This presentation is all about finding the percentile, decile and quartile of a grouped data. an example is provided in each type of measure of positions.
This presentation is all about finding the percentile, decile and quartile of a grouped data. an example is provided in each type of measure of positions.
The steps in computing the median are similar to that of Q1 and Q3
. In finding the median,
we need first to determine the median class. The Q1 class is the class interval where
the π
4
th score is contained, while the class interval that contains the 3π
4
π‘β
score is the Q3 class.
Formula :ππ = LB +
ππ
4
βπππ
πππ
π
LB = lower boundary of the of the ππ class
N = total frequency
πππ= cumulative frequency of the class before the ππ class
πππ
= frequency of the ππ class
i = size of the class interval
k = the value of quartile being asked
The interquartile range describes the middle 50% of values when
ordered from lowest to highest. To find the interquartile range (IQR),
first find the median (middle value) of the upper and the lower half of
the data. These values are Q1 and Q3
. The IQR is the difference
between Q3 and Q1
.
Interquartile Range (IQR) = Q3 β Q1
The quartile deviation or semi-interquartile range is one-half the
difference between the third and the first quartile.
Quartile Deviation (QD) =
π3βπ1
2
The formula in finding the kth decile of a distribution is
π·π = ππππ +
(
π
10)π β ππ
ππ·π
π
πΏπ΅ππ β πΏππ€ππ π΅ππ’πππππ¦ ππ π‘βπ ππ‘β ππππππ
π β π‘ππ‘ππ ππ’ππππ ππ πππππ’ππππππ
ππ β ππ’πππ’πππ‘ππ£π πππππ’ππππ¦ ππππππ π‘βπ ππ‘β ππππππ
πΉππ β πππππ’ππππ¦ ππ π‘βπ ππ‘β ππππππ
π β ππππ π π ππ§π
The steps in computing the median are similar to that of Q1 and Q3
. In finding the median,
we need first to determine the median class. The Q1 class is the class interval where
the π
4
th score is contained, while the class interval that contains the 3π
4
π‘β
score is the Q3 class.
Formula :ππ = LB +
ππ
4
βπππ
πππ
π
LB = lower boundary of the of the ππ class
N = total frequency
πππ= cumulative frequency of the class before the ππ class
πππ
= frequency of the ππ class
i = size of the class interval
k = the value of quartile being asked
The interquartile range describes the middle 50% of values when
ordered from lowest to highest. To find the interquartile range (IQR),
first find the median (middle value) of the upper and the lower half of
the data. These values are Q1 and Q3
. The IQR is the difference
between Q3 and Q1
.
Interquartile Range (IQR) = Q3 β Q1
The quartile deviation or semi-interquartile range is one-half the
difference between the third and the first quartile.
Quartile Deviation (QD) =
π3βπ1
2
The formula in finding the kth decile of a distribution is
π·π = ππππ +
(
π
10)π β ππ
ππ·π
π
πΏπ΅ππ β πΏππ€ππ π΅ππ’πππππ¦ ππ π‘βπ ππ‘β ππππππ
π β π‘ππ‘ππ ππ’ππππ ππ πππππ’ππππππ
ππ β ππ’πππ’πππ‘ππ£π πππππ’ππππ¦ ππππππ π‘βπ ππ‘β ππππππ
πΉππ β πππππ’ππππ¦ ππ π‘βπ ππ‘β ππππππ
π β ππππ π π ππ§π
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2. MEASURES OF CENTRAL
TENDENCIES
provides a very convenient way of describing a set
of scores with a single score with a single number
that describes the performance of the group.
15,586
Average COVID-19 cases daily for the
month of August
4. Mean (x
Μ )
Mean is the most commonly used measure of
the center data.
Also known as arithmetic average.
5. 1. Mean (ungrouped data)
Population mean
ΞΌ = Ξ£x
N
Sample mean
x
Μ = Ξ£x
N
Weighted mean
x
Μ = Ξ£ wi xi
wi
6. Example 1
The score of 15 students in a Quiz in Statistics consist of 25
items. The highest score is 25 and the lowest score is 10.
Here are the scores 25, 20, 18, 18, 17, 15, 15, 15, 14, 14,
13, 12, 12, 10, 10. Find the mean of the scores.
7. Example 1
Ξ£x = 25+20+18+18+17+15+15+15+14+14+13+12+12+10+10
Step 1: Get the summation of scores (Ξ£x)
Ξ£x = 228
Step 2: Determine the number of scores (N)
N = 15
Step 3: Divide the Ξ£x by N
x
Μ = Ξ£x
N
x
Μ = 228
15
x
Μ = 15.20
8. Example 1
Step 4. Analysis
The average performance of 15 students who participated in the quiz
consisting of 25 items is 15.2. The implication of this is that student who
got scores below 15.2 did not perform well in the said examination.
Students who got scores higher than 15.2 performed well in the
examination compared to the performance of the whole class.
9. Example 2
Find the General Weighted Average (GWA) of Francisco for the first
semester of the school year 2020-2021. Use the table below:
Subjects Grade (xi) Units (wi) (xi) (wi)
DSLM 103 1.25 3
DSLM 303 1.00 3
DSLM 100 1.25 3
Dissertation 1 1.50 6
Ξ£wi = Ξ£(xi)(wi) =
10. Example 2
Subjects Grade (xi) Units (wi) (xi) (wi)
DSLM 103 1.25 3
DSLM 303 1.00 3
DSLM 100 1.25 3
Dissertation 1 1.50 6
Ξ£wi = Ξ£(xi)(wi) =
Step 1: Multiply each value to its weight or unit.
11. Example 2
Subjects Grade (xi) Units (wi) (xi) (wi)
DSLM 103 1.25 3 3.75
DSLM 303 1.00 3 3.00
DSLM 100 1.25 3 3.75
Dissertation 1 1.50 6 9.00
Ξ£wi = Ξ£(xi)(wi) =
Step 2: Get the sum of units (Ξ£wi).
12. Example 2
Subjects Grade (xi) Units (wi) (xi) (wi)
DSLM 103 1.25 3 3.75
DSLM 303 1.00 3 3.00
DSLM 100 1.25 3 3.75
Dissertation 1 1.50 6 9.00
Ξ£wi = 15 Ξ£(xi)(wi) =
Step 3: Get the sum of the product of grade and units Ξ£(xi)(wi)
13. Example 2
Step 4: Use the formula x
Μ = Ξ£wixi
Ξ£wi
x
Μ = Ξ£wixi
Ξ£wi
x
Μ = 19.5
15
x
Μ = 1.30
14. 2. Mean (grouped data)
Grouped data are data or scores that are arranged in a
frequency distribution.
15. Frequency distribution is the arrangement of scores
according to category of classes including frequency.
17. Also known as class mark method. Obtained by
using the midpoint of each class or category
Midpoint method
Mean (grouped)
x
Μ = Ξ£fXm
n
f =
Xm =
n =
18. Example 3
The scores of 40 students in a Science quiz consisted of 60 items are
tabulated below.
X f Xm fXm
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n = 40
19. Example 3
X f Xm fXm
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n = 40
Step 1: Find the midpoint or class mark (Xm) of each class or category
using the formula Xm = LL+UL
2
20. Example 3
X f Xm fXm
10-14 5 12
15-19 2 17
20-24 3 22
25-29 5 27
30-34 2 32
35-39 9 37
40-44 6 42
45-49 3 47
50-54 5 52
n = 40 Ξ£fXm =
Step 2: Multiply the frequency and the corresponding class mark fXm
21. Example 3
X f Xm fXm
10-14 5 12 60
15-19 2 17 34
20-24 3 22 66
25-29 5 27 135
30-34 2 32 64
35-39 9 37 333
40-44 6 42 252
45-49 3 47 141
50-54 5 52 260
n = 40 Ξ£fXm =1345
Step 3: Find the sum of Step 2 (Ξ£fXm)
22. Example 3
Step 4: Solve the mean using the formula x
Μ = Ξ£fXm
n
x
Μ = 1345
40
x
Μ = 33.63
23. Example 3
Step 5: Analysis
The mean performance of 40 students in Science quiz is 33.63.
Those students who got scores below 33.63 did not perform well
in the said quiz while those who got scores above 33.63
performed well.
25. Properties of Mean
1. It measures stability. Mean is the most stable among other
measures of central tendency because every score contributes to
value of the mean.
2. It is easily affected by extreme scores.
3. It may not be an actual score in the distribution.
4. It can be applied to level of measurement.
5. It is very easy to compute.
26. Median (xΜ)
Mean is what divides the scores in the
distribution into two equal parts.
Also known as the middle score or 50th
percentile.
27. 1. Median (ungrouped data)
Find the median score of 7 students in English Class.
Example 1
x (score)
17
19
5
16
10
2
15
29. Example 1
x (score)
19
17
16
15
10
5
2
Step 2: Determine the middle most score in a distribution if n is an odd
number. But if the n is even, get the average of the two middle most
score,
We can use this formula too:
xΜ = n+1
2
xΜ = 7+1
2
xΜ = 4 (4th score)
30. Example 1
Step 3: Analysis
The median score is 15. Fifty percent (50%) or three of the scores
are above 15 (19,17,16) and 50% or three scores are below
(10,5,2).
31. 1. Median (ungrouped data)
Find the median score of 8 students in an English Class.
Example 2
x (score)
19
30
17
5
15
10
2
16
33. Example 2
x (score)
30
19
17
16
15
10
5
2
We can use this formula too:
xΜ = n+1
2
xΜ = 8+1
2
xΜ = 4.5 (4th and 5th score)
Step 2: Determine the middle most score in a distribution if n is an odd
number. But if the n is even, get the average of the two middle most
score,
35. Example 2
Step 4: Analysis
The median score is 15.5. Fifty percent (50%) or three of the
scores are above 15.5 those are 15, 10, and 5; and 50% are
greater than 15.5, those are 30,19,17,16 which mean four scores
are below 15.5 and four scores are bove 15.5.
37. Scores of 40 students in a Science class consist of 60 items and they are
tabulated below. The highest score is 54 and the lowest score is 10.
Example 3
X f cf
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
38. Formula:
Example 3
xΜ = LB + n/2 β cfp c.i.
fm
xΜ = median value
MC = median class is the class with the smallest cumulative frequency greater
than or equal to Β½ of the total frequencies.
- a category containing the n/2 (locator)
LB = lower boundary of the median class (MC)
Cfp = cumulative frequency before the median class if the scores are
arranged from lowest to highest value.
fm = Frequency of median class
c.i. size of the class interval
39. Example 3
X f cf
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
Step 1: Complete the table for cumulative frequency by adding each frequency from a
frequency distribution table to the sum of its predecessors
40. Example 3
X f cf
10-14 5 5
15-19 2 7
20-24 3 10
25-29 5 15
30-34 2 17
35-39 9 26
40-44 6 32
45-49 3 35
50-54 5 40
n=40
Step 1: Complete the table for cumulative frequency by adding each frequency from a
frequency distribution table to the sum of its predecessors
41. Example 3
Step 2: Get n/2 of the scores in the distribution so that you can identify
MC.
n = 40 = 20
2 2
Locator of median class MC.
Choose the first cumulative
frequency greater than n/2.
42. Example 3
X f cf
10-14 5 5
15-19 2 7
20-24 3 10
25-29 5 15
30-34 2 17
35-39 9 26
40-44 6 32
45-49 3 35
50-54 5 40
n=40
MC
43. Example 3
X f cf
10-14 5 5
15-19 2 7
20-24 3 10
25-29 5 15
30-34 2 17
35-39 9 26
40-44 6 32
45-49 3 35
50-54 5 40
n=40
Step 3: Determine the LB, cfp, fm and c.i.
MC fm
cfp
LB = 34.5
c.i. = 5
44. Example 3
Step 4: Solve the median class using the formula
xΜ = LB + n/2 β cfp c.i.
fm
45. Example 3
Step 4: Solve the median class using the formula
xΜ = LB + n/2 β cfp c.i.
fm
xΜ = 34.5 + 40/2 β 17 5
9
xΜ = 34.5 + 20 β 17 5
9
xΜ = 34.5 + 3 5
9
xΜ = 34.5 + 1.67
xΜ = 36.17
46. Example 2
Step 5: Analysis
The median value is 36.17, which means that 50% or 20 scores
are less than 36.17.
47. Properties of Median
1. It may not be an actual observation in the data set.
2. It can be applied I ordinal level.
3. It is not affected by extreme values because median is a
positional measure.
48. Mode (xΜ)
Mode or modal score is a score/s that
occurred most in the distribution.
49. Classification of Mode (xΜ)
Unimodal Bimodal Multimodal
1 mode in the
distribution
2 modes in the
distribution
3 or more
modes in the
distribution
25
20
20
20
16
12
7
5
25
20
20
18
16
16
7
5
25
25
20
20
16
16
7
5
51. Example 1
Scores of 40 students in a science class consist of 60 items and they are
tabulated below.
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
52. Example 1
Scores of 40 students in a science class consist of 60 items and they are
tabulated below.
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
53. Example 1
Formula
XΜ = LB + d1 c.i.
d1 + d2
LB = lower boundary of the modal class
MC = modal class
d1 = difference between the frequency of the modal class and the
frequency above it, when scores are arranged from lowest to
highest.
d2 = difference between the frequency of the modal class and the
frequency below it, when scores are arranged from lowest to
highest.
c.i. = size of class interval
54. Example 1
Step 1: Determine the modal class by looking at the class with highest
frequency.
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
55. Example 1
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
d1 = 9-2 = 7
Step 2: Determine the LB, d1, d2 and c.i.
56. Example 1
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
d2 = 9-6 = 3
Step 2: Determine the LB, d1, d2 and c.i.
57. Example 1
Step 2: Determine the LB, d1, d2 and c.i.
X f
10-14 5
15-19 2
20-24 3
25-29 5
30-34 2
35-39 9
40-44 6
45-49 3
50-54 5
n=40
LB = 34.5 c.i. = 5
58. Example 1
Step 3: Solve the mode using the formula
Formula
XΜ = LB + d1 c.i.
d1 + d2
= 34.4+ 7 5
7 + 3
= 34.4 + 3.5
= 38
59. Example 1
Step 4: Analysis
The mode of the score distribution that consists of 40 students is 38,
because 38 occurred several times.
60. Properties of Mode
1. It can be used when the data are qualitative as well as
quantitative.
2. It may not be unique.
3. It is not affected by extreme values.
4. It may not exist.
