Measure of central tendency or location is single v value about which the set of
observations tend to cluster. It is also called the average.
The Arithmetic Mean or simply the mean, denoted by X, is the sum of all the
observations divided by the total number of all observations divided by the total number
of observations. In the symbols, if we let Xi = the value of the ith observation and n =
number of observation, then the mean is given by
X = ƩX
N
For grouped data,
X = ƩfXi where: Xi is the Midpoint
N f is the frequency
Another formula in computing the grouped mean of a frequency distribution is by coding.
X = Midpt + C ( Ʃfd)
N
Here, Midpoint is an assumed mean, which is chosen from the class marks, C is the class size or the class interval and d is the deviation of the ith from X computed as
d = Xi -XO where ; Xo is the Mdpt
c
Properties of Mean
- The mean of the deviations of the observations from the mean is always equal to zero. Ʃ(Xi – X)
- the sum of the squared deviations of the observations from the mean is smallest,
Ʃ(Xi – X)2 = minimum
- The mean reflects the magnitude of every observation, since every observation contributes to the value of the mean.
- It is easily affected by the presence of extreme values, and hence not a good measure of central tendency when extremes observations do occur.
- Means of subgroups maybe combined when properly weighted. Combined mean is called the weighted arithmetic mean.
The Median
The median, denoted by Md, is a single value that divides an array of observations into two equal parts, such that, fifty percent of the observations fall below it and fifty percent fall above it. The formula given below:
For ungrouped data:
Md = X ( N+1) if N is odd.
2
Md = Xn if N is even
2
For grouped data:
Md = Lmd + c ( N/2 – Fb)
fmd
where:
Lmd = lower True class boundary (TCB) of the median class, where the median class is a class whose less than Cumulative Frequency ( <CF) is greater than or equal to ½ of N.
C = class size
Fb = “less than” cumulative frequency of the class immediately preceding the median calss
Fmd = frequency of the median class
Properties of the Median
- It is positional value and hence is not affected by the presence of the extreme values unlike the mean.
- The sum of the absolute deviations from a point, say a, is smallest when a is equal to the median,
Ʃ/ Xi – Md/ is minimum
For group data, the equivalent expression is
Ʃ fi /Xi – Mdg/ is minimum
- The median is not suitable further computations and hence median of subgroups cannot be combined in the same manner as the mean.
- The median data can, the mode is calculated even with open-ended intervals provide the median class in no open- ended.
The Mode
The mode, denoted by Mo, is the value that occurs most frequently in the given data set. For grouped data, the mode is calculated using the formula:
fmo – fb
Mo = LMO + c 2fmo – fb - fa
Where:
Lmo = Lower TCB of the model class, where the modal class is the class with the highest frequency
c = class size
fmo = frequency of modal class
fb = frequency of the class immediately preceding the modal class
fa = frequency of the class immediately following the modal class
Properties of the Mode
- The mode is determined by the frequency and not by the values of the observations.
- It cannot be manipulated algebraically and hence modes of subgroups cannot be combined.
- The mode can be defined with qualitative and quantitative variables.
- The mode is very much affected by the method of grouping data.
- It cannot be computed with open-ended intervals provided the modal class is not open-ended.
Example: Solve the mean of the ungrouped Data.
Scores of 10 students of a 50-item test in English 1
X= 19, 15, 20, 31, 42, 31, 36, 18, 28,46
Solve mean ( X )
X = ƩX/N
= 286/10
= 28.6
Solve the Median (Md) Md = 10 +1 Md =5.5th
If N is even 2 5.5th of X
Md = N+1
2
Arrange the score from the lowest to highest or from the highest to the lowest.
X = 15, 18. 19, 20, 28, 31, 31, 36, 42, 46
Here N is even.
Md is the half way between 28 and 31. Add these two scores, the divide by two.
Md = 28 + 31
2
= 59/2
= 29.5
If N is odd.
Example:
X = 23, 28, 35, 38, 41, 46, 50
Determine the halfway.
Md = N + 1
2 So, 4th score from left to the right or from the right to the left is 38.
= 7+1 Therefore the is Md = 38.
2
= 8/2
= 4th score
Solve the mode.
Determine the mode of 10 students’ score in a 50-item test in English 1.
Arrange the scores from the lowest to the highest.
X = 15, 18, 19, 20, 28, 31, 31, 36, 42, 46
Look for a score that appeared most.
Answer: Mo = 31 Since 31 occurred twice
Solve Mean, Median, Mode for grouped data
Example:
A frequency distribution of 65 students’ score in 100-item test in Mathematics 1.
Class interval | Mdpt(X) | f | fx | c | cf | |
82-86 | 84 | 2 | 168 | 6 | 12 | |
77-81 | 79 | 4 | 316 | 5 | 20 | |
72-76 | 74 | 1 | 74 | 4 | 4 | |
67-71 | 69 | 0 | 0 | 3 | 0 | |
62-66 | 64 | 11 | 704 | 2 | 22 | |
56-61 | 59 | 8 | 472 | 1 | 8 | |
52-56 | 54 | 15 | 810 | 0 | 0 | |
47-51 | 49 | 10 | 490 | -1 | -10 | |
42-46 | 44 | 4 | 176 | -2 | -8 | |
37-41 | 39 | 6 | 234 | -3 | -18 | |
32-36 | 34 | 4 | 136 | -4 | -16 | |
Total | 65 | 3580 | 11 | 14 | ||
X = Mdpt + c (Ʃfd/N) Md = LL + c(N/2 –Ʃcf<) Mo = LL+c/2 (f1-f2)
= 54 + 5 (11/65) fc 2fo-f2-f1
= 54.85 = 52 + 5(65/2 – 24) = 52+5/2(8-10)
15 2(15-10-8)
= 52 + 5(32.5 – 24) = 52+2.5(-2)
15 -6
= 52 + 5 (8.5) = 52.5+ 0.83
15 = 53.33
= 52 + 2.83
= 54.83
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