Measures of Central Tendency

andMeasures of variability

Lyra Vanessa SalesMa. Theresa C. SenoLiezel Vitto

Measures of Central Tendency

In simple terms, a measure of central tendency is either a midpoint an average, or the most frequent score in distribution of scores. The most common measures of central tendency are:MedianMeanMode

The Median

The median is a point in a scale which divides the scale into two equally.

Characteristic of the Median

1. The median computed from the ungrouped scores is called a crude, rough or counting median; from a class frequency distribution refined median.

2. The median is a midpoint of a scale and so it theoretically divides a scaled group into two equal subgroups.

3. The median is the most stable measures of central tendency because it is not affected by the meaningful of the scores. For instance, the series 5, 6, 7, 8, 9, the median is 7. Even if 9 becomes 50 the median is still 7.

Characteristic of the Median

Characteristic of the Median

4. The value of the median depends upon the magnitude of the middlemost scores or scores. If the middlemost sore is high the median is high, if it is low the median is low.

Uses of MedianThe median is computed and used when1.The exact midpoint of a scale or scaled

distribution is needed.2.A group with a quantified scaled characteristic,

for example ability, is to be divided into two equal subgroups.

3. A stable measure of central tendency is needed.

4. A group is too heterogeneous, that is, there are extremely high scores of the group and there are also extremely low scores. For instance, if the scores are 8, 10, 15, 25, and 35 the median is preferably used.

5. A class frequency distribution is badly skewed.

Uses of Median

Uses of Median

6. A scaled distribution is open-ended.X F

90 & ABOVE 385 680 875 5

70 & BELOW 4 N = 26

Computation of the median from Ungrouped Scores

Scale

Is a succession of numbers, steps, classes, degrees, gradations, or categories with a fixed interval.

Computation of the median from Ungrouped Scores

When there are few scores, say less than 30, the ordered arrangement of scores may be used to determine the median. The method follows:

A. Data Given: The following test scores in arithmetic:Odd number of scores: 56 88 90 76 72 82 73 63 75

63Even numbers of scores: 84 77 82 86 78 79 60 71

75 59 82

B. Procedure:1. Arrange the scores in a descending

order, that is, from the highest to lowest score.

Examples:

Example 19088827675737263636256

N =11

The median is 73, the middlemost score.

Example 2868482827978777571706059

N= 12

The middlemost scores are 78 and

77.

Mdn = 78 + 77 2

= 155 2 = 77.5

2. If the number of scores is odd, the middlemost score is the median. In example 1, the middlemost score is 73. Hence the median is 73.

3. If the number of scores is even, the median is the average of the two middle most scores. In example 2, the two middlemost scores are 78 and 77. The average is (78 + 77)/2=155/2=77.5. Hence, the median is 77.5.

Computation of the Median from a class Frequency Distribution

When there are many scores, say more than 30, class frequency distribution may be utilized in computing the median. The procedures follows:

A. Data given: The following test scores in English7 4 5 6 8 8 9 0 7 6 7 7 7 2 8 2 7 5 6 3 6 2 7 5 6 3

8 48 2 8 6 7 6 7 7 6 0 7 1 8 2 7 5 7 0 5 9 8 1 8 3 7 9

7 36 7 6 4 6 8 6 9 7 1 7 8 7 5 7 3 7 0 6 6 6 9 7 7 8 0

B. Procedure1. Group the scores into a class frequency

description using eight classes.2. Compute the cumulative frequencies upward

from the lowest class frequency to the highest class frequency. Table 10

Table 10

TABLE 10CLASS FREQUENCY DISTRIBUTION

X / CLASS TALLIES FREQUENCY CUMULATIVE FREQUENCY

90 – 94 I 1 4185 – 89 II 2 4080 – 84 IIIII-II 7 3875 – 79 IIIII-IIIII-I 11 3170 – 74 IIIII-III 8 2065 - 69 IIIII 5 1260 – 64 IIIII 5 755 - 59 II 2 2TOTAL N = 41

FREQUENCY CUMULATIVE FREQUENCY

1 412 407 3811 318 205 125 72 2

N = 41

The highest cumulative frequency

checks with the number of scores, N.

They should be equal.

B. Procedure3. Use the formula Mdn = E1 + in whichMdn = the medianE1 = the exact lower limit of the median classN = the number of cases or scoresCf = the cumulative frequency equal to or next lower than N/2F = the frequency of the median classI = the interval

4. Find the values of the symbol I the formula starting with N/2 as follows:Mdn = E1 + a. = = 20.5b. Cf = 20, the cumulative frequency lower than 20.5c. E1= 74.5 the exact lower limit of the median class, the

class which is just above the Cf of 20, 75-79d. F = 11, the frequency of the median class.e. I = 5, the interval which is the difference between two

adjacent lower limits.

5. Substitute the values for the symbols in the formula and solve.a. Substract 20 from 20.5

= .5Mdn = 74. 5 + Mdn = 74.5 +

b. Multiply .5 by 5Mdn = 74.5 +

c. Divide 2.5 by 11 to 3 decimal places

Mdn = 74.5 + .227

d. Add 74.5 and .227Mdn = 74.727

e. Round off 74.727 to 2 decimal places

Mdn = 74.73

Example:

Table II

X F Cf87 2 6784 4 6581 7 6178 9 5475 11 4572 12 3469 7 2266 8 1563 5 760 1 257 1 1

N=67

Mdn = E1 +

The Arithmetic Mean

Arithmetic mean is the average of a group of scores. Like the median, it is a measure of central tendency.

Characteristic of the mean

1.The arithmetic mean or simply mean, being the average of the scores, is the center of gravity or balance point of the scores.

2.The mean is easily affected by the magnitudes of the scores. For onstance, the mean of the scores 1, 2, 3, 4 and 5 is 3. If 5 becomes 6, the mean becomes 3.2.

3.The mean is the most reliable among the measures of central tendency.

Uses of the MeanThe mean is computed and used when1.The average of the scores is wanted.2.The center of gravity of the scores is desired.3.It is desired that every score has an effect

upon the measure of central tendency.4.The most reliable measure of central tendency

is needed.

5. The group from whom the scores have been derived is more or less homogeneous. The mean is not realistic if the group is heterogeneous. For instance, the mean of the scores 2, 3, 5, 7 and 50 is 13.4 which is very far from all the scores.

6. Other statistical measures in which the mean is involved such as the standard deviation, t—ratio, critical ratio, etc., are computed.

Computation of Mean by Averaging

The mean is composed by averaging when the scores are not too many, and they are not grouped or classified.

The method of averaging (long or absolute method) follows:

A.Data given : The following are the test scores in arithmetic:25 38 41 68 71 52 64 30 45 35 58

B. Procedure:1. Use the formula M =

In which M = the meanTX = the total of the scoresN = the number of cases or scores

X2538416871526430453558

TX= 527

2. Find the values of the symbol in the formula as follows:

a. Add the scores to find their total, TX.TX = 527

3. Substitute the values of the symbols in the formula and solve:

M = a. Divide 527 by 11 to 3 decimal placesM = 47. 909b. Round off 47. 91

Computation of the Mean by the Midpoint Method

The mean may be computed by the midpoint method if the scores are many, say more than 30, and are grouped into a class frequency distribution. The method follows:

A.Data Given : The following are scores in a test in grammar

50 91 45 60 51 54 83 56 68 71 55 73 85 66 74 65

87 45 84 70 69 59 61 67 68 77 72 80 67 71 74

57 54 64 64 75 62 69 56 77 79 66 65 63 70

B. Procedure1. Use the formulaM =

In whichM = the meanTFMp = the total of the products of the class midpoints and their corresponding frequencies.N = the number of cases or scores

2. Find the value of the symbols in the formula as follows:

a. Group the scores into a class frequency distribution

X Tallies F Mp FMp90 – 94 I 1 92 9285 – 89 II 2 87 17480 – 84 III 3 82 24675 – 79 IIII 4 77 30870 – 74 IIIII- III 8 72 57665 – 69 IIIII -IIIII 10 67 67060 - 64 IIIII -II 7 62 43455 – 59 IIIII 5 57 28550 – 54 IIII 4 52 20845 - 44 II 2 47 94

N = 46 TFMp= 3087

b. Determine the midpoints of the classes. Use the formula Mp = E1 +

c. Multiply the midpoints, Mp, by their corresponding frequencies to find Fmp92 x 1 =9287 x 3 = 246d. Add the FMp’s to find TFMp.TFMp = 3067

e. Add the frequencies to find N. N=463. Substitute the values for the symbol in the formula and solve:M = M = M = 67.108 / 67.11

Another Method

By the use of lower limit1.Use the formula M= In which,M = the meanTFL = the total of the products of the class lower limits and their corresponding frequencies.

I = the intervalN = the number of cases or scores

2. Find the values of the symbols in the formula as follows:

a. Multiply the lower limits of the classes by their corresponding frequencies.

b. Add the FL’s to find TFL

Table 13Class Frequency Distribution

X F FL90 1 9085 2 17080 3 24075 4 30070 8 56065 10 65060 7 42055 5 27550 4 20045 2 90

N= 46 TFL = 2995

M=

The ModeThe mode may be defined as the score

occurring the most number of times. It is the score with the highest frequency.

Determining the Mode from an Ordered Arrangement of scores

The mode determined from an ordered arrangement of scores is a rough or crude mode. The method of determining the mode from an ordered arrangement of scores follows:

A. Data Given: The following are the test scores in Geometry:

25 30 37 52 52 30 37 30 42 37

B. Procedure

1.Arrange the scores in a descending order, that is, from the highest to the lowest score.

2.Look for the scores that occurs the most number of times. This is the crude mode.

ExampleX525242413737373730303025

Mode = 37

Determining the Crude Mode from a Score Frequency Distribution

The method of determining the rough or crude mode from a score frequency distribution follows:

A. Data Given: The following are the test scores in Geometry:

25 30 37 52 52 30 37 30 42 37

B. Procedure

25 30 37 52 52 30 37 30 42 371.Arrange the scores in a descending order, but write the

score only once although the score occurs several times.2.Tally the scores. For each score, write a tally or short bar

opposite the score to which it is equal. Count the tallies and the score with the highest number of tallies is the rough mode.

X Tally F52 II 242 I 141 I 137 IIII 430 III 325 I 1

Determining the Crude Mode from a Class Frequency Distribution

The method of determining the crude mode from a class frequency distribution follows:

x F95 290 585 1080 1275 970 465 2

B. Procedure

1. Use the formula Mo = E1 + In which:

Mo = the modeE1 = the exact lower limit of the class with the highest frequencyI = interval

2. Look for the class with the highest frequency.3. The interval of the classes is 5.4. Substitute the values for their corresponding symbols in the formula and solve.

Mo = 79.5 + = 79.5 + 2.5

Mo = 82

Computation of the Refined Mode

1.When the refined mode, called the Pearson theoretical mode, is to be computed, use the formula

Mo = 3Mdn – 2MIn whichMo = modeMdn = the medianM = the mean

2. If the median and the mean of a frequency distribution are 84.5 and 83.75 respectively the values for their corresponding symbols in the formula and solve:Mo = 3Mdn – 2M

= 3 (84.5) – 2(83.75)= 167.50 – 253.50

Mo = 86

Computation of the Percentile

Percentile are points which divide a scale into 100 equal parts. A percentile indicates that a certain percent of scores under consideration are at or below the point and a certain percent above.

The methods of computing the percentile follows:

A. Data Given:x F CF95 2 5090 4 4885 7 4480 9 3775 10 2870 8 1865 6 1060 3 455 1 1

N = 50

B. Procedure Use the general formulaPp = E1 + I In whichN = the number of scoresPp = the percentile desired, p indicates the percentile rank desiredCf = the cumulative frequencyjust below Pn/100E1 = the exact lower limit of the class just above Cf, the path percentile classF= the frequency of the class containing the E1 or the pth percentile classI = the interval, the difference between any two adjacent classes

2. Find the values of the symbols in the formula. Suppose the desired percentile is: 35

Pp = 35N=50 = Cf = 10E1 = 69.5F = 8 I = 5

x F CF95 2 5090 4 4885 7 4480 9 3775 10 2870 8 1865 6 1060 3 455 1 1

N = 50

Substitute:Pp = E1 + IP35 = 69.5 + P35 = 69.5 + P35 = 69.5 + P35 = 69.5 + 4.687P35 = 74.187 / 74.19

Example: Suppose P87 is desired

P87 = E1 + IN = 50Cf = 37E1 = 84.50F = 7I = 5

P87 = E1 + I

P87 = 84.50 + P87 = 84.50 + P87 = 84.50 + P87 = 84.50 + 4.642P87 = 89.14