Measures of Central Tendency - Mean

Central Tendency is the point about which the scores tend to cluster, a sort of average in the series. It is the center of concentration of scores in any set of data. It is a single number which represents the general level of performance of a group.

Three (3) measures of Central Tendency

a.) Mean – The mean on arithmetic mean, or arithmetic average is defined as the sum of the values in the data group divided by the no. of values.

When to use the mean
1. When the scores are distributed symmetrically around a central point.
2. When the measure of central tendency having the greatest stability is wanted.

3. When other statistics like standard deviation, coefficient of correlation, etc. are to be computer later, since these statistics are based upon the mean.

Finding the Mean from Ungrouped Data

Where: x = score or measure
X = ∑X /N N = No. of scores or measures
∑ = summation of

Example:

1.) Last year the five sales counselors of Pacific Plans Inc. sold the following number f educators plans; 24,16,35,13,25. Find the mean.

Solution:
X = (24+16+35+13+25)/5= 22.6

Finding the mean from Grouped Data:

Long method:

X = ∑f M / N

Where: f = class frequency
M = class midpoint
N = sum of the frequencies

By the “ Assumed mean” or short method :

X=AM+(∑f X / N)c

Where: AM = assumed mean
c = class size
x = deviation

Example: Scores of 50 students on a college algebra test

Class internal, Midpoint, fx, Class frequency f, x, fM
scores
45-47 ,46 ,18 ,3 ,6, 138
42-44 ,43, 20, 4 ,5 ,172
39-41 ,40, 16, 4, 4, 160
36-38 ,37 ,12, 4, 3 ,148
33-35 ,34, 4, 2 ,2 ,68
30-32 ,31, 3 ,3 ,1 ,93
27-29 ,28 ,-0 ,13, 0,364
24-26 ,25 ,-8, 8 ,-1, 200
21-23 ,22, -6, 3, -2, 66
18-20, 19, -9, 3, -3,57
15-17, 16, 0, 0 ,-4 ,0
12-14 ,13, -10, 2, -5, 26
9-11 ,10, -6, 1, -6 ,10
34 50 1502
Long method :


X = ∑f M/ N = 1502/50=30.04


Assumed mean or short method : Steps :

1. Prepare 4 column ( class interval,f,x,fx ).
2. Select the interval to contain the assumed mean (AM). For the assumed mean, one may take the midpoint of the interval near the center of the distribution, or the midpoint of the interval with the highest frequency.
3. Determine the x column starting with 0, number each class interval positive up to the highest class interval; negative up to the lowest class interval.
4. Multiply f by x to determine the fx column.
5. Find the algebraic sum of the positive fx’s and the negative fx’s to get ∑fx.

Short method :

X=AM+(∑f X / N)c=28+(34/50)3=30.04

Weighted Arithmetic mean or Combined mean

Where: w = weight of x
∑wx = sum of the weight of
∑w = sum of the weight of x

Example: The same test was administered to fourth year high school students in 3 schools. Each school had computed its own mean using internal width of 3 as shown below.

School A

Class Interval
Scores f, x, fx

39-41, 1, 4 ,4
36-38,2 ,3 ,6
33-35, 4 ,2, 8
30-32, 4, 1 ,4
27-29 2 0 0
24-26, 3, -1, -3
21-23 ,4 ,-2, -8
18-20,2,-3,-6
22 5
School B
42-44, 1, 6, 6
39-41, 0, 5,0
36-38, 2, 4 ,8
33-35 ,5 ,3, 15
30-32 ,6, 2, 12
27-29, 7, 1, 7
24-26 3 0 0
21-23, 4, -1, -4
18-20, 2, -2, -4
15-17 ,2, -3 ,-6
12-14 ,1, -4, -4
9-11, 2, -5, -10
35 20
School C
39-41, 1, 3, 3
36-38, 2 ,2 ,4
33-35 ,10, 1, 10
30-32, 6, 0 ,0
27-29 ,7, -1 ,-7
24-26, 2, -2, -4
21-23, 1, -3, -3
18-20, 0 ,-4, 0
15-17 ,1, -5, -5
30 -2

To find the weighted mean of the 3 schools, follow the procedures below:

1. Find the highest and the lowest scores of the schools.
2. Prepare the step intervals column for the combined distribution.
3. Write the frequencies for each steps interval for the three schools.
4. Find the total frequency for each steps interval for the total combined distribution.
5. Compute the mean from this distribution.

Class interval
Scores , sch.A (f) ,sch.B (f), sch.c(f) ,total f, x, fx

42-44, 0, 1, 0, 1, 5, 5
39-41 ,1, 0 , 1, 2 ,4,8
36-38 2,2,2,6,3,18
33-35 ,4, 5 ,10 ,19, 2,38
30-32 ,4, 6 ,6, 16, 1 ,16
27-29, 2, 7 ,7, 16, 0 ,0
24-26 ,3, 3, 2, 8, -1, -8
21-23, 4, 4, 1, 9, -2 ,-18
18-20, 2, 2, 0 ,4 ,-3 ,-12
15-17, 0, 2 ,1, 3 ,-4, -12
12-14 ,0 ,1, 0 ,1 ,-5, -5
9-11 ,0, 2, 0, 2 ,-6, -2
22, 35, 30, ,18

X=AM+(∑f X / N)c=28+(18/87)3=28.6

WX = ∑wx / ∑w =22(28.69)+35(26.71)+30(30.8)/87=28.62