Measures of Central Tendency
1. Mean,
2. Median and
3. Mode
Population mean:
μ=
Σ𝑥
𝑁
Sample mean:
𝑥̅ =
Σ𝑥
𝑛
Properties of the Mean
• The sum of all deviations of all measurements in a set from the mean is zero
• It can be calculated for any set of numerical data, so it always exists.
• Any set of numerical data has one and only one mean.
• It lends itself to higher statistical treatment.
• It is the most reliable since it takes into account every item in the set of data
• It is greatly affected by extreme values
• It is used only if the data are interval or ratio and when normally distributed
Median - is the value of the middle item in an ordered arrangement of data
Properties of the Median
• It is a point in the distribution with 50% of the scores on each side of it. It is the midpoint of the
distribution.
• It is not affected by extreme values
• It is appropriate to use when there are extreme values
• It is used when the data is ordinal
• It exists in both quantitative and qualitative data
Mode is the score or scores which occur most often
Properties of the Mode
• is used when you want to find the value that occurs most often
• It is a quick approximation of the average
• It is an inspection average
• It is the most unreliable among the three measures of central tendency because its value is
undefined in some observations
Ungrouped Data
Suppose we have the data
17, 22, 19, 18, 19
Find the mean, median and mode
To determine the mean, we get the sum of the items and divide by the number of items.
Symbol for mean = µ for population and 𝑥̅ for the sample
95
µ or 𝑥̅ = 5 = 19
To determine the median, let us arrange the items in either ascending or descending order. The data
becomes
17, 18, 19, 19, 22
The middle term is 19.
It is easier to determine the median if the number of items is odd, if it is even, the median is the mean of
the two middle terms. For example, using the data
17, 18, 19, 19, 20, 21
The two middle terms are both 19, so the median is also 19.
To determine the mode, it is the data that occur most often. The mode is also 19.
Weighted Mean
There are times when some values are given more importance than others. A very good example is the
computation of the General Weighted Average (GWA). In this case, weight is the number of units
Compute the General Weighted Average of a particular student whose grades are as follows:
Subject
Grade
Unit
Physics
3.0
4
English
2.0
3
Calculus
1.5
5
P. E.
1.0
2
3.0(4)+2.0(3)+1.5(5)+1.0(2)
𝑋̅𝐺𝑊𝐴 =
4+3+5+2
𝑋̅𝐺𝑊𝐴 = 1.96
Another application of weighted mean is in getting the mean responses in a Likert-type question. Likerttype question is used when the researcher wants to know the opinions of the respondents regarding any
topic or issues of interest.
Grouped Data: Mean
𝑥̅ =
𝛴𝑓𝑥
𝑛
Using the distribution:
Class Interval 𝑓
171 – 179
2
162 – 170
4
153 – 161
5
144 – 152
12
135 – 143
9
126 – 134
5
117 – 125
3
𝛴f = 40
𝑥 (class mark)
175
166
157
148
139
130
121
𝑓𝑥
350
664
785
1776
1251
650
363
𝛴fx = 5839
𝑥̅ =
𝛴𝑓𝑥
𝑛
5839
𝑥̅ = 40
𝑥̅ = 146
Median
Md = 𝐿𝑚 +
𝑁
2
( −𝑐𝑓)(𝑖)
𝑓𝑚
Where: 𝐿𝑚 = lower limit boundary of the median class
𝑁 = total frequency
𝑐𝑓 = cumulative frequency before median class
𝑓𝑚 = frequency of median class
𝑓
2
4
5
12
9
5
3
Class Interval
171 – 179
162 – 170
153 – 161
144 – 152
135 – 143
126 – 134
117 – 125
𝑐𝑓
40
38
34
29 ==> median class
17
8
3
Determination of Median
STEP 1: Determine median class by dividing the total frequency by 2 and locating the quotient in the cf.
In this distribution, it is in the 30 – 34 class interval.
STEP 2: Substitute the values in the formula. The lower limit boundary is the mean of the lower limit of
the median class and the upper limit of the class below it. In this case, the lower limit boundary is
144+143
= 143.5.
2
20−17
Md = 143.5 + 12 (9)
= 143.5 + 2.25
Md = 146
MODE: Mo = 𝐿𝑚𝑜 +
∆1
(𝑖)
∆1+ ∆2
Where:
𝐿𝑚𝑜 = lower limit boundary of the modal class
Δ1 = difference between the highest frequency and the frequency just below it
Δ2 = difference between the highest frequency and the frequency just above it
Modal Class - class interval with the highest number of frequency.
Class Interval
171 – 179
162 – 170
153 – 161
144 – 152
135 – 143
126 – 134
117 – 125
f
2
4
5
12 ==>modal class
9
5
3
3
(9)
Mo = 143.5 +
3+7
= 143.5 + 2.7
Mo = 146
Quartiles
The distribution is divided into four parts
First Quartile
𝑄1 = 𝐿𝑄1 +
𝑁
4
( −𝑐𝑓𝑄1 )(𝑖)
𝑓𝑄1
Where:
𝐿𝑄1 = lower limit boundary of the Q1 class
𝑁 = total frequency
𝑐𝑓𝑄1 = cumulative frequency before Q1 class
𝑓𝑄1 = frequency of Q1 class
Determination of Q1 class
STEP 1: Determine Q1 class by dividing the total frequency by 4 and locating the quotient in the cf. In this
distribution, it is in the 135 – 143 class interval.
STEP 2: Substitute the values in the formula. The lower limit boundary is the mean of the lower limit of
134+135
the Q1 class and the upper limit of the class below it. In this case, the lower limit boundary is
=
2
134.5.
Q2 or the second quartile also the median
Q3 or the third quartile
𝑄3 = 𝐿𝑄3 +
(
3𝑁
−𝑐𝑓𝑄3 )(𝑖)
4
𝑓𝑄3
Where: 𝐿𝑄3 = lower limit boundary of the Q3 class
𝑁 = total frequency
𝑐𝑓𝑄3 = cumulative frequency before Q3 class
𝑓𝑄3 = frequency of Q3 class
Determination of Q3 class
3
STEP 1: Determine Q3 class by multiplying the total frequency by 4 and locating the quotient in the cf. In
this distribution, it is in the 153 – 161 class interval.
STEP 2: Substitute the values in the formula. The lower limit boundary is the mean of the lower limit of
152+153
the Q3 class and the upper limit of the class below it. In this case, the lower limit boundary is
=
2
152.5
𝑓
2
4
5
12
9
5
3
Class Interval
171 – 179
162 – 170
153 – 161
144 – 152
135 – 143
126 – 134
117 – 125
𝑐𝑓
40
38
34 Q3 Class
29
17 Q1 Class
8
3
40
( −8)(9)
Q1 = 134.5 + 4
= 134.5 + 2
Q1 = 136.5
(
9
3[40]
Q3 = 152.5 + 4
= 152.5 + 1.8
Q3 = 154.3
−29)(9)
5
Deciles - The distribution is divided into 10 equal parts
D1 or the first decile
𝐷1 = 𝐿𝐷1 +
𝑁
10
( −𝑐𝑓𝐷1 )(𝑖)
𝑓𝐷1
Where:
𝐿𝐷1 = lower limit boundary of the D1 class
𝑁 = total frequency
𝑐𝑓𝐷 = cumulative frequency before D1 class
𝑓𝐷1 = frequency of D1 class
Percentile (% ile) - the distribution is divided into 100 equal parts
Based on the formula for Quartile and Decile, how would the formula of percentile look like?
𝑃1 = 𝐿𝑃1 +
(
𝑁
−𝑐𝑓𝑃1 )(𝑖)
100
𝑓𝑃1