Zulueta, Judie Mae T.

Cluster D

Assessment of Student’s Learning

A. MEAN

3.) Find the mean of the data: 25, 33, 31, 32, 26, 33, 28, 40, 27, 35, 34, 20
μ = 25+33+31+32+26+33+28+40+27+35+34+20
12

μ = 364 = 30.3333
12
4.) Find the mean of the data: 120, 117, 107, 128, 132, 143, 125, 138, 120, 117, 119
μ = 120+117+107+128+132+143+125+138+120+117+119
11

μ = 1366 = 124.1818
11
5.) Find the mean of the scores: 14, 16, 20, 26, 35, 43, 28, 35, 29, 37, 40, 38, 18, 20, 28, 30, 42,
24, 50, 33
μ = 14+16+20+26+35+43+28+35+29+37+40+38+18+20+28+30+42+24+50+33
20

μ = 606= 30.3
20
B. MEDIAN
3.) Find the median of the data: 25, 33, 31, 32, 26, 33, 28, 40, 27, 35, 34, 20

40, 35, 34, 33, 33, 32, 31, 28, 27, 26, 25, 20
X 12 + X 12 X 6 + X 6 +1 32+ 31 63
+1
Md = 2 2 = = = = 31.5
2 2 2 2

4.) Find the median of the data: 120, 117, 107, 128, 132, 143, 125, 138, 120, 117, 119

143, 138, 132, 128, 125, 120, 120, 119, 117, 117, 107

Md = X 11+1 = X 6 =¿120
2

C. MODE
5.) Find the mode of the data: 18, 17, 18, 21, 16, 20, 19, 18, 14, 20

21, 20, 20, 19, 18, 18, 18, 17, 16, 14

Mo = 18
Measures of Central Tendency of Grouped Data
A. MEAN(μ)
b.)
CI f CM
2–6 5 4
7 – 11 4 9
12 – 7 14
16
17 – 5 19
21
22 – 5 24
26
27 – 4 29
31

µ = 5(4) + 4(9) + 7 (14) + 5(19) + 5(24) + 4(29)

5+4+7+5+5+4
= 20+36+98+95+120+116
30
= 485
30
= 16.16

c.)
CI f CM
13 – 23 2 18
24 – 34 3 29
35 – 45 7 40
46 – 56 11 51
57 – 67 11 62
68 – 78 9 73
79 – 89 6 84
90 – 100 1 95

µ = 2(18) + 3(29) + 7(40) + 11(51) + 11(62) + 9(73) + 6(84) + 1(95)

2+3+7+11+11+9+6+1
= 36+87+280+561+682+657+504+95
50
= 2902
50
= 58.0

d.)
 CI f CM
90 – 99 1 94.5
80 – 89 4 84.5
70 – 79 9 74.5
60 – 69 10 64.5
50 – 59 12 54.5
40 – 49 7 44.5
30 – 39 4 34.5
20 – 29 1 24.5
10 – 19 2 14.5

µ = 1(94.5) + 4(8.5) + 9(74.5) + 10(64.5) + 12(54.5) + 7(44.5) + 4(34.5) + 1(24.5) + 2(14.5)

1+4+9+10+12+7+4+1+2
= 94.5+338+670.5+645+654+311.5+138+24.5+29
50
= 2905
50
= 58.1

B. MEDIAN (Md)
b.)
CI f <CF
2–6 5 5
7 – 11 4 9
12 – 7
16 16
17 – 5
21 21
22 – 5
26 26
27 – 4
31 30
30
*** median class is 12-16 since it contains term or 15th term
2
30
Md = 11. 5 + 5 2
7 ( )
−9 6
= 11. 5 + 5 ( ¿ = 11. 5 + 5 ( 0.857 ) = 11.5 + 4.285 = 15.785
7

c.)
CI f <CF
13 – 23 2 2
24 – 34 3 5
35 – 45 7 12
46 – 56 11 23
57 – 67 11 34
68 – 78 9 43
79 – 89 6 49
 90 – 100 1 50
50
*** median class is 57-67 since it contains term or 25th term
2
50
Md =56.. 5 + 11 2
( )
−23
11
= 56. 5 + 11 (
2
11
¿ = 56. 5 + 11( .1818 ) = 56.5 + 2= 58.5

d.)
CI f <CF
90 – 99 1 1
80 – 89 4 5
70 – 79 9 14
60 – 69 10 24
50 – 59 12 36
40 – 49 7 43
30 – 39 4 47
20 – 29 1 48
10 – 19 2 50
50
*** median class is 50-59 since it contains term or 25th term
2
50
Md =49. 5 + 10 2
10 ( )
−24
= 49.5 + 10 (
1
10
¿ = 49. 5 + 10(.1 ) = 49.5 + 1=50.5

C. MODE (Mo )
b.)
CI f
2–6 5
7 – 11 4
12 – 16 7
17 – 21 5
22 – 26 5
27 – 31 4

*** modal class is 12-16 since it contains the highest frequency which is 7

7−4 3
Mo = 11. 5 + 5 ( 2 (7 )−4−5 )
= 11. 5 + 5
5 ()
= 11. 5 + 3 = 15. 5

c.)
CI f
13 – 23 2
24 – 34 3
35 – 45 7
46 – 56 11
57 – 67 11
68 – 78 9
79 – 89 6
 90 – 100 1
11−7
Mo = 56. 5 + 11 ( 2 (11 )−7−11 )
= 56. 5 + 5 (1) = 56. 5 + 5 = 61.5

d.)
CI f
90 – 99 1
80 – 89 4
70 – 79 9
60 – 69 10
50 – 59 12
40 – 49 7
30 – 39 4
20 – 29 1
10 – 19 2

*** modal class is 50-59 since it contains the highest frequency which is 12

Mo = 49. 5 + 10 ( 2 (1212−10
) −10−7 ) 2
= 49. 5 + 10( ) = 49. 5 + 2.86 = 52.36
7

UNGROUPED DATA
2. Given the data: 11, 20, 27, 18, 17, 16, 15, 20, 24,
Find Q1, Q3, D5, D9, P15, P20
DATA: 11, 15, 16, 17, 18, 20, 20, 24, 27
X
a.) Q 1 = 1 (9) = X2.25 = X 2 + .25(X3 –X2) = 15 + .25(16-15) =15.25
4

X 3 (9)
b.) Q 3 = = X6.75 = X 6 + .75(X7 –X6) = 20 + .75(20-20) =20
4

X
c.) D5 = 5 (9) = X4.5 = X 4+ .5(X5 –X4) = 17 + .5(18-17) =17.5
10

X
d.) D9 = 9(9) = X 81 = X8.1 = X8 + .1 (X9 –X8) = 24 + .1 (27-24) =24 + .01(3) = 24.3
10 10

P15=X 15 ( 9) =X 135
e.) 100
= X1.35 = X1 + .35 (X2 –X 1) = 11+.35(15-11) = 11+.35(4) = 12.4
100

P20=X 20( 9) = X 180

f.) 100
= X1.8 = X1 + .8 (X2 –X 1) =11+.8 (15-11) = 11+.8(4) = 14.2
100

3. Given the data: 91, 61, 46, 62, 54, 62, 93, 90, 99, 76, 48, 83, 59, 96, 66, 94, 52, 51, 59, 62, 89
N 100, 92, 70, 59, 91, 73, 68, 49, 54, 85, 43, 78, 50, 45, 98, 69, 77, 42, 46
Find Q1, Q2, D9, D1, P20, P25
DATA: 42, 43, 46, 46, 48, 49, 50, 51, 52, 54, 54, 59, 59, 59, 61, 62, 62, 62, 66, 68, 68, 70,
N 73, 76, 77, 78, 83, 85, 89, 90, 91, 91,92, 93, 94, 96, 98, 99, 100
X =X 10=52
a.) Q 1 = 1 (40)
4
 X 2 (40)
b.) Q 2 = = X20 = 66
4

X
c.) D9 = 9(40) = X36 = 94
10

X
d.) D1 = 1 (40) = X4 = 46
10

P20=X 20( 40)

e.) = X8 = 50
100

P25=X 25( 40)

f.) = X10= 52
100

GROUPED DATA

2. Given the data on the scores of the fifty students in a history class,
Find Q1, Q2, D3, D9, P15, P28

C.I f TCB <CF

49 - 45 2 44.5 – 49.5 50
40 – 44 6 39.5 – 44.5 48
35 – 39 11 34.5 – 39.5 42
30 – 34 10 29.5 – 34.5 31
25 – 29 12 24.5 – 29.5 21
20 – 24 5 19.5 – 24.5 9
15 – 19 4 14.5 – 19.5 4

1(50)
a.) Q1= L +c 4
f [
−F
] = 24.5 + 5 [ 12.5−9
12 ] ´ ) = 24.5 + 1.458 =
= 24.5 + 5(0.291666

25.958
2(50)
b.) Q2= L +c 4
f [
−F
]
= 29.5 + 5
25−21
10 [ ]
= 29.5 + 5 (0.4) = 29.5 + 2 = 31.5

3(50)
c.) D3 = L +c 10
f [
−F
]
= 24. 5 + 5
15−9
12 [
= 24. 5 + 5 ( 0.5) = 27]
9(50)
d.) D9 = L +c 10
f [
−F
]
= 39. 5 + 5
45−42
5 [
= 39. 5 + 5 ( 0.6) = 42.5 ]
15(50)
e.) P15 = L +c 100
f [
−F
= 19. 5 + 5
]7.5−4
5 [
= 19. 5 + 5( 0.7) = 23 ]
 28(50)
f.) P28 = L +c 100
f
−F
[ ]
= 19. 5 + 5
14−9
12 [ ] ´ ) = 21.58
= 19. 5 + 5 (0.41666

MEASURES OF VARIABILITY OR DISPERSION

B. Semi – interquartile Range (Q) OR Quartile Deviation (QD)

3.) Find the QD of the scores in Math 118: 34, 45, 42, 38, 20, 39, 56, 29
-arrange the data: 20, 29, 34, 38, 39, 42, 45, 56
X 1 (8 )
Q1 = = X2 = 29
4

X 3 (8)
Q3 = = X6 = 42
4

42−29 13
QD = = =6.5
2 2
C. Abolute Mean Deviation or Average Mean Deviation ( MD)
2.) Find the MD of the scores in Math 118: 34, 45, 42, 38, 20, 39, 56, 29
34+ 45+ 42+38+20+39+56+ 29 303
μ= = = 37.875
8 8
MD =

|34−37.875|+|45−37.875|+|42−37.875|+|38−37.875|+|20−37.875|+|39−37.875|+|56−37.875|
8
+|29−37.875|
=
8
3.875+7.125+4.125+0.125+ 17.875+ 1.125+ 18.125+ 8.875 61.25
MD = = =7.65625
8 8
D. VARIANCE (2 or SD2 ) AND STANDARD DEVIATION ( or SD)
3.) Find the variance and standard deviation of the data: 2, 4, 6, 8, 10, 12, 7, 9, 13, 20
2+ 4+ 6+8+10+ 12+7 +9+13+20 91
μ= = = 9.1
10 10
2 =

(2−9.1)2 +(4−9.1)2+(6−9.1)2+(8−9.1)2+(10−9.1)2 + ( 12−9.1 )2+(7−9.1)2 +(9−9.1)2 +(13−9.1)2 + ¿ ¿

8
(20−9.1)2 234.9 =23.49
= =
8 10
 = √ 2 = √ 23.49 = 4.846648 4.85
E. COEFFICIENT OF VARIATION (C.V)
3.) Find the coefficient of variation of the data in example number 3 under variance and
standard deviation.
σ 4.85
CV = x 100% = x 100 %=53.30 %
μ 9.1
Measures of Variability For Grouped data
A. Range
2.)
C.I TCB Range = 98.5 – 68.5 = 30
94 – 98 93.5 – 98.5
89 – 93 88.5 – 93.5
84 – 88 83.5 – 88.5
79 – 83 78.5 – 83.5
74 – 78 73.5 – 78.5
69 – 73 68.5 – 73.5
C. Abolute Mean Deviation or Average Mean Deviation ( MD)
3.) Find MD of the data below:
C.I f CM
6 ( 8 ) + ( 10 ) ( 17 ) + ( 8 ) ( 26 ) + ( 5 ) ( 35 ) + ( 1 )( 44 )
4 – 12 6 8 μ= = 21.5
30
13 – 21 10 17
22 – 30 8 26
31 – 39 5 35
40 – 48 1 44
k

∑ f i ∨CM i−μ∨¿ 6|8 – 21.5|+10|17 – 21.5|+ 8|26 – 21.5|+5|35 – 22.6|+1|44 – 21.5|

i=1
MD = k
= ¿
6+10+ 8+5+1
∑ fi
i=1

6 ( 13.5 )+ 10 ( 4.5 ) +8 ( 4.5 )+5 ( 13.5 ) +1 ( 22.5 ) 252

= = = 8.4
30 30
D. VARIANCE (2 or SD2 ) AND STANDARD DEVIATION ( or SD)
3. Find the Variance and Standard Deviation of the sample data below:
C.I f CM
6 ( 8 ) + ( 10 ) ( 17 ) + ( 8 ) ( 26 ) + ( 5 ) ( 35 ) + ( 1 )( 44 )
4 – 12 6 8 μ= = 21.5
30
13 – 21 10 17
22 – 30 8 26
31 – 39 5 35
40 – 48 1 44
k
2
σ = ∑ f i ¿¿¿
i=1
 6(8−21.5)2+10 (17−21.5)2 +8( 26−21.5)2 +5(35−21.5)2 +1( 44−21.5)2
=
6+10+8+5+1
= 6¿¿
1093.5+ 202.5+162+ 911.25+506.25 2875.5
σ2 = = =95.85
30 30
σ = √ 95.85 = 9.790301 9.80

E. COEFFICIENT OF VARIATION (C.V)

σ 9.80
3. CV = x 100% = x 100 %=45.58 %
μ 21.5

V. ASSESSMENT:

A. EXERCISES IN MEASURES OF CENTRAL TENDENCY

1. Given the data below put it in a frequency distribution table and then compute for the
Mean, Median, and Mode of the grouped data.

48 12 18 33 39 42 45 15 23 19 25 19 43 63

52 14 17 28 31 53 26 40 37 23 22 20 49

65 26 29 18 35 30 21 27 44 34 18 37 58

1. R = 65−¿12 = 53
2. k = √ 40 = 6.32 = 6
R 53
3. c = = = 8.83 = 9
k 6

C1 Tally Freq TCB CM < CF > CF RF

12-20 11111-11111 10 11.5-20.5 16 10 40 25
21-29 11111-11111 10 20.5-29.5 25 20 30 25
30-38 11111-11 7 29.5-38.5 34 27 20 17.5
39-47 11111-1 6 38.5-47.5 43 33 13 15
48-56 1111 4 47.5-56.5 52 37 7 10
57-65 111 3 56.5-65.5 61 40 3 7.5
Total 40 100%

MEAN
µ=

10 ( 16 )+ 10 ( 25 )+7 (34 ) +6 ( 43 )+ 4 ( 52 ) +3(31) 160+ 250+238+258+208+183 1297

= = =32.4 25
10+10+7+6+ 4+ 3 40 40
MEDIAN
40
Md = 20. 5 + 9 2
( )
−10
10
= 20. 5 + 9 (
10
10
¿ = 20. 5 + 9 ( 1) = 29.5

MODE

Mo = 20. 5 + 9 ( 2 (1010−10
)−10−7 )
= 20.5 + 9( 0 ) = 20. 5 + 0 = 20.5

B.
1. Find Q3, D8, D3, P30, P70
C.I f TCB <CF
4 – 12 6 3.5 – 12.5 6
13 – 21 10 12.5 – 21.5 16
22 – 30 8 21.5 – 30.5 24
31 – 39 5 30.5 – 39.5 29
40 – 48 1 39.5 – 48.5 30
3(30)

[
a.) Q3= L +c 4
f
−F
]
= 21.5 + 9
22.5−16
8 [ ]
= 21.5 +9 (0.8125) = 21.5 + 7.3125 =

28.8125
8(30)
b.) D8 = L +c
[ 10
f
−F
]
= 21. 5 + 9
24−16
8 [ ]
= 21. 5 + 9 ( 1) = 30.5

3(30)

[
c.) D3 = L +c 10
f
−F
]
= 12. 5 + 9
9−6
10 [ ]
= 12. 5 + 9 (0.3) = 12.5 + 2.7 = 15.2

30(30)

[
d.) P30 = L +c 100
f
−F
= 12. 5 + 9
]9−6
10 [ ]
= 12. 5 + 9 (0.3) = 12.5 + 2.7 = 15.2

70(30)

[
e.) P30 = L +c 100
f
−F
= 21. 5 + 9
]21−16
8 [ ]
= 21. 5 + 9 ( 0.625) = 12.5 + 5.625 =

18.125
2. Solve for Q1, D7, P81, P12,

C.I f TCB <CF

93 – 98 5 93.5 – 98.5 55
89 – 93 5 88.5 – 93.5 50
84 – 88 13 83.5 – 88.5 45
79 – 83 7 78.5 – 83.5 32
 74 – 78 16 73.5 – 78.5 25
69 – 73 9 68.5 – 73.5 9
1(55)
a.) Q1= L +c 4
[
f
−F
= 73.5 + 5
]
13.75−9
16 [ ]
= 73.5 + 5 (0.296875) = 73.5 + 1.4843=

74.9843
7(55)

[
b.) D7 = L +c 10
f
−F
]
= 83. 5 + 5
38.5−32
13 [ ]
= 83. 5 + 5 (0.5) = 83.5 + 2.5 = 86

81(55)

[
c.) P81 = L +c 100
f
−F
= 83. 5 + 5
]
44.55−32
13 [ ]
= 83. 5 + 5 (0.965) = 83.5 + 4.825 =

88.325
12(55)
d.) P12 = L +c 100
f[ −F
= 68.5+ 5
6.6
9 ] [ ]
= 68. 5 + 5 (0.733333) = 68.5 + 3.6666 =

72.16666

3. Solve for Q3, D3, D6, P51, P83

C.I f TCB <CF

15.2 – 19.6 4 15.15 – 19.65 4
19.7 – 24.1 5 19.65 – 24.15 9
24.2 – 28.6 10 24.15 – 28.65 19
28.7 – 33.1 6 28.65 – 33.15 25
33.2 – 37.6 5 33.15 – 37.65 30
1(30)
a.) Q3= L +c 4
[
f
−F
]
= 19.65 + 5.4
7.5−4
5 [ ]
= 19.65 + 5.4 (0.7) = 19.65 + 3.78 = 23.43

3(30)
b.) D3 = L +c 10
[f
−F
]
= 19.65 + 5.4
9−4
5 [ ]
= 19.65 + 5.4 (1) = 19.65 + 5.4 = 25.05

6(30)
c.) D6 = L +c
[ 10
f
−F
]
= 24.15 + 5.4
18−9
10 [ ]
= 24.15 + 5.4 (0.9) = 24.15 + 4.86 = 29.01
 51(30)
d.) P51 = L +c 100
f[ −F
= 24.15 + 5.4
]15.3−9
10 [ ]
= 24.15 + 5.4 (0.63) = 24.15 + 3.402 =

27.552
83(30)
e.) P83 = L+c 100
f[ −F
= 28.65+5.4
]
24.9−19
6 [ ]
= 28.65+5.4(0.98333)=28.65+5.3098=

33.9598
C. Find MD of the data below:

C.I f CM
27 – 31 4 29
22 – 26 5 24
17 – 21 5 19
12 – 16 7 14
7 – 11 4 9
2–6 5 4.5

( 4 )( 29 ) + ( 5 )( 24 )+ ( 5 )( 19 ) + ( 7 )( 14 )+ ( 4 )( 9 )+(5)(4.5)
μ= = 16.25
30

4|29 – 16.25|+5|24 – 16.25|+5|19 – 16.25|+7|14 – 16.25|+ 4|9 – 16.25|+5|4.5 – 16.25|

MD ¿
4+5+5+ 7+4 +5
4 ( 12.75 ) +5 ( 7.75 ) +5 ( 2.75 ) +7 ( 2.25 ) +4 ( 7.25 )+ 5(11.75) 207
= = = 6.9
30 30