Frequency Table - Elementary Statistics

Construction of
Frequency Table
STATS&PROB
SerJhob

STATS&PROB
Quantitative Frequency Distribution Table
• a table where the data are tabulated based on
numerical classes or interval.
Qualitative Frequency Distribution Table
• a table where the data are tabulated based on
description.
SerJhob

STATS&PROB
CLASS INTERVAL F
98 – 99 9
96 – 97 9
94 – 95 13
92 – 93 13
90 – 91 15
88 – 89 6
86-87 2
TOTAL N = 67
Quantitative Frequency Distribution Table
SerJhob

STATS&PROB
Courses No. of Students
BSEd Mathematics 30
BSEd Biological Science 28
BSEd Physical Science 15
Engineering 50
Nursing 80
TOTAL N = 203
Qualitative Frequency Distribution Table
SerJhob

STATS&PROB
Class Limits
•This pertains to the lowest and the highest value that can
go in each class.
Lower Class Limit
• This refers to the lowest value that can be entered in a
class.
Upper Class Limit
•This pertains to the highest value that can be entered in a
class.
SerJhob

STATS&PROB
Class Boundary
•This is considered as the true limit or real limit. The class
boundary can be obtained by simply adding 0.5 to the
upper limit and subtracting 0.5 from the lower limit.
Class Width or Class Size
• Getting the difference between the boundaries
• Getting the difference between two successive lower
limits or two successive upper limits
Frequency it pertains to the number of values that fall in a
certain class.
SerJhob

STATS&PROB
Midpoint (Class Mark)
•This is a value that acts as representative of a certain class. It
can be obtained by adding the lower and upper limit divided
by two.
Less than Cumulative Frequency Distribution
• This refers to the distribution whose frequencies are less
than or below the upper class boundary they correspond to.
Greater than Cumulative Frequency Distribution
• This refers to the distribution whose frequencies are greater
than or above the upper class boundary they correspond to.
SerJhob

STATS&PROB
Example 1
•The following are the raw scores of 67 students in a
Mathematics test (out of 100 items). Construct a cumulative
frequency distribution table with corresponding frequencies,
midpoint, less than and greater than cumulative frequency.
The number of classes is 7.
91 98 92 94 90 97 90 97 99 88 92 97
98 95 98 98 92 89 92 99 97 95 89 92
97 92 93 95 99 90 94 94 97 94 90 97
90 90 93 91 93 93 90 95 95 90 89 86
98 94 95 94 98 93 90 89 87 97 88 92
90 94 90 93 90 97 91
SerJhob

STATS&PROB
Step 1 – Get the highest and the lowest score or value.
Step 2 – Get the Range
Step 3 – Determine the number of classes. The decided
number of classes (rows) is 7
Step 4 – Determine the size of the class intervals.
Step 5 – Construct the class. The lower limit of the lowest class
interval should be a multiple of two.
SerJhob

STATS&PROB
The frequency table will look like the table below.
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
86 – 87
88 – 89
90 – 91
92 – 93
94 – 95
96 – 97
98 – 99

STATS&PROB
The frequency table will look like the table below.
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
86 – 87 85.5 – 87.5 II 2 86.5 2 67
88 – 89 87.5 – 89.5 IIIII – I 6 88.5 8 65
90 – 91 89.5 – 91.5
IIIII – IIIII –
IIIII
15 90.5 23 59
92 – 93 91.5 – 93.5
IIIII – IIIII –
III
13 92.5 36 44
94 – 95 93.5 – 95.5
IIIII – IIIII –
III
13 94.5 49 31
96 – 97 95.5 – 97.5 IIIII – IIII 9 96.5 58 18
98 – 99 97.5 – 99.5 IIIII – IIII 9 98.5 67 9

STATS&PROB
Example 2
•The following are the areas of land (in hectare) owned by 60
families. Construct a quantitative frequency distribution table
with corresponding frequencies, midpoint, less than and
greater than cumulative frequency. The desired classes is 10.
24 34 32 46 56 49 25 35 33 47 57
50 26 36 35 48 58 53 27 37 37 49
59 61 28 38 39 50 60 62 29 39 40
51 46 63 30 40 41 52 47 61 31 41
42 53 49 62 32 27 43 54 50 63 33
28 44 55 53 45
SerJhob

STATS&PROB
Example 2
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
60 – 63
56 – 59
52 – 55
48 – 51
44 – 47
40 – 43
36 – 39
32 – 35
28 – 31
24 – 27
TOTAL

STATS&PROB
Example 2
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
60 – 63 59.5 – 63.5 IIIII – II 7 61.5 60 7
56 – 59 55.5 – 59.5 IIII 4 57.5 53 11
52 – 55 51.5 – 55.5 IIIII – I 6 53.5 49 17
48 – 51 47.5 – 51.5 IIIII – III 8 49.5 43 25
44 – 47 43.5 – 47.5 IIIII – I 6 45.5 35 31
40 – 43 39.5 – 43.5 IIIII – I 6 41.5 29 37
36 – 39 35.5 – 39.5 IIIII – I 6 37.5 23 43
32 – 35 31.5 – 35.5 IIIII – II 7 33.5 17 50
28 – 31 27.5 – 31.5 IIIII 5 29.5 10 55
24 – 27 23.5 – 27.5 IIIII 5 25.5 5 60
TOTAL N = 60

STATS&PROB
PROBLEM 1
•Forty Super B batteries were tested to determine how long
they would last. The results, to the nearest minute, were
recorded as follows:
432 324 332 326 756 632 225 435 353 427
501 226 436 345 481 583 453 227 347 737
359 261 428 638 739 650 320 662 629 339
512 546 462 430 743 641 452 247 461 321
Construct a quantitative frequency distribution table. The
desired number of classes is 10.
SerJhob

STATS&PROB
OBLEM 2
he following are the daily allowance (in pesos) of the 35
udents
45.42 70.35 55.32 56.34 75.48 65.76 87.34
37.49 54.23 98.45 78.45 78.34 98.34 87.43
56.23 76.44 89.22 68.51 87.43 67.24 45.49
57.34 98.50 56.34 23.45 32.45 34.45 45.67
45.87 78.53 45.76 50.47 76.45 78.34 56.52
Construct a quantitative frequency distribution table. The
sired number of classes is 10.
SerJhob

STATS&PROB
PROBLEM 1 (ANSWER)
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
TOTAL
224 - 276
277 - 329
330 - 382
383 - 435
436 - 488
489 - 541
542 - 594
595 - 647
648 - 700
701 and above
223.5 – 276.5
276.5 – 329.5
329.5 – 382.5
382.5 – 435.5
435.5 – 488.5
488.5 – 541.5
541.5 – 594.5
594.5 – 647.5
647.5 – 700.5
700.5 and above
IIIII
IIII
IIIII-I
IIIII
IIIII-I
II
II
IIII
II
IIII
5
4
6
5
6
2
2
4
2
4
250
303
356
409
462
515
568
621
674
727
5
9
15
20
26
28
30
34
36
40
40
35
31
25
20
14
12
10
6
4
40 4885
XL

STATS&PROB
PROBLEM 2 (ANSWER)
SerJhob
CLASS
INTERVAL
CLASS
BOUNDARIES
TALLY F X <CF >CF
TOTAL
22 - 29
30 - 37
38 - 45
46 - 53
54 - 61
62 - 69
70 - 77
78 - 85
86 - 93
94 - 101
21.5 – 29.5
29.5 – 37.5
37.5 – 45.5
45.5 – 53.5
53.5 – 61.5
61.5 – 69.5
69.5 – 77.5
77.5 – 85.5
85.5 – 93.5
93.5 – 101.5
I
III
II
IIII
IIIII-II
III
IIII
IIII
IIII
III
1
3
2
4
7
3
4
4
4
3
25.5
33.5
41.5
49.5
57.5
65.5
73.5
81.5
89.5
97.5
1
4
6
10
17
20
24
28
32
35
35
34
31
29
25
18
15
11
7
3
35 615
XXXV

Measures of
Central Tendency
STATS&PROB
SerJhob
Midterm – Lesson 1 & 2

STATS&PROB
Objectives
After studying this lesson, you are expected to:
•Define measures of central tendency;
•Define mean, median and mode;
•Enumerate the advantages and disadvantages of each
measure;
•Calculate mean, median and mode for group and ungrouped
data; and
•Apply mean, median and mode in various fields of science
and specialization.
SerJhob

STATS&PROB
Measures of Central Tendency
•A statistical measure that gives information regarding the
central location of the score relative to the entire set of data.
Thus, this measure is designed to give numerical description
about the set of data gathered not on making numerical
inferences. Three commonly used Measures of Central
Tendency are the mean, median and mode.
SerJhob

STATS&PROB
The Mean
•The most popular measure of central tendency and
sometimes layman calls it average.
Population Mean
•The sum of all the values taken from the population (𝑁)
divided by the total number of units distributed. (𝜇) mu
Sample Mean
•The sum of all the values taken from the sample (n) divided
by the total number of units considered. (x)
SerJhob

STATS&PROB
ADVANTAGES
• Mean can be easily computed and is subject to less error
• Every observation contributes to the value of the mean
• This is the best measure for regular distribution.
• This is the most reliable because all observations contribute to the
value of the mean.
DISADVANTAGES
• Mean does not give information regarding the homogeneity and
heterogeneity of the distribution.
• It can be easily affected by extreme values/outliers.
• The more the distribution becomes heterogeneous the lower
satisfactory is the mean.
SerJhob

STATS&PROB
Formula for Ungrouped Data
𝑥 =
𝑖=1
𝑛
𝑋𝑖
𝑛
=
𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛
𝑛
Where: 𝑥=Mean 𝑛=number of samples x=raw score
Formula for Grouped Data
𝑥 =
𝑓𝑋𝑚
𝑛
Where: 𝑥=Mean 𝑛=number of samples f=frequency
Xm = midpoint
SerJhob

STATS&PROB
PROBLEM 1
•The following are the scores of 10 students in an arithmetic
test.
79 89 90 90 87 85 100 91 98 83
Find the mean.
Computation: To compute for the mean, substitute the raw
score and divide it by the total number of units considered.
SerJhob

STATS&PROB
PROBLEM 1
test.
79 89 90 90 87 85 100 91 98 83
Mean: 𝑥 = 𝑖=1
𝑛
𝑋𝑖
𝑛
=
𝑋1+𝑋2+⋯+𝑋𝑛
𝑛
=
79+89+90+90+87+85+100+91+98+83
10
=
892
10
= 89.2
SerJhob

STATS&PROB
PROBLEM 2
•The enrollment in a Barangay Elementary School in last 7
years was 365, 400, 452, 500, 525, 607, and 802. What was
the average enrollment per year?
•To compute for the mean, add the number of enrollees per
year and divide it by 7 years.
Mean: 𝑥 = 𝑖=1
𝑛
𝑋𝑖
𝑛
=
𝑋1+𝑋2+⋯+𝑋𝑛
𝑛
=
365+400+452+500+525+607+802
7
=
3651
7
= 521.57 or 522
SerJhob

STATS&PROB
EXERCISES
•The following are the grades of students in their subjects.
Find the average grades rounded to nearest tenths:
•Student A: 89, 90, 91, 89, 89, 98, 85, 89, 93, 94
•Student B: 74, 83, 83, 88, 84, 76, 75, 84, 75, 91
•Student C: 88, 83, 84, 85, 88, 83, 85, 82, 85, 82
•Student D: 89, 92, 91, 85, 87, 95, 96, 90
•Student E: 77, 77, 77, 77, 77, 77, 77, 77, 77
•Student F: 98.5, 87.6, 91.2, 83.2, 95.4, 98.8, 85.9
•Student G: 86.76, 95.43, 92.32, 85.34, 78.54, 76.81
SerJhob
Mean: 90.7
Mean: 81.3
Mean: 84.5
Mean: 90.6
Mean: 77
Mean: 91.5
Mean: 85.9

STATS&PROB
PROBLEM 3
• The following is a cumulative frequency distribution table of the scores of 67
students in a 100-item Mathematics test, showing the classes and its
corresponding frequency.
SerJhob
Score f Xm fXm
98 – 99 9
96 – 97 9
94 – 95 13
92 – 93 13
90 – 91 15
88 – 89 6
86 – 87 2
TOTAL N = 67
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
98.5
96.5
94.5
92.5
90.5
88.5
86.5
886.5
868.5
1228.5
1202.5
1357.5
531
173
6247.5
Mean =
𝟔𝟐𝟒𝟕.𝟓
𝟔𝟕
= 𝟗𝟑. 𝟐𝟓

STATS&PROB
PROBLEM 4
• The following scores are gathered from 55 students in a 50-item quiz, showing
the classes and its corresponding frequency.
SerJhob
Score F Xm fXm
45 & above 9
39 – 44 8
33– 38 11
27 – 32 13
21 – 26 10
15 – 20 2
9 – 14 2
TOTAL N = 55
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
47.5
41.5
35.5
29.5
23.5
17.5
11.5
427.5
332
390.5
383.5
235
35
23
1826.5
Mean =
𝟏𝟖𝟐𝟔.𝟓
𝟓𝟓
= 𝟑𝟑. 𝟐𝟏

STATS&PROB
EXERCISES
• You grew fifty baby carrots using special soil. You dig them up and measure
their lengths (to the nearest mm) and group the results:
SerJhob
Lengths F Xm fXm
150 – 154 12
155 – 159 10
160 – 164 8
165 – 169 7
170 – 174 8
175 – 179 3
180 – 184 2
TOTAL N = 50
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
152
157
162
167
172
177
182
1824
1570
1296
1169
1376
531
364
8130
Mean =
𝟖𝟏𝟑𝟎
𝟓𝟎
= 𝟏𝟔𝟐. 𝟔

STATS&PROB
The Median
•The weakness of the mean leads us to examine another
measure of central tendency which is not affected by few
extreme values – the median.
•It is a positional measure that divides the distribution into
two equal parts when the data is arranged in order. 𝑀𝑑
•The median is considered the most stable measure of central
tendency of the distribution is irregular.
SerJhob

STATS&PROB
ADVANTAGES
•This is not easily influenced by extreme values.
•This is the best measure for irregular distribution.
DISADVANTAGES
•It requires arrangement of the data according to
their sizes.
•It is erroneous if the data do not cluster in the
middle of the distribution.
SerJhob

STATS&PROB
Procedure for Ungrouped Data
1. Arrange the numerical data in ascending or descending order.
2. When the observation is odd, get the middle most value to be the
median of the class.
3. When the observation is even, add the two middle most values and
divide it by two.
𝑀𝑑 = 𝐿𝑙 +
𝑛
2
− 𝐹
𝑓𝑤
𝑖
Where: 𝑀𝑑=Median, 𝐿𝑙=Lower True Limit, 𝐹=Less than cumulative
frequency, 𝑓𝑤= frequency of the class, 𝑖= interval, and 𝑛=sample size.
SerJhob

STATS&PROB
PROBLEM 5
test.
79 89 90 90 87 85 100 91 98
Find the mean & median score.
To get the median, arrange the score in an ascending (or
descending) order.
79 85 87 89 90 90 91 98 100
Since the total number of scores is odd (9 scores), get the
middle score.
SerJhob

STATS&PROB
PROBLEM 5
•The median score is 90.
•The result implies that 50% of the scores in the test are below
90 while the other 50% of the scores are above 90.
•The mean score is 89.89
•The result implies that the average score of all the test takers
is 89.89
SerJhob

STATS&PROB
PROBLEM 6
•The points scored by a volleyball team in series of matches
are as follows:
24 15 17 10 21 14 8 25 11 13
To get the median, arrange the score in an ascending (or
descending) order.
8 10 11 13 14 15 17 21 24 25
Since the total number of scores is even (10 scores), get the
sum of the middle most value and divide it by two.
SerJhob
14.5

STATS&PROB
PROBLEM 6
•The median score is 14.5.
•The result implies that 50% of their scores in series of
matches are below 15 while the other 50% of the scores are
above 15.
•The mean score is 15.8.
•The result implies that the average points scored by a
volleyball team is 16.
SerJhob

STATS&PROB
EXERCISES (Review for Median)
Find the median score.
•Student A: 89, 90, 91, 89, 89, 98, 85, 89, 93, 94
•Student B: 74, 83, 83, 88, 84, 76, 75, 84, 75, 91
•Student C: 88, 83, 84, 85, 88, 83, 85, 82, 85, 82
•Student D: 89, 92, 91, 85, 87, 95, 96, 90
•Student E: 77, 77, 77, 77, 77, 77, 77, 77, 77
•Student F: 98.5, 83.2, 91.2, 83.2, 95.4, 98.8, 85.9
•Student G: 86.76, 95.43, 92.32, 85.34, 78.54, 86.76
SerJhob
𝑀𝑑: 89.5
𝑀𝑑: 83
𝑀𝑑: 84.5
𝑀𝑑: 90.5
𝑀𝑑: 77
𝑀𝑑: 91.2
𝑀𝑑: 86.76

STATS&PROB
PROBLEM 7
SerJhob
Score f F
98 – 99 9
96 – 97 9
94 – 95 13
92 – 93 13
90 – 91 15
88 – 89 6
86 – 87 2
TOTAL N = 67
Formula:
𝑴𝒅 = 𝑳𝒍 +
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
67
58
49
36
23
8
2
Median = 𝟗𝟏. 𝟓 +
𝟑𝟑.𝟓−𝟐𝟑
𝟏𝟑
2
𝐧
𝟐
=
𝟔𝟕
𝟐
= 𝟑𝟑. 𝟓
= 𝟗𝟏. 𝟓 +
𝟐𝟏
𝟏𝟑
= 𝟗𝟑. 𝟏𝟐

STATS&PROB
PROBLEM 8
SerJhob
Score f F
45 & above 9
39 – 44 8
33– 38 11
27 – 32 13
21 – 26 10
15 – 20 2
9 – 14 2
TOTAL N = 55
Formula:
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
55
46
38
27
14
4
2
Mode = 𝟑𝟐. 𝟓 +
𝟐𝟕.𝟓−𝟐𝟕
𝟏𝟏
6
𝐧
𝟐
=
𝟓𝟓
𝟐
= 𝟐𝟕. 𝟓
= 𝟑𝟐. 𝟓 +
𝟑
𝟏𝟏
= 𝟑𝟐. 𝟕𝟕

STATS&PROB
EXERCISES
• You grew fifty baby carrots using special soil. You dig them up and measure
their lengths (to the nearest mm) and group the results:
SerJhob
Score f F
150 – 154 12
155 – 159 10
160 – 164 8
165 – 169 7
170 – 174 8
175 – 179 3
180 – 184 2
TOTAL N = 50
Formula:
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
12
22
30
37
45
48
50
Mode = 𝟏𝟓𝟗. 𝟓 +
𝟑𝟎−𝟐𝟐
𝟖
5
𝐧
𝟐
=
𝟓𝟎
𝟐
= 𝟐𝟓
= 𝟏𝟓𝟗. 𝟓 +
𝟒𝟎
𝟖
= 𝟏𝟔𝟒. 𝟓

STATS&PROB
The Mode
•It is the observation that occurs most frequently in the data
and the observation that occurs more than once.
•It is very effective in describing nominal data.
•The mode of the distribution may or may not exist.
SerJhob

STATS&PROB
ADVANTAGES
•It is always a real value since it does not fall on zero.
•It does not require calculation and is simple to approximate.
•It does not need Mathematical manipulation if the data is
ungrouped.
•It is not easily affected by extreme values.
DISADVANTAGES
•It gives limited information about the distribution.
•It is sometimes misleading and subject to error.
SerJhob

STATS&PROB
𝑀𝑜𝑑𝑒 = 3𝑀𝑑 − 2𝑋
Where: 𝑀𝑑=Median, 𝑋=Mean.
SerJhob

STATS&PROB
•KINDS OF MODE
•Unimodal – if the distribution has one mode.
•Bimodal – if the distribution has two scores that are most
frequently occurring.
•Multimodal – if the distribution has two or more scores that
are equally occurring in terms of frequency (also known as
polymodal).
SerJhob

STATS&PROB
PROBLEM 9
•Find the mode of the data.
1 2 2 1 3 2 3 5
Just by looking at the data, we find the frequency of 2 is 3
and it is more than the frequency of all the other scores.
So, the mode of the data is 2, or modal score is 2.
The kind of mode is unimodal.
SerJhob
Score f
1 Two
2 Three
3 Two
5 One

STATS&PROB
PROBLEM 10
•Below are the scores (out of 20 items) earned by the student
in 8 different examinations:
9 6 12 15 7 19 8 12
Find the modal score
Just by looking at the data, we find the frequency of 12 is 2
and it is more than the frequency of all the other scores.
So, the mode of the data is 12, or modal score is 12.
The kind of the mode is unimodal.
SerJhob
Score f
6 One
7 One
8 One
12 Two
15 One
19 One

STATS&PROB
PROBLEM 11
•Find the mode of the data, given the following scores:
1 1 2 4 2 3 4 3 5 4 5 1
Just by looking at the data, we find the frequencies of 4 and
1 are both 3 and it is more than the frequency of all the
other scores.
So, the mode of the data is 1 and 4.
The kind of mode is bimodal.
SerJhob
Score F
1 Three
2 Two
3 Two
4 Three
5 Two

STATS&PROB
PROBLEM 12
•Find the mode of the data, given the following scores:
10 10 12 11 9 7 5 7 2 1 3 11
Just by looking at the data, we find the frequencies of 10, 11
and 7 have 2 and it is more than the frequency of all the
other scores.
So, the mode of the data is 10, 11 and 2.
The kind of mode is multimodal / polymodal.
SerJhob

STATS&PROB
EXERCISES
Find the modal grades.
•Student A: 89, 90, 91, 89, 89, 98, 85, 89, 93, 94
•Student B: 74, 83, 83, 88, 84, 76, 75, 81, 75, 91
•Student C: 88, 83, 84, 85, 88, 83, 85, 82, 85, 82
•Student D: 89, 92, 91, 85, 87, 95, 96, 90
•Student E: 77, 77, 77, 77, 77, 77, 77, 77, 77
•Student F: 98.5, 87.6, 91.2, 83.2, 95.4, 98.5, 85.9
•Student G: 86.76, 95.43, 86.762, 86.74, 86.75, 86.761
SerJhob
𝑀𝑜: 89
𝑀𝑜: 83 & 75
𝑀𝑜: 85
𝑀𝑜: 𝑛𝑜𝑛𝑒
𝑀𝑜: 77
𝑀𝑜: 98.5
𝑀𝑜: 𝑛𝑜𝑛𝑒

STATS&PROB
PROBLEM 13
SerJhob
Score f Xm fXm
98 – 99 9
96 – 97 9
94 – 95 13
92 – 93 13
90 – 91 15
88 – 89 6
86 – 87 2
TOTAL N = 67
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
98.5
96.5
94.5
92.5
90.5
88.5
86.5
886.5
868.5
1228.5
1202.5
1357.5
531
173
6247.5
Mean =
𝟔𝟐𝟒𝟕.𝟓
𝟔𝟕
= 𝟗𝟑. 𝟐𝟓

STATS&PROB
PROBLEM 13
SerJhob
Score f F
98 – 99 9
96 – 97 9
94 – 95 13
92 – 93 13
90 – 91 15
88 – 89 6
86 – 87 2
TOTAL N = 67
Formula:
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
67
58
49
36
23
8
2
Mode = 𝟗𝟏. 𝟓 +
𝟑𝟑.𝟓−𝟐𝟑
𝟏𝟑
2
𝐧
𝟐
=
𝟔𝟕
𝟐
= 𝟑𝟑. 𝟓
= 𝟗𝟏. 𝟓 +
𝟐𝟏
𝟏𝟑
= 𝟗𝟑. 𝟏𝟐

STATS&PROB
PROBLEM 13
SerJhob
Formula:
Median = 𝟗𝟑. 𝟏𝟐
Mean = 𝟗𝟑. 𝟐𝟓
= 3 93.12 − 2 93.25
= 279.36 − 186.5
= 𝟗𝟐. 𝟖𝟔 𝑜𝑟 𝟗𝟑

STATS&PROB
PROBLEM 14
SerJhob
Score f Xm fXm
45 & above 9
39 – 44 8
33– 38 11
27 – 32 13
21 – 26 10
15 – 20 2
9 – 14 2
TOTAL N = 55
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
47.5
41.5
35.5
29.5
23.5
17.5
11.5
427.5
332
390.5
383.5
235
35
23
1815
Mean =
𝟏𝟖𝟐𝟔.𝟓
𝟓𝟓
= 𝟑𝟑. 𝟐𝟏

STATS&PROB
PROBLEM 14
SerJhob
Score f F
45 & above 9
39 – 44 8
33– 38 11
27 – 32 13
21 – 26 10
15 – 20 2
9 – 14 2
TOTAL N = 55
Formula:
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
55
46
38
27
14
4
2
Mode = 𝟑𝟐. 𝟓 +
𝟐𝟕.𝟓−𝟐𝟕
𝟏𝟏
6
𝐧
𝟐
=
𝟓𝟓
𝟐
= 𝟐𝟕. 𝟓
= 𝟑𝟐. 𝟓 +
𝟑
𝟏𝟏
= 𝟑𝟐. 𝟕𝟕

STATS&PROB
PROBLEM 14
SerJhob
Formula:
Median = 𝟑𝟐. 𝟕𝟕
Mean = 𝟑𝟑. 𝟐𝟏
= 3(32.77) − 2(33.21)
= 98.31 − 66.42
= 𝟑𝟏. 𝟖𝟗 𝑜𝑟 𝟑𝟐

STATS&PROB
SEATWORK 1
The following sales of movie tickets (in million pesos) were recorded during the
15-day showing period nationwide.
51 53 53 53 55
55 56 57 57 57
59 60 61 62 62
68 68 71 73 74
78 80 82 84 87
88 88 90 94 95
95 96 98 100 103
Compute and compare the measures of central tendency for ungrouped and
grouped data with desired number of classes of 7.
SerJhob

STATS&PROB
SEATWORK 1
51 53 53 53 55
55 56 57 57 57
59 60 61 62 62
68 68 71 73 74
78 80 82 84 87
88 88 90 94 95
95 96 98 100 103
• Mean = 2563 / 35 = 73.23
• Median = 71
• Mode = 53 and 57 (bimodal)

STATS&PROB
SEATWORK 1
SerJhob
Score f Xm fXm
92 and above 7 665
85 – 91 4 352
78 – 84 4 324
71 – 77 3 222
64 – 70 2 134
57 – 63 8 480
50 – 56 7
TOTAL 35
Formula:
𝑥 =
𝑓𝑋𝑚
𝑛
95
88
81
74
67
60
53 371
2548
Mean =
𝟐𝟓𝟒𝟖
𝟑𝟓
= 𝟕𝟐. 𝟖

7
STATS&PROB
SEATWORK 1
SerJhob
35
28
Score f Xm
92 and above 7
85 – 91 4
78 – 84 4
71 – 77 3
64 – 70 2
57 – 63 8
50 – 56 7
TOTAL 35
24
20
17
15
Formula:
𝒏
𝟐
− 𝑭
𝒇𝒘
𝒊
Mode = 𝟔𝟑. 𝟓 +
𝟏𝟕.𝟓−𝟏𝟓
𝟐
𝟕
𝐧
𝟐
=
𝟑𝟓
𝟐
= 𝟏𝟕. 𝟓
= 𝟔𝟑. 𝟓 +
𝟏𝟕.𝟓
𝟐
= 𝟕𝟐. 𝟐𝟓

STATS&PROB
SEATWORK 1
SerJhob
Formula:
Median = 𝟕𝟐. 𝟐𝟓
Mean = 𝟕𝟐. 𝟖𝟎
= 3(72.25) − 2(72.80)
= 216.75 − 145.60
= 𝟕𝟏. 𝟏𝟓 𝑜𝑟 𝟕𝟏

STATS&PROB
Assignment
•Write the names of random selected 50 students in your
college and measure their height, prepare a frequency table
and compute the mean, median, and mode (grouped and
ungrouped). The decided number of classes is 10.
SerJhob

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Frequency Table - Elementary Statistics

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