Assessment of Learning 2 (Overview)

What is Frequency?
• Frequency is how often
something occurs.

Frequency Distribution
- is a tabular arrangement of data into
appropriate categories showing the number
of observations in each category or group.

NASCAR Nextel Cup Drivers’ Ages

Parts of a frequency
distribution :
• Class limit
• Class size
• Class boundaries
• Class marks

Class limit
- Is the groupings defined by the
lower and upper limits.
Example: LL – UL
10 – 14
15 – 19
20 – 24

Lower class limit (LL)
- smallest number in each group
Upper class limit (UL)
- highest number in each group

Class size
- Is the width of each class interval.
Example: LL – UL
10 – 14
15 – 19
20 – 24
What is the class size of the score
distribution above?
Answer: 5 

Class boundaries
- Are the numbers used to separate
each category in the distribution .
Example: LL – UL 𝐿 𝑐𝑏 − 𝑈𝑐𝑏
10 – 14 9.5 – 14.5
15 – 19 14.5 – 19.5
20 – 24 19.5 – 24.5

Class marks
- Are the midpoint of the lower and
upper class limits.
- Formula is 𝑋 𝑚=
𝐿𝐿 −𝑈𝐿
2
Example: LL – UL 𝑋 𝑚
10 – 14 -
15 – 19 -
20 – 24 -

Class marks
- Are the midpoint of the lower and
upper class limits.
- Formula is 𝑋 𝑚=
𝐿𝐿 −𝑈𝐿
2
Example: LL – UL 𝑋 𝑚
10 – 14 12
15 – 19 17
20 – 24 22

Steps in constructing
Frequency Distribution

Raw scores of 40 students in a 50-
item mathematics quiz.
17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32

Compute
the Range;
R=HS-LS
Determine
the class
size.
Set up the
class
limits.
Make a
tally.
Count the
frequency
(f).

Computing the Range
The range is the difference between the
highest score and the lowest score.
Range = HS - LS

17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Lowest score

17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32
Lowest score
Highest score

Computing the Range
The range is the difference between
the highest score and the lowest score.
Range = HS – LS
R= 50 – 15
R= -

Computing the Range
The range is the difference between
the highest score and the lowest score.
Range = HS – LS
R= 50 – 15
R= 35

Solve the value of k:
k = 1 + 3.3 log n
Determine the class size (c.i).
formula: 𝑐. 𝑖 = 𝑅
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓𝑐𝑙𝑎𝑠𝑠𝑒𝑠
or 𝑐. 𝑖 = 𝑅
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6
𝑘
Find the class size:
c.i = 𝑅
𝑘

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6
𝑘
c.i = 𝑅
𝑘
c.i = 35
6

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6
𝑘
c.i = 𝑅
𝑘
c.i = 35
6
c.i = 5.833

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6
𝑘
c.i = 𝑅
𝑘
c.i = 35
6
c.i = 5.833
c.i = 6
From Assessment of Learning 1, Yonardo Gabuyo

k = 𝑛
𝑘

k = 𝑛
k = 40
𝑘

k = 𝑛
k = 40
k = 6.32455532
𝑘

k = 𝑛
k = 40
k = 6.32455532
k = 6
𝑘
c.i = 𝑅
𝑘
c.i = 35
6
c.i = 5.833
c.i = 6

Set up the class limits.
X
15 – 20
21 – 26
27 – 32
33 – 38
39 – 44
45 - 50

Make a tally
X Tally
15 – 20 / / / / /
21 – 26 / / / / / / / / /
27 – 32 / / / /
33 – 38 / / / / / / / / / /
39 – 44 / / / /
45 - 50 / / / / / / / /

Indicate the frequency.
X Tally Frequency (f)
15 – 20 / / / / / 5
21 – 26 / / / / / / / / / 9
27 – 32 / / / / 4
33 – 38 / / / / / / / / / / 10
39 – 44 / / / / 4
45 - 50 / / / / / / / / 8
n = 40

Frequency distribution
X Tally f 𝑿 𝒎 cf
15 – 20 / / / / / 5 17.5 5
21 – 26 / / / / / / / / / 9 23.5 14
27 – 32 / / / / 4 29.5 18
33 – 38 / / / / / / / / / / 10 35.5 28
39 – 44 / / / / 4 41.5 32
45 - 50 / / / / / / / / 8 47.5 40
n = 40

Graphical Representation of
Scores in Frequency
Distribution

There are different methods of
graphing frequency distribution:
• Bar graph
• Histogram
• Frequency polygon
• Pie graph

Bar graph
When data is in categories (countries,
movies, music, etc.), this type of graph is
usually used.
0%
10%
20%
30%
40%
50%
60%
Rodents Fish Cats Dogs Rabbits
percentage
pets
Pet Preference
Pet Preference

0
2
4
6
8
10
12
15 - 20 21 - 26 27 - 32 33 - 38 39 - 44 45 -50
frequency
Classes
Scores
Scores

Histogram
Groups numbers into ranges and is a great
way to show results of continuous data.

0
2
4
6
8
5 15 25 35 45 55 65
Frequency
Histogram: Height of Red Trees
Class Midpoints
Histogram Example
Class
10 but less than 20 15 3
FrequencyClass Midpoint

Frequency distribution
X Tally f Class Boundaries
f cf
15 – 20 / / / / /
5 14.5 – 20.5 5 5
21 – 26 / / / / / / / / /
9 20.5 – 26.5 9 14
27 – 32 / / / /
4 26.5 – 32.5 4 18
33 – 38 / / / / / / / / / /
10 32.5 – 38.5 10 28
39 – 44 / / / /
4 38.5 – 44.5 4 32
45 - 50 / / / / / / / /
8 44.5 – 50.5 8 40
n = 40

14.5 - 20.5
20.5 - 26.5
26.5 - 32.5
32.5 - 38.5
38.5 - 44.5
44.5 - 50.5
0
2
4
6
8
10
12
Class boundaries
Frequency

Frequency polygon
- Is constructed by plotting the class
marks against the class frequencies.
0
1
2
3
4
5
6
7
8
9
10
5 10 15 20 25
Frequency
Midpoints
Scores

0
2
4
6
8
10
12
17.5 23.5 29.5 35.5 41.5 47.5
Frequency
Midpoints
Scores

Pie graph
This displays data in an easy-to-read ‘pie-
slice’ format with varying slice sizes.
48%
19%
9%
9%
10% 5%
Monthly Budget
Rent Food Utilities Fun Clothes Phone

12%
23%
10%
25%
10%
20%
Scores
15 - 20
21 - 26
27 - 32
33 - 38
39 - 44
45 - 50
A PIE graph

Equation :
Where:
Σ𝑥 − the sum of the individual
values.
N- total number of values.

1. All values are used.
2. Most sensitive measures of central tendency
to use for ratio data.
3. Influenced by extreme scores.
4. The sum of the deviations from
the mean is 0.

Equation:
𝜇 =
Σ𝑥
𝑁
𝜇 =
Σ𝑥
𝑁
𝜇- the population mean.
N- the number of values in the population.
x- any particular value.
Σ𝑥- the sum of the x values in the population.

Faculty members of (5) colleges:
16
25
20
40
32

𝜇 =
Σ𝑥
𝑁
𝜇 =
16+25+20+40+32
5
𝜇 =
133
5
𝜇 =26.6
𝜇 = 27

Equation:
Where:
Σx- the sum of all data values
n- the number of data items in sample

The following are the ages of samples of 8
children in a city:
9 8 1 3 4 5 6 7

x =
Σ𝑥 𝑖
𝑛
x =
9+8 +1+ 3+ 4+ 5 +6 +7
8
x=
43
8
x= 5.4
x= 5
__
__
__
__
__

Equation::
Where,:
Σx- sum of the individual value.
n- total number of values.
n
x
x

:

Raw scores of 7 students in a 60 item science
examination.
Scores: 56, 45, 40, 34, 34, 32, 25

x =
Σ𝑥
𝑛
x =
56+45+40+34+34+32+25
7
x=
266
7
x = 38
__
__
__
__

Equation:
x =
Σ𝑓(𝑥 𝑚)
𝑛
where:
m- the midpoint of the interval,
f- the frequency for the interval,
Σ𝑓(𝑥 𝑚) – sum of the product of the midpoint
and the frequency,
n- the number of values.

17 25 30 33 25 45 23 19
27 35 45 48 20 38 39 18
44 22 46 26 36 29 15 21
50 47 34 26 37 25 33 49
22 33 44 38 46 41 37 32

Class Interval f xm fxm
15 –20 5 17.5 87.5
21 –26 9 23.5 211.5
27 –32 4 29.5 118
33 –38 10 35.5 355
39 –44 4 41.5 166
45 - 50 8 47.5 380
n=40 f(xm)=1318

Class Interval f 𝒙 𝒎 f𝒙 𝒎
12 –17 2 14.5 29
18 –23 7 20.5 143.5
24 –29 7 26.5 185.5
30–35 7 32.5 227.5
36–41 7 38.5 269.5
42-47 7 44.5 311.5
48-53 3 50.5 1151.5

x =
Σ𝑓(𝑥 𝑚)
𝑛
x=
1366
40
x = 34.15
__
__
__

 WEIGHTED MEAN
 COMBINE MEAN

 is a measurement of central
tendency. It represents the
average of a given data.
 is similar to arithmetic mean or
sample mean.

1. Multiply the
numbers in
your data set
by the weights.
2. Add the numbers
in Step 1 up. Set this
number aside for a
moment.
3. Add up all
the weights
4.Divide the
numbers you found
in Step 2 by the
number you found in
Step 3

The marks obtain by 20 students in a
Mathematics test are 12,15,18, 10, 20,
and the corresponding frequencies are
3,6,5,4,2.
Compute the average marks obtained
by the students.

Solution :
xi fi xi fi
12
15
18
10
20

xi fi xi fi
12 3
15 6
18 5
10 4
20 2
∑fi = 20

xi fi fixi
12 3 36
15 6 90
18 5 90
10 4 40
20 2 40
∑fi = 20 ∑ fixi = 296

Substitute the given values
Answer : 14.8
×=
𝑓𝑖 ×𝑖
𝑓𝑖

 The number of students absent in a on few
subsequent days are as follows:
1,3,4,4,1,7,5,2,4,3,7,3,4,5,3,5,7,7.

 Find the average of absent students using the
weighted mean formula.

x i f i
1 2
2 1
3 4
4 4
5 3
7 4

xi fi fixi
1 2 2
2 1 2
3 4 12
4 4 16
5 3 15
7 4 28
∑fi = 18 ∑fixi = 75

Weighted Average with Percentage
 A student is enrolled in a
biology course where the final
grade is determined based on the
following categories: tests 40%,
final exam 25%, quizzes 25%, and
homework 10%.

 The student has earned the following scores
for each category: tests-83, final exam-75,
quizzes-90, homework-100.
 We need to calculate the student's overall
grade.

 Three sections of a Statistics class containing 35,
40, and 45 students averaged 80, 85 and 69
respectively on the same final examination.
 What is the combined mean for all the three
sections .

Number of
Students (n)
Average
Grade (×)
n×
35 80 2,800
40 85 3,400
45 89 4,005
= 120 = 10,205

× =
𝑋 𝑁
2
+
𝑋 𝑁+2
2
2 ,if N is even
× = 𝑋1 + N
2 , if N is odd
_________________
_________________

40 40
34 34
34 34
45 45
56 56
32 32
25 25
60
25= 𝑋1 25= 𝑋1
32= 𝑋2 32= 𝑋2
34= 𝑋3 34= 𝑋3
34= 𝑋4 45= 𝑋4
40= 𝑋5 56= 𝑋5
45= 𝑋6 32= 𝑋6
56= 𝑋7 25= 𝑋7
60= 𝑋8

×= 𝑿 𝟏 + 𝐍
2
= 𝑋1 + 7
2
= 𝑋8
2
=𝑋4
×=12

×=
𝑿 𝑵
𝟐
+
𝑿 𝑵+𝟐
𝟐
2
=
𝑋8
2
+
𝑋 𝑁+2
2
2
= 𝑋4+
𝑋8+2
2
2
=𝑋4+
𝑋10
2
2
= 𝑋4+ 5
2
=
𝑋9
2
= 𝑋4.5
= 34.5×

15-20 4 4
21-26 9 13
27-32 3 16
33-38 10 26
39-44 4 30
45-50 10 40
N=40

L +
𝑵
𝟐
- F
fm
(i)=
Where,
L = exact lower limit of the interval
containing the median class
F = the sum of all frequencies below L.
fm= frequency of interval containing
the median class.
N= total number of cases
i = class interval
×

×= 32.5 +
40
2
- 16
10
= 32.5 + 20 – 16
10
= 32.5 + 4
10
= 32.5 + 24
10
= 32.5 + 2.4
× = 34.9
L = 32.5
F = 16
Fm = 10
N = 40
I = 6
_______
_______
__
(6)
____
(6)
(6)

 The value that appears most in a
given data.
 French expression “a la mode”
meaning fashionable.

 Unimodal
 Bimodal
 Trimodal or Multimodal

Scores of Section
A
Scores of Section
B
Scores of Section
C
25 25 25
24 24 25
24 24 25
20 20 22
20 18 21
20 18 21
16 17 21
12 10 18
10 9 18
7 6 18

 Most typical value of a
distribution.
 Can be used to describe
qualitative distribution.

 Value of mode cannot always be
determined.
 Value of mode is not based on each
and every item of the series.
 It does not always exist.

The number of points scored in a
series of football game listed
below.
7, 13, 18, 24, 9, 3, 18

Order the scores from least to
greatest.
3, 7, 9, 13, 18, 18, 24
Answer:

In a crash test, 11 cases were tested
to determine the impact speed
required to obtain minimal bumper
damage.
24, 15, 18, 20, 18, 22, 24, 26, 18, 26,
24

A marathon race was completed by 5
marathon participants.
2.7, 8.3, 3.5, 5.1, 4.9

On a cold winter day in January, the
temperature for 9 North American cities is
recorded in Fahrenheit.
-8, 0, -3, 4, 12, 0, 5, -1, 0

The following is the number of problems
that Ms. Matty assigned for homework on
10 different days.
8, 11, 9, 14, 11, 9, 15, 9, 18, 11

The number points scored in a series of
basketball games is listed below.
8, 19, 14, 19, 14, 24, 8

Raw scores of 40 students in a 50-item
mathematics quiz.

Computing the Range
The range is the difference between the highest score and
the lowest score.
Range = HS – LS
R= 50 – 15
R= 35

k = 1 + 3.3 log n
k = 1 + 3.3 log 40
k = 1 + 3.3(1.60205991)
k = 1 + 5.286797971
k = 6.286797971
k = 6

Is the groupings defined by
the lower and upper limits.
Example: LL – UL
15 – 20
21 – 26
27 – 32

1. Determine the modal class.
2. Get the value of 𝒅 𝟏.
3. Get the value of 𝒅 𝟐.
4. Get the lower boundary of the
modal class.
5. Apply the formula.

𝒙 = 𝑳 𝑩 +
𝒅 𝟏
𝒅 𝟏+𝒅 𝟐
c.i.

• 𝒙 - Mode
• 𝑳 𝑩 - lower boundary of modal
class
• Modal Class (MC) - category
containing highest frequency

• 𝒅 𝟏 - difference between frequency of
modal class and frequency above it
• 𝒅 𝟐 - difference between frequency of
modal class and frequency below it
• c.i. - size of class interval

.
MC: 33-38
𝒅 𝟏: 10-4 = 6
𝒅 𝟐: 10-4 = 6
𝑳 𝑩: 33-0.5 = 32.5
c.i.: 6
Formula:
𝒙 = 𝑳 𝑩 +
𝒅 𝟏
𝒅 𝟏+𝒅 𝟐
c.i.

𝒙 = 32.5 +
6
6+6
6
= 32.5 +
6
12
6
= 32.5 + 0.5 6
= 32.5 + 3
𝒙 = 35.5 or 36

𝑴 𝒐 = 𝑳 𝑩 +
𝒅 𝟏
𝒅 𝟏+𝒅 𝟐
c.i.

• Mo - Mode
• LB - lower boundary of modal
class
• Modal Class (MC) - category
containing highest frequency

• 𝑑1 - difference between frequency
of modal class and frequency above
it
• 𝑑2 - difference between frequency
of modal class and frequency below
it
• c.i. - size of class interval

• Measures of 40 mango leaves in cm.
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 3
n= 40

MEASURES OF POSITION
• QUARTILE
• DECILE
• PERCENTILE

These are
values that
divide the set
of data into
four equal
parts.
QUARTILE PERCENTAGE
𝑄1 25%
𝑄2 50%
𝑄3 75%

Qk =is the indicated
quartile
q = 1, 2, 3
n = number of cases

25, 32, 34, 34, 40, 45, 56
(arrange the scores from highest to lowest)
1st 3rd 4th2nd 5th 6th 7th

𝑄2 =
2
4
7 + 1 −
2
4
nth score
=
14
4
+
1
2
= 3.5 + 0.5
= 4th score
𝑄2 = 34

25, 32, 34, 34, 40, 45, 56
𝑄2/ 4th score

Qk = the indicated quartile
k = 1, 2, and 3
Lb = lower boundary of the
quartile class
cfp= cumulative frequency
fq = frequency of the
indicated decile class
ci = size of the class interval

SCORES (X) TALLY FREQUENCY CF <
12- 17 II 2 2
18- 23 IIII- II 7 9
24- 29 IIII- II 7 16
30- 35 IIII- II 7 23
36- 41 IIII -II 7 30
42- 47 IIII - II 7 37
48- 53 III 3 40
n = 40

𝑄2 = 29.5 +
20 −16
7
6 Given: LB = 29.5
= 29.5 +
4
7
6
2(𝑛)
4
= 20
= 29.5 + 3.4 cf = 16
𝑄2 = 32.9 f = 7
ci = 6

12- 17 II 2 2
18- 23 IIII- II 7 9
24- 29 IIII- II 7 16
30- 35 IIII- II 7 23
36- 41 IIII -II 7 30
42- 47 IIII - II 7 37
48- 53 III 3 40
n = 40
𝑄2

These are values that
divide a set of
observations into 10
equal parts. Usually
denoted by
D1,D2,D3,… D9.
DECILE PERCENTAGE
𝐷1 10%
𝐷2 20%
𝐷3 30%
𝐷4 40%
𝐷5 50%
𝐷6 60%
𝐷7 70%
𝐷8 80%
𝐷9 90%

𝐷 𝐾= is the indicated
decile
k = 1, 2, 3, 4, 5, 6, 7, 8,
9
n = number of cases

𝐷5=
5
10
7 + 1 −
5
10
=
35
10
+
1
2
= 3.5 + 0.5
= 4th score
𝐷5 = 34

25, 32, 34, 34, 40, 45, 56
𝐷5/ 4th score

𝐷 𝑘= indicated decile
k = 1, 2,3,…9
𝐿 𝐵 =lower boundary of the
fd = frequency of the

𝐷5 = 𝐿 𝐵𝐷5
+
5(𝑛)
10
−𝑐𝑓𝐷5
𝑓𝐷5
ci Given:
= 29.5 +
20−16
7
6 LB = 29.5
= 29.5 +
4
7
6
5(𝑛)
10
= 20
= 29.5 + 3.4 cf = 16
𝐷5 = 32.9 f = 7
ci = 6

12- 17 II 2 2
18- 23 IIII- II 7 9
24- 29 IIII- II 7 16
30- 35 IIII- II 7 23
36- 41 IIII -II 7 30
42- 47 IIII - II 7 37
48- 53 III 3 40
n = 40
𝐷5

These are values
that divide the set
of data into 100
equal parts.
Usually denoted
by 𝑃1,𝑃2,𝑃3,…𝑃99.
PERCENTILE PERCENTAGE
𝑃1 1%
𝑃2 2%
𝑃3 3%
𝑃99 99%

𝑃 𝐾 =is the indicated
percentile
k = 1, 2, 3,…, 99
n = number of cases

𝑃50=
50
100
7 + 1 −
50
100
=
350
100
+
1
2
= 3.5 + 0.5
= 4th score
𝑃50= 34

𝑃𝑘= indicated percentile
K = 1, 2,3,…99
𝐿 𝐵 =lower boundary of the
fd = frequency of the indicated
decile class

𝑃50=𝐿 𝐵 𝑃50
50 𝑛
100
−𝑐𝑓𝑃50
𝑓𝑃50
ci Given:
= 29.5 +
20−16
7
6 LB = 29.5
= 29.5 +
4
7
6
50 𝑛
100
= 20
= 29.5 + 3.4 cf = 16
𝑃50= 32.9 f = 7
ci = 6

12- 17 II 2 2
18- 23 IIII- II 7 9
24- 29 IIII- II 7 16
30- 35 IIII- II 7 23
36- 41 IIII -II 7 30
42- 47 IIII - II 7 37
48- 53 III 3 40
n = 40
𝑃50

GRACIAS!
HOPE YOU’VE LEARNED. 

Mean
No Variation in Cash Flow
Variation in Cash Flow
homogeneous
heterogeneous

a. Range for Ungrouped Data
R= HS-LS
Where,
R- range value
HS- highest score
LS- lowest score

Example:
Find the range of the two groups
of score distribution.
GROUP A GROUP B
25 20
32 30
34 34
34 35
40 43
45 46
56 60

Example:
GROUP A GROUP B
25 20
32 30
34 34
34 35
40 43
45 46

Example:
GROUP A GROUP B
32 30
34 34
34 35
40 43
45 46

Range for Ungrouped Data
For Group A
RA = HS-LS
= 56-25
RA = 31

Range for Ungrouped Data
For Group B
RB = HS-LS
= 60-20
RB = 40

b. Range for Grouped Data
R= HSUB-LSLB
Where,
R - range value
HSUB - upper boundary of
highest score
LSLB - lower boundary of
lowest score

Example:
Find the value of range of the
scores of 40 students in Science
examination test. .
x f
15-20 4
21-26 9
27-32 3
35-38 10
39-44 4
45-50 10
n= 40

Range for Grouped Data
R= HSUB-LSLB
= 50.5-14.5
R= 36

a. Mean Deviation for Ungrouped Data
MD=
×−×
𝐧
Where,
MD- mean deviation value
x- individual score
x- sample mean
n- number of scores
_

Example:
Find the mean deviation of the scores
of 7 students in a Science test.
The scores are 40, 34, 34, 45, 56, 32 and 25

x x-x
40
34
34
45
56
32
25
×−×
__

x= ×
n
x= 40+34+34+45+56+32+25
n
= 266
7
x= 38
________
_
_
_

x x-x
40 2
34 -4
34 -4
45 7
56 18
32 -6
25 -13
x= 266
×−×
__

x x-x
40 2 2
34 -4 4
34 -4 4
45 7 7
56 18 18
32 -6 6
25 -13 13
x= 266 ×−× =
54
×−×
__
__

MD= ×−×
n
= 54
7
MD= 7.71
−

b. Mean Deviation for Grouped Data
MD=
n_________
𝒇 × 𝒎 − ×

Where,
MD- mean deviation value
f- class frequency
xm- class mark or midpoint
of each category
x- mean value
n- number of cases
_

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5
21-26 9
27-32 4
33-38 10
39-44 4
45-50 8
n=40
Example:
Find the mean deviation of the given
scores below.

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5 17.5
21-26 9 23.5
27-32 4 29.5
33-38 10 35.5
39-44 4 41.5
45-50 8 47.5
n=40

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5 17.5 87.5
21-26 9 23.5 211.5
27-32 4 29.5 118
33-38 10 35.5 355
39-44 4 41.5 166
45-50 8 47.5 380
n=40 f × 𝒎=
𝟏𝟑𝟏𝟖

×= 𝒇 × 𝒎
n
= 1318
40
×= 32.95

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5 17.5 87.5 -15.45
21-26 9 23.5 211.5 -9.45
27-32 4 29.5 118 -3.45
33-38 10 35.5 355 2.55
39-44 4 41.5 166 8.55
45-50 8 47.5 380 14.55
n=40 f × 𝒎=
𝟏𝟑𝟏𝟖

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5 17.5 87.5 -15.45 15.45
21-26 9 23.5 211.5 -9.45 9.45
27-32 4 29.5 118 -3.45 3.45
33-38 10 35.5 355 2.55 2.55
39-44 4 41.5 166 8.55 8.55
45-50 8 47.5 380 14.55 14.55
n=40 f × 𝒎=
𝟏𝟑𝟏𝟖

× f × 𝒎 f× 𝒎 × 𝒎 − × × 𝒎−× f × 𝒎−×
15-20 5 17.5 87.5 -15.45 15.45 77.25
21-26 9 23.5 211.5 -9.45 9.45 85.05
27-32 4 29.5 118 -3.45 3.45 13.8
33-38 10 35.5 355 2.55 2.55 25.5
39-44 4 41.5 166 8.55 8.55 34.2
45-50 8 47.5 380 14.55 14.55 116.4
n=40 f × 𝒎=
𝟏𝟑𝟏𝟖
f × 𝒎−×
= 352.2

MD= f × 𝒎−×
n
= 352.2
40
MD= 8.805

a. Variance of Ungrouped Data
Population variance
𝝈 𝟐=
×−𝝁 𝟐
𝐍
Sample variance
𝒔 𝟐=
×−× 𝟐
𝐧−𝟏

Example:
Using the data, find the variance and standard deviation
of the scores of 7 students in a science quiz.
x × − × × − × 𝟐
40
34
34
45
56
32
25
× = 266

x × − × × − × 𝟐
40 2
34 -4
34 -4
45 7
56 18
32 -6
25 -13
× = 266
×= 38

x × − × × − × 𝟐
40 2 4
34 -4 16
34 -4 16
45 7 49
56 18 324
32 -6 36
25 -13 169
× = 266
×= 38
× − × 𝟐
= 614

Population variance
𝝈 𝟐
=
×−𝝁 𝟐
𝐍
= 614
7
𝝈 𝟐
=87.71

Sample variance
𝒔 𝟐
=
×−× 𝟐
𝐧−𝟏
= 614
7-1
= 614
6
𝒔 𝟐
= 102.33

b.Variance of Grouped Data
Population variance
𝝈 𝟐
=
𝒇 × 𝒎−𝝁 𝟐
𝐍
Sample variance
𝒔 𝟐
=
𝒇 × 𝒎−× 𝟐
𝐧−𝟏

Score distribution of the test results of 40 students in
a science class consisting of 50 items. Solve the
variance and standard deviation.
× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐 f × 𝒎 − × 𝟐
15-20 5
21-26 9
27-32 4
33-38 10
39-44 4
45-50 8
N=40

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐 f × 𝒎 − × 𝟐
15-20 5 17.5
21-26 9 23.5
27-32 4 29.5
33-38 10 35.5
39-44 4 41.5
45-50 8 47.5
N=40

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐 f × 𝒎 − × 𝟐
15-20 5 17.5 87.5
21-26 9 23.5 211.5
27-32 4 29.5 118
33-38 10 35.5 355
39-44 4 41.5 166
45-50 8 47.5 380
N=40
f × 𝒎=
1318

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐
f × 𝒎 − × 𝟐
15-20 5 17.5 87.5 32.95
21-26 9 23.5 211.5 32.95
27-32 4 29.5 118 32.95
33-38 10 35.5 355 32.95
39-44 4 41.5 166 32.95
45-50 8 47.5 380 32.95
N=40
f × 𝒎=
1318

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐
f × 𝒎 − × 𝟐
15-20 5 17.5 87.5 32.95 -15.45
21-26 9 23.5 211.5 32.95 -9.45
27-32 4 29.5 118 32.95 -3.45
33-38 10 35.5 355 32.95 2.55
39-44 4 41.5 166 32.95 8.55
45-50 8 47.5 380 32.95 14.55
N=40
f × 𝒎=
1318

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐
f × 𝒎 − × 𝟐
15-20 5 17.5 87.5 32.95 -15.45 238.70
21-26 9 23.5 211.5 32.95 -9.45 89.30
27-32 4 29.5 118 32.95 -3.45 11.90
33-38 10 35.5 355 32.95 2.55 6.50
39-44 4 41.5 166 32.95 8.55 73.10
45-50 8 47.5 380 32.95 14.55 211.70
N=40
f × 𝒎=
1318

× f × 𝒎 f× 𝒎 × × 𝒎-× × 𝒎 − × 𝟐 f × 𝒎 − × 𝟐
15-20 5 17.5 87.5 32.95 -15.45 238.70 1193.5
21-26 9 23.5 211.5 32.95 -9.45 89.30 803.7
27-32 4 29.5 118 32.95 -3.45 11.90 97.6
33-38 10 35.5 355 32.95 2.55 6.50 65
39-44 4 41.5 166 32.95 8.55 73.10 292.4
45-50 8 47.5 380 32.95 14.55 211.70 1693.6
N=40
f × 𝒎=
1318
f × 𝒎 − × 𝟐
= 4145.8

Population variance
𝝈 𝟐
=
𝒇 × 𝒎−𝝁 𝟐
𝐍
= 4145.8
40
𝝈 𝟐
=103.645

Sample variance
𝒔 𝟐
=
𝒇 × 𝒎−× 𝟐
𝐧−𝟏
= 4145.8
40-1
= 4145.8
39
𝒔 𝟐
= 106.30

a. Standard Deviation of Ungrouped Data
Population Standard Deviation
𝝈=
×−𝝁 𝟐
𝐍
Sample Standard Deviation
s=
×−× 𝟐
𝐧−𝟏

𝝈=
×−𝝁 𝟐
𝐍
=
𝟔𝟏𝟒
𝟕
= 𝟖𝟕. 𝟕𝟏
𝝈 = 9.37

s=
×−× 𝟐
𝐧−𝟏
=
𝟔𝟏𝟒
𝟔
= 𝟏𝟎𝟐. 𝟑𝟑
s =10.12

b. Standard Deviation of Grouped Data
𝝈= 𝒇 × 𝒎−𝝁 𝟐
𝐍
s=
𝒇 × 𝒎−×
𝟐
𝐧−𝟏

𝝈=
𝒇 × 𝒎−𝝁 𝟐
𝐍
=
𝟒𝟏𝟒𝟓.𝟖
𝟒𝟎
= 𝟏𝟎𝟑. 𝟔𝟒𝟓
𝝈 =10.18

s=
𝒇 × 𝒎−× 𝟐
𝐧−𝟏
=
𝟒𝟏𝟒𝟓.𝟖
𝟑𝟗
= 𝟏𝟎𝟔. 𝟑𝟎
s =10.31

Section A Section B Section C
12 12 12
12 12 12
14 12 12
15 13 12
17 13 12
18 14 12
18 17 13
18 20 26
19 20 26
23 28 26
23 28 26
30 30 30
X = 18.25 X = 18.25 X = 18.25
S= 5.15 S= 6.92 S= 7.63

GROUP 1: ×= 156cm , s= 6
GROUP 2: ×= 156cm , s=10
more varied

BOYS : ×= 160 lbs
s= 8
GIRLS : ×= 100 lbs
s= 9

b. Formula for population
𝐶𝑉 =
𝑆
𝜇
× 100%
Where,
s- standard variation
𝜇- mean

a. Formula for population
𝐶𝑉 =
𝑆
𝜇
× 100%
GROUP 𝜇 s CV
A 38 9.37
B 38.29 11.86

𝐶𝑉 =
𝑆
𝜇
× 100%
Group A
𝐶𝑉 =
9.37
38
× 100%
= 0.25 × 100%
𝐶𝑉 = 25%
Group B
𝐶𝑉 =
11.86
38.29
× 100%
= 0.31 × 100%
𝐶𝑉 = 31%

a. Formula for population
𝐶𝑉 =
𝑆
𝜇
× 100%
GROUP 𝜇 𝑠 CV
A 38 9.37 25 %
B 38.29 11.86 31 %

b. Formula for sample
𝐶𝑉 = 𝑆
×
× 100%
Where,
s- standard variation
×- mean

a. Formula for sample
𝐶𝑉 = 𝑆
×
× 100%
GROUP × 𝑠 CV
A 38 10.12
B 38.29 12.82

𝐶𝑉 = 𝑆
×
× 100%
Group A
𝐶𝑉 =
10.12
38
× 100%
= 0.27 × 100%
𝐶𝑉 = 27%
Group B
𝐶𝑉 =
12.82
38.29
× 100%
= 0.33 × 100%
𝐶𝑉 = 33%

a. Formula for sample
𝐶𝑉 = 𝑆
×
× 100%
GROUP × 𝑠 CV
A 38 10.12 27%
B 38.29 12.82 33%

Formula:
𝑸𝑫 =
𝑸 𝟑− 𝑸 𝟏
𝟐
Where,
𝑸𝑫 - quartile deviation
𝑸 𝟑 - 3rd quartile
𝑸 𝟏 - 1st quartile

For ungrouped data
Find the quartile deviation of the scores of
7 students in a Science test.
The scores are 40, 34, 34, 45, 56, 32 and 25

Scores: 25 32 34 34 40 45 56
𝑄3=
3𝑁
4
𝑡ℎ
𝑄3=
3(7)
4
𝑡ℎ
𝑄3=
21
4
𝑡ℎ
𝑄3=5.25 𝑡ℎ item
Thus, 𝑄3 = 45

Scores: 25 32 34 34 40 45 56
𝑄3=
3𝑁
4
𝑡ℎ 𝑄1=
𝑁
4
𝑡ℎ
𝑄3=
3(7)
4
𝑡ℎ
𝑄3=
21
4
𝑡ℎ
Thus, 𝑄3 = 45
𝑄1=
7
4
𝑡ℎ
Thus, 𝑄1 = 32

Scores: 25 32 34 34 40 45 56
𝑄3 = 45
𝑄1 = 32
Formula: 𝑸𝑫 =
𝑸 𝟑− 𝑸 𝟏
𝟐
𝑄𝐷 =
45−32
2
=
13
2
𝑸𝑫 = 6.5

Class interval f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
Find the quartile deviation of the given scores below.

𝑄3=
3𝑁
4
𝑡ℎ
𝑄3=
3(40)
4
𝑡ℎ
𝑄3=
120
4
𝑡ℎ
𝑸 𝟑= 𝟑𝟎 𝒕𝒉 item
Formula : 𝑄3= 𝐿 𝑄3
+
𝑄3−𝑐𝑓 𝑏
𝑓 𝑄3
𝑖

Class interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
+
𝑓 𝑄3
𝑖

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =
cf 𝑏 =
𝑓𝑄3 =
i=

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5
cf 𝑏 =23

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5
cf 𝑏 =23
𝑓𝑄3 =7
i= 6

+
𝑓 𝑄3
𝑖
𝑄3=35.5 + 30−23
7
6
𝑄3=35.5 + 7
7
6
𝑄3=35.5 + 1 6
𝑄3=35.5 + 6
𝑸 𝟑= 𝟒𝟏. 𝟓

𝑄1=
𝑁
4
𝑡ℎ
𝑄1=
40
4
𝑡ℎ
𝑸 𝟑= 𝟏𝟎 𝒕𝒉 item
+
𝑓 𝑄1
𝑖

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=10
Hence,
𝑙 𝑄3 =
cf 𝑏 =
𝑓𝑄3 =
i=

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5
cf 𝑏 =9

Class
interval
f × 𝑚 cf
Class
boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5
cf 𝑏 =9
𝑓𝑄1 =7
i= 6

+
𝑓 𝑄1
𝑖
𝑄1=23.5 + 10−9
7
6
𝑄1=35.5 + 1
7
6
𝑄1=35.5 + 0.14 6
𝑄1=23.5 + 0.84
𝑸 𝟏= 𝟐𝟒. 𝟑𝟒

𝑄3 = 41.5
𝑄1 = 24.34
Formula: 𝑸𝑫 =
𝑸 𝟑− 𝑸 𝟏
𝟐
𝑄𝐷 =
41.5−24.34
2
=
17.16
2
𝑸𝑫 = 8.58

Formula :
PR= 𝑷 𝟗𝟎− 𝑷 𝟏𝟎
Where,
PR - percentile range
𝑷 𝟗𝟎 -90th percentile
𝑷 𝟏𝟎 - 10th percentile

item science quiz.
15 17 18 19 20 21 22 22
23 25 25 25 26 26 27 29
30 32 33 33 33 34 35 36
37 37 38 38 39 41 44 44
45 45 46 46 47 48 49 50

𝑷
𝟗𝟎= 𝟗𝟎(𝟒𝟎)
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟗𝟎= 𝟑𝟔𝟎𝟎
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟗𝟎= 𝟑𝟔 𝒕𝒉 item ,
which is
𝑷 𝟗𝟎= 𝟗𝟎𝑵
𝟏𝟎𝟎
𝒕𝒉

15 17 18 19 20 21 22 22
23 25 25 25 26 26 27 29
30 32 33 33 33 34 35 36
37 37 38 38 39 41 44 44
45 45 46 46 47 48 49 50
36th item

𝑷
𝟗𝟎= 𝟗𝟎(𝟒𝟎)
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟗𝟎= 𝟑𝟔𝟎𝟎
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟗𝟎= 𝟑𝟔 𝒕𝒉 item ,
which is
𝑷 𝟗𝟎= 𝟗𝟎𝑵
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟏𝟎= 𝟏𝟎𝑵
𝟏𝟎𝟎
𝒕𝒉
𝑷
𝟏𝟎= 𝟏𝟎(𝟒𝟎)
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟏𝟎= 𝟒𝟎𝟎
𝟏𝟎𝟎
𝒕𝒉
𝑷 𝟏𝟎= 𝟒 𝒕𝒉 item ,
which is

15 17 18 19 20 21 22 22
23 25 25 25 26 26 27 29
30 32 33 33 33 34 35 36
37 37 38 38 39 41 44 44
45 45 46 46 47 48 49 50
36th item
4th item

𝑃90=46
𝑃10 = 19
Formula:
PR= 𝑷 𝟗𝟎− 𝑷 𝟏𝟎
= 46-19
PR=27

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5
cf 𝑏 =23

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄3=30
Hence,
𝑙 𝑄3 =35.5
cf 𝑏 =23
𝑓𝑄3 =7
i= 6

Class interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
+
𝑓 𝑄1
𝑖

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5
cf 𝑏 =9

Class
interval
f × 𝑚 cf
Class boundaries
Lower Upper
12-17 2 14.5 2 11.5 17.5
18-23 7 20.5 9 17.5 23.5
24-29 7 26.5 16 23.5 29.5
30-35 7 32.5 23 29.5 35.5
36-41 7 38.5 30 35.5 41.5
42-47 7 44.5 37 41.5 47.5
48-53 3 50.5 40 47.5 53.5
N=40
𝑄1=10
Hence,
𝑙 𝑄1 =23.5
cf 𝑏 =9
𝑓𝑄1 =7
i= 6

Assessment of Learning 2 (Overview)

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