mean.
Now, let us illustrate the procedures to find the mean scores using coding method.
Example 6:8
A research measure in millimeters of 50 bolts gave the following frequency distribution in Table 6.15
Length x (mm) | 20.2 | 20.4 | 20.6 | 20.8 | 21.0 | 21.2 | 21.4 |
Frequency (f) | 2 | 6 | 10 | 15 | 9 | 7 | 1 |
Solution
\\boldsymbol{L}\\boldsymbol{e}\\boldsymbol{n}\\boldsymbol{g}\\boldsymbol{t}\\boldsymbol{h}
(\\boldsymbol{m}\\boldsymbol{m}) |
\\boldsymbol{D}\\boldsymbol{e}\\boldsymbol{v}\\boldsymbol{i}\\boldsymbol{a}\\boldsymbol{t}\\boldsymbol{i}\\boldsymbol{o}\\boldsymbol{n}\\boldsymbol{f}\\boldsymbol{r}\\boldsymbol{o}\\boldsymbol{m}
\\boldsymbol{c}\\boldsymbol{h}\\boldsymbol{o}\\boldsymbol{s}\\boldsymbol{e}\\boldsymbol{n}\\boldsymbol{v}\\boldsymbol{a}\\boldsymbol{l}\\boldsymbol{u}\\boldsymbol{e}
(\\boldsymbol{x}\\hbox{--}20.8) | Units is 0.2mn
\\boldsymbol{x}_{\\boldsymbol{u}}=\\frac{\\boldsymbol{x}\\hbox{--}20.8}{0.2} |
(\\boldsymbol{f}) |
\\boldsymbol{x}_{\\boldsymbol{u}}\\boldsymbol{f} | ||
20.2 | –0.6 | –3 | 2 | –6 | ||
20.4 | –0.4 | –2 | 6 | –12 | ||
20.6 | –0.2 | –1 | 10 | –10 | ||
20.8 | 0 | 0 | 15 | 0 | ||
21.0 | 0.2 | 1 | 9 | 9 | ||
21.2 | 0.4 | 2 | 7 | 14 | ||
21.4 | 0.6 | 3 | 1 | 3 | ||
\\therefore \\overline{\\boldsymbol{x}}=\\frac{\\hbox{--}2}{50}=\\hbox{--}0.04 |
\\boldsymbol{n}=\\sum \\boldsymbol{f}=50 |
\\sum \\boldsymbol{x}_{\\boldsymbol{u}}\\boldsymbol{f}=\\hbox{--}2.0 | ||||
If \\overline{x}_{u}=\\hbox{--}0.04, we can return back to the original unit of x
Decoding
In the coding step, the last step was divided by 0.2, the reverse is to multiply by 0.2 return to the correct units of(x\\hbox{--}20.8). Therefore:
\\overline{x}\\hbox{--}20.8=\\hbox{--}0.04\\times 0.2=\\hbox{--}0.08
Add 20.8 to both sides, you have
\\overline{x}=20.8\\hbox{--}0.8=20.72=\\overline{x}=20.72
6.8General Observation about Measure of Central Tendency
The mean can be affected by extreme values in the distribution but not the Median. For example: Take look into this set of scores.
x\\colon 6,8,9,10
\\overline{x}=\\frac{6+7+8+9}{5}=\\frac{40}{5}=8.
While the Median=8i.e, Mean = Median.
If the set of scores is changed to be 6, 7, 8, 9, 20
The mean = \\overline{x}=\\frac{6+7+8+9}{5}=\\frac{50}{5}=10 but the Median remains 8.
The mode may not exist in a distribution and is not always unique, but mean median always exists and has uniqueness.
\\boldsymbol{E}\\boldsymbol{x}\\boldsymbol{a}\\boldsymbol{m}\\boldsymbol{p}\\boldsymbol{l}\\boldsymbol{e}\\colon The set of scores 6, 7, 8, 9, 10 has no Mode 6, 6, 7, 7, 9 is bimodal. And the mode is not unique because it may have more than one and two.
The addition of the deviation from the mean always equal zero
The four basics on measure of location by a constant lead to the following definitions.
Addition and Subtraction
For any distribution, if a constant is added to (or subtracted from) each observation, the corresponding measure of central tendency changes by the same value or score.
\\mathbf{E}\\mathbf{x}\\mathbf{a}\\mathbf{m}\\mathbf{p}\\mathbf{l}\\mathbf{e}\\colon Look at the following set of scores in a test: 3, 8, 9, 10, 10.
Original data | Plus 3 | Minus 3 |
3 | 3 | 0 |
8 | 6 | 5 |
9 | 11 | 6 |
10 | 13 | 7 |
10 | 13 | 7 |
40 | 55 | 25 |
Mean = | 8 | 11 | 5 |
Median = | 9 | 12 | 6 |
Mode = | 10 | 13 | 7 |
b. Multiplication or Division
If each observation is multiplied (or divided) by a constant, the corresponding measure of central tendency must also be multiplied or divided by the same value or score.
\\boldsymbol{E}\\boldsymbol{x}\\boldsymbol{a}\\boldsymbol{m}\\boldsymbol{p}\\boldsymbol{l}\\boldsymbol{e}\\colon Using the Data above
Original Data | Multiply by 3 | Divide by 3 |
3 | 9 | 1 |
8 | 24 | 2.7 |
9 | 27 | 3 |
10 | 30 | 3.3 |
10 | 30 | 3.3 |
40 | 120 | 13.3 |
Mean \\overline{X}= | 5 | 24 | 2.7 |
Median= | 9 | 27 | 3 |
Mode = | 10 | 30 | 3.3 |
Student Activity
The following data are the summary of response of 50 secondary school teachers to a questionnaire on factors affect teaching efficiency.
S/N | Factors | Strongly Agree 4 | Agree 3 | Disagree 2 | Strongly Disagree 1 |
Teacher’s qualification | 20 | 10 | 15 | 5 | |
Appropriateness of teaching instruction | 5 | 15 | 25 | 5 | |
Lack of Textbooks | 30 | 10 | 5 | 5 | |
Using of Learning Aids | 20 | 20 | 5 | 5 | |
Lack of interest by students | 25 | 10 | 5 | 5 |
Using the data above:
Calculate the teachers’ mean responses for each factor.
On the basis of the statistics calculated, rank the factors.
Analyse the data to identify which factor affect the teaching in primary schools
Find the mean, median and mode for the data obtained in the frequency table.
Class Interval | Frequency |
34.9—39.9 | 5 |
39.9—44.9 | 8 |
44.9—49.9 | 13 |
49.9—54.9 | 5 |
– 59.9 | 1 |
Enumerate and explain what you know about measures of central tendency.
List major properties of mean, mode and median.
Compute the mean, mode median for the following data.
27, 21, 19, 32, 33, 25, 27.
4, 3, 1, 8, 8, 1, 4.
Gams (mm) | Frequency |
40—49 | 7 |
50—59 | 10 |
60—69 | 5 |
70—79 | 8 |
80—89 | 18 |
90—99 | 10 |
100—109 | 12 |
Using the assumed mean of 84.5, calculate the gam size of the distribution.
References
Adu, D.B (1998). Comprehensive Mathematics for Senior Secondary Schools, Ltd.
Awotunde, P. O. & Ugodulunwa (2002). An Introduction to Statistical Methods in Education. Printed and published in Nigeria by Fab Anieh (Nig) Ltd.
Razaq, B. & Ajayi, O. S. (n.d). Research methods & Statistical Analysis.