1. VICTOR MANALO CHRISTIAN L. MOTOL
MAED – EDUCATIONAL MANAGEMENT MAED – ENGLISH LANGUAGE
TEACHING
JENRICK ROSARIE JAYSEL D. ROSARIO
MAED – ENGLISH LANGUAGE TEACHING MAED - MATHEMATICS
Mean, Median, and Mode of
Ungrouped & Grouped Data
MEASURES OF CENTRAL TENDENCY:
MINDORO STATE COLLEGE OF AGRICULTURE AND TECHNOLOGY
∞ GRADUATE SCHOOL ∞
STAT 200: STATISTICS AND EVALUATION OF EDUCATION
2. CHRISTIAN L. MOTOL
MAED – ENGLISH LANGUAGE TEACHING
Mean of
Ungrouped
Data
MEASURES OF
CENTRAL
TENDENCY:
MINDORO STATE COLLEGE OF
AGRICULTURE AND
TECHNOLOGY
∞ GRADUATE SCHOOL ∞
STAT 200: STATISTICS AND
EVALUATION OF EDUCATION
4. What is MEAN?
• The mean of data indicates how the
data are distributed around the
central part of the distribution
• The mean, often called the average, of a
numerical set of data, is simply the sum
of the data values divided by the
number of values. This is also
referred to as the arithmetic mean.
5. Mean of Raw Data
The mean (or arithmetic mean) of n
observations (variates) x1, x2, x3, x4,
....., xn is given by…
9. A student scored 80%, 72%, 50%, 64% and 74%
marks in five subjects in an examination. Find the mean
percentage of marks obtained by him.
Solution:
Here, observations in percentage are: x1= 80, x2 = 72, x3 = 50, x4 = 64, x5, = 74.
Therefore, their mean A = x1+x2+x3+x4+x5
5
= 80+72+50+64+74
5
= 340
5
= 68
Therefore, mean percentage of marks
obtained by the student was 68%.
10. Chris Jacobs scores the following runs in six innings of a series.
45, 2, 78, 20, 116, 55.
Find the mean of the runs scored by the batsman in the series.
Solution:
Here, the observations are x1 = 45, x2 = 2, x3 = 78, x4 = 20, x5 = 116, x6 = 55.
Therefore, their mean A = x1+x2+x3+x4+x5 +x6
6
= 45+2+78+20+116+55
6
= 316
6
= 52.7
Therefore, mean of the runs scored by
the batsman in the series was 52.7