Polytechnic University of the Philippines
San Juan Branch
Problem Set
Data Management
Name: Irish May L. Quilente
Course/Year & Section: BSA 1-2
Date: 07/07/2022
Do as indicated. Round off your final answer to the nearest hundredths.
1. Find the mean, median and mode, given the scores of 10 students in their Mathematics activity.
1 27 16 7 31 7 30 3 21
Answer: 1 3 7 7
16
21 27 30 31
n=9
Mean
Median
Mode
The summation of all the scores divided by n
143/9= 15.89
16
7
Mean
Median
Mode
∑x
n
143
x̅ =
9
16
7
It is an unimodal data.
With Solution
x̅ =
x̅ = 𝟏𝟓. 𝟖𝟗
2. Consider the following Mathematics test scores of Grade 12 Students of PUP San Juan: Find the mean median and mode.
Stems
4
5
6
7
8
9
Leaves
5 7 9
3 6 8 8
0 4 7
1 2 3 6 8
4 5 6 8
4 6
Answer: 45 47 49 53 56 58 58 60 64 67
n=21
Mean
Median
The summation of all scores divided by n
1460/21= 69.52
71
With Solution
Mode
58
71
72 73 76 78 84 85 86 88 94 96
Mean
Median
Mode
∑x
x̅ =
n
1460
x̅ =
21
71
58
x̅ = 𝟔𝟗. 𝟓𝟐
3. The following scores obtained by Grade 10 students in Mathematics:
12
20
14
22
14
23
17
25
17
18
25
19
25
19
25
19
26
19
29
25
29
29
31
31
32
32
33
34
36
41
n=30
Construct its frequency distribution table and find its measures of central tendency. Note: 8 class interval
a. Range: HV-LV= 41-12= 29
b. Class size: 29/8= 3.63
Class Interval
12-15
16-19
20-23
24-27
28-31
32-35
36-39
40-43
i=4
Tally
III
IIIII II
III
IIIII I
IIIII
IIII
I
I
Frequency
(𝑓)
3
7
3
6
5
4
1
1
n= 30
Boundaries Aries/
Class Bond
11.5-15.5
15.5-19.5
19.5-23.5
23.5-27.5
27.5-31.5
31.5-35.5
35.5-39.5
39.5-43.5
Class Mark
(𝑥𝑚 )
13.5
17.5
21.5
25.5
29.5
33.5
37.5
41.5
𝑓𝑋𝑚
40.5
122.5
64.5
153
147.5
134
37.5
41.5
Less than cumulative frequency
(< 𝑐𝑓)
3
10
13
19
24
28
29
30
∑fXm=741
greater than cumulative frequency
(> 𝑐𝑓)
30
27
20
17
11
6
2
1
Solution:
Mean
Median
Mode
𝑛
−𝐶𝑓𝑃
2
The summation of all scores divided by n
741/30= 24.7
𝑥̅ = 𝐿𝐵𝑥̅ + (
𝑓𝑚
𝑥̂ = 𝐿𝐵𝑥̂ + (𝑑
)ⅈ
𝑑1 = f-fp
= 7-4
=3
𝑑2 = 7-3
=4
=n/2 =30/2
=15
𝛴𝑓𝑥
𝑥̅ = 𝑛 𝑚
=741/30
=24.7
15−13
𝑥̅ = 23.5 + ( 6 )4
=23.5 + (0.33) 4
=23.5 +1.32
24.82
4
=15.5 +( )4
415
=15.5 = (0.5)4
=17.5
4. Find the measures of variance of the given data.
Scores
81
82
82
83
85
85
86
87
91
92
n= 10
̅
𝒙−𝒙
-4
-3
-3
-2
0
0
1
2
6
7
𝑑1 1
1 +𝑑2
|𝒙 − 𝒙
̅|
4
3
3
2
0
0
1
2
6
7
∑|𝒙 − 𝒙
̅| = 28
̅)𝟐
(𝒙 − 𝒙
16
9
9
4
0
0
1
4
36
49
∑(𝒙 − 𝒙
̅)𝟐 = 128
)⋅𝐿
Solution:
Mean
Range
∑x
x̅ =
n
854
x̅ =
10
Mean Absolute Deviation
∑|𝑥−𝑥̅ |
MAD=
𝑛
=HV – LV
=92-81
=11
MAD = 28/10
MAD = 2.8
Variance
Standard Deviation
∑f(xm − x̅)z
s2 =
n−1
128
2
s =
10 − 1
∑(x − x̅)2
s=√
n−1
s 2 = 𝟏𝟒. 𝟐𝟐
s = √14.22
s = 𝟑. 𝟕𝟕
x̅ = 𝟖𝟓. 𝟒
128
s=√
9
5. Find the measures of variance of the given data.
Class
Interval
Frequency
𝒇
Class
Mark
𝑿𝒎
𝑓𝑋𝑚
Deviation
̅
𝑿𝒎 − 𝒙
(𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏)
̅̅̅
𝒇(𝑿𝒎 − 𝒙)
̅|
𝒇|𝒙𝒎 − 𝒙
Square deviation
̅)𝟐
(𝒙𝒎 − 𝒙
𝒇(𝒔𝒒𝒖𝒂𝒓𝒆 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏)
̅)𝟐
𝒇(𝒙𝒎 − 𝒙
1 – 10
11 – 20
21 – 30
31 – 40
3
5
5
4
5.5
15.5
25.5
35.5
16.5
77.5
127.5
142
-33.33
-23.33
-13.33
3.33
-99.99
-166.65
-66.65
-13.32
99.99
166.65
66.65
13.32
1110.09
544.29
177.69
11.09
3330.27
2721.45
888.45
44.36
41 – 50
51 – 60
61 – 70
71 – 80
ⅈ =10
2
3
6
2
n=30
45.5
55.5
65.5
75.5
91
166.5
393
151
∑ 𝑓𝑋𝑚 =
1165
6.67
16.67
26.67
36.67
13.34
50.01
160.02
73.34
13.34
50.01
160.02
73.34
∑ 𝒇|𝒙𝒎 − 𝒙
̅| =
593.32
44.49
277.89
711.29
1344.69
88.98
833.67
4267.74
2689.38
̅)𝟐 =
∑ 𝒇(𝒙𝒎 − 𝒙
14864.3
Solution:
Range
𝑅 = 𝑥ℎ𝑢𝑏 − 𝑥𝑢𝑏
R= 80.5-0.5
=80
Mean
Mean= 1,165/30
Mean Absolute Deviation
Variance
Standard Deviation
14,863.3
30−1
MAD= 593.32/30
𝑠 2 =14,864.3/30-1
S=√
=19.78
=14,864.3/29
=√
=512.56
=√512.56
=22.64
=38.83
14,863.3
29