Mode

Mode = is that single measure or score which occurs most frequently. When data are grouped into a frequency distribution, the crude mode is usually taken to be the midpoint of that interval which contains the largest frequency.

When to use the mode:

1. When a quick and approximate measure of central tendency is all that is wanted.
2. When the measure of central tendency should be the most typical value.

Finding mode from the ungrouped data:
Example:

1. A set of numbers 11, 12, 13, 16, 16, 16, 19, 20 has 16 as the mode.
2. A set of numbers 45, 49, 52, 55, 58 has no mode.
3. A set of numbers 4, 4, 6, 8, 8, 8, 9, 9, 9, 10 has modes of 8 and 9 and is called bimodal.

Mode of grouped data
To determine the mode of grouped data we have to find first the modal class. In a frequency distribution, the modal class can be easily determine by inspection as it is the class with the highest frequency.

Mo = Lmo + [ d1/d1 + d2 ] c


Where: Lmo = lower boundery of the modal class
d1 = difference between the frequency of the modal class and the frequency of the class next lower in value.
d2 = difference between the frequency of the modal class and the frequency of the class next higher in value.
C = class size

Find the mode of example 3.8. table 3.1

Weekly wage ( in peso) f Lower class boundary
P 870-899 4 869.5
900-929 6 899.5
930-959 10 929.5
960-989 13 959.5
990-1019 8 989.5
1020-1049 7 1019.5
1050-1079 2 1049.5


Mo = 959.5 + [ 3/3+5 ] 30 = P 970.75

Median

b.) Median = in the midpoint of the distribution. Half of the values in a distribution fall below median and the other half fall above it.

When to use the median:
1. When the exact midpoint f the distribution is wanted the, 50% point.
2. When there are extreme scores which would markedly affect the mean. Extreme scores do not disturb the median.
3. When it is desired that certain scores should influence the central tendency, but all that is known about them is that they are above or below the median.
- determine of whether the cases fall within the lower halve or the upper halve of a distribution (appropriate locator of central tendency).

Finding the median from ungrouped data:
1. When N is odd, the median is the middle score.
Ex. 20 15 13 11 8 7 6
There are 7 scores and the median is 11

2. When N is even, the median is the average of the two middle score.
Ex. 21 18 15 14 11 8 8 7
There are 8 scores and the median is (14+11)/2 = 12.5

3. When several scores have the same value as the midscore.
Ex. 15 15 14 11 9 9 9 6 5
Median is 9
Ex. 1. Find the median of the following set of observations.
8 4 1 3 & 7
Sol. Array the set of observations and find the median
1 3 4 7 8
4 is the middle item

2. Compute for the median from the following set of data
12 9 6 10 7 & 14
Array the data and computer for the median
6 7 9 10 12 14
Median = (9 + 10 )/2 = 9.5

Finding the Median from the Grouped Data

Md = L + [N/2 – F2 / f2] C

Where: L = lower class boundary of the interval where the median lies
N = No. of scores or sum of frequency
F2 = cumulative frequency less than up to the class immediately preceding
the median class (F<)
f2 = frequency of the median class
C= class size
Steps:
1. Prepare 3 columns (Class intervals, class frequency and cumulative frequency less than)
2. Determine the Median class. The median class is that class interval where n/2 lies.
3. Substitute the data to the formula.

Ex. Find the median of the frequency distribution

Weekly wages,No. of Workers(f) ,F< (cumulative frequency less than)
(in peso)
870 – 899, 4, 4
900 – 929 ,6, 10
930 – 959 ,10, 20 – F<
960 – 989, 13, 33 Median class
990 – 1019 ,8 , 41
1020 – 1049, 7, 48
1050 – 1079, 2, 50
N=50
To determine the median class:

Solve for N/2 = 50/2 = 25th

25th items fall in 960 – 989 class interval therefore it is the median class

Md = L + [(N/2 – F2)/f2] C

= 959.5 + [(50/2 – 20) /13] 30

Md = Php 971.04

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

Graphical Method of Presenting Data and Frequency

1. Histogram. Class boundaries (x) vs. class frequency (y)
2. Frequency Polygon. Class Mark (x) vs. class frequency (y)
3. Less than ogive, upper class limit (x) vs. less than cumulative frequency (y)
4. Greater than ogive. Lower class limit (x) vs. greater than cumulative frequency (y)

Other Definition of Terms

Array – This is the arrangement of data from the highest to lowest or from lowest to highest.

Range, R - is the difference between the highest and the lowest number.

Number of class- it depends on the size and nature of or class interval distribution. The no. of classes is determined into which the range will be divided. Usually, an effective no. of classes is somewhere between 4 and 20.

No. of classes = range / class size or class width +1

Note:
a.) If series contains less than 50 cases, 10 cases or less are just enough
b.) If series contains 50 to 100 cases, 10 to 15 classes are recommended
c.) If more than 100 cases, 15 or more classes are good

Class Limit – the end number of a class. It is the highest and the lowest values that can go into each class.

Class Size – the width of each class interval

Class Boundaries – are the “true” class limits defined by lower and upper boundaries. The lower boundaries can be determined by getting the average of the upper limit of a class and the lower limit of the next class. They can also be obtained by simply adding of a unit (0.5) to the upper limit and subtracting the same to the lower limit of each class.

Class Mark, M – also known as class Midpoint. It is the average of the lower and upper limits or boundaries of each class.

Class Interval – The range of values used in defining a class. It is simply the length of a class. It is the difference or distance between the upper and lower class boundaries of each class and is affected by the nature of the data and by the number of classes. It is a good practice to set up uniform class interval whenever possible for easier computation and interpretation.