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1 
PART I 
INTRODUCTION TO 
STATISTICS
2 
CHAPTER I INTRODUCTION TO STATISTICS 
Nature 
Statistics is derived from the Latin word status which means “state”. In the word 
statistics, it refers to the actual numbers derived from data and a method of analysis. 
Definition 
Statistics is a branch of mathematics which concerned with the methods for 
collecting, organizing, presenting, analyzing and interpreting quantitative data to aid in 
drawing valid conclusions and in decision making. 
TYPES OF STATISTICS 
Descriptive statistics 
 A type of statistics, which focuses on the collecting, summarizing, and presenting a 
mass of data so as to yield meaningful information. 
Examples: 
1. A math teacher wants to determine the percentage of students who passed the 
examination. 
2. Lula is a bowler wants to find her bowling average for the past 10 games. 
Inferential statistic 
 A type of statistics, that deals with making generalizations and analyzing sample 
data to draw conclusions about a population. This is a process of obtaining 
information about a large group from the study of a smaller group. 
Examples: 
1. Kim is a basketball player wants to estimate his chance of winning the MVP 
award based on his current season averages and averages of his opponents. 
2. A manager would like to predict base on previous year’s sales, the sales 
performance of company for the next six years. 
BASIC STATISTICAL TERMS 
1. Population is consisting of all elements such as events, objects, and individuals whose 
characteristic is being studied. 
Example: 
The researcher would like to determine the number of male BSE students in 
CvSU-Imus campus. 
2. Variable is a characteristic of an item or individual that will be analyzed using statistics. 
Variables are usually denoted by any capital letter. 
3. Sample- is a portion of population selected for study.
3 
4. Parameter- a numerical measure that describes a characteristic of a population. 
5. Statistics- a numerical measure that describes a characteristic of a sample. 
TYPES OF VARIABLE 
Qualitative variable- a variable, that cannot be measured numerically but can be classified 
into different categories. 
Examples: Names, Gender, Hair color, subjects enrolled in a semester, species. 
Quantitative variable- consists of variable that can be measured numerically. 
Examples: price of a house, height, gross sales, numbers of cars own. 
CLASSIFICATION OF QUANTITATIVE VARIABLE 
Discrete Quantitative Variable – a variable results from either a finite number of possible 
values or a countable number of possible values. In other words, a discrete variable can 
assume only certain values no intermediate values. 
Examples: number of patient, number of sold cars, number of book 
Continuous Quantitative Variable- a variable results from infinitely many possible 
values that can be associated with points on a continuous scale in such a way that there are 
no gaps of interruptions. 
Examples: time, height of a person, weight 
LEVELS OF MEASUREMENT 
The level of measurement of data determines the algebraic operations that can be 
performed and th statistical tools that can be applied to the data set. 
LEVEL 1. Nominal is characterized by data that consist of names, labels, or categories 
only. 
Examples: gender, marital status, employment, religion, address, degree program 
LEVEL 2. Ordinal it involves data that may be arranged in some order, but differences 
between data values either cannot be determined or are meaningless. 
The data measured can be ordered or rank. 
Examples: grades of the students, military rank, job position, year level 
LEVEL 3. Interval In this level has precise differences between measures but there is no 
true zero. 
Examples: temperature, IQ score 
LEVEL 4. Ratio is the interval level modified to include the inherent zero starting point. 
For values at this level, differences and ratios are meaningful. 
Examples: weekly allowance, area, and volume
4 
Some Commonly Used Symbols in Statistics 
Σ capital letter sigma denotes the sum of, and summation of 
f small letter f denotes as frequencies 
F capital letter F denotes cummulative frequencies 
n small letter n denotes sample size 
i small lettre i denotes interval 
N capital letter N denotes population size 
X capital letter X denotes independent variable 
Y capital letter Y denotes dependent variable 
푥̅ mean of the sample 
μ capiltal letter m denotes population mean
5 
CHAPTER II PRESENTATION AND GRAPHS 
 Mass or bulk of data that is collected from population sample based on the 
observations, from primary or secondary source is still considered a raw data. 
 Raw data is the data recorded in the sequence in which they are collected and 
before they are processed or ranked. This decision cannot be derive easily if the 
results are not yet organized because it does not give a clear idea or presentation of 
what has gathered 
Three different forms of presentation 
 Textual form of presentation 
- These forms of presentation make use of words, sentences and paragraph 
in presenting data collected. 
For example 
 Attitudes to women with children working vary depending on the stage of the 
youngest child, with support for working outside the home increasing, as the 
children get older. 
 When the youngest child is pre-school age the majority of respondents, (57%) 
thought the woman should stay at home. About a third thought she should work 
part-time, and there was little agreement with her working full-time outside the 
home (2%). 
 Once the youngest child is at school, most people think women should work part-time 
(71%) and there is little support for women not working outside the home at all 
(6%). 
 Once the children have left home, the majority of people think women should be in 
paid work full-time (64%). 
 Tabular form of presentation 
- It shows the data in a more concise and systematic manner. A way of classifying 
related numerical facts in horizontal arrays called lines or rows and vertical arrays called 
columns. The space common to a particular raw and column called cell. 
Example
6 
 Graphical presentation 
-it is the most interesting and most effective means of organizing and presenting 
statistical data. Graphs tell a story with visuals rather than in words or numbers and can 
help readers or students to understand and interpret the findings of the data collected. 
Kinds of Graphs that commonly used in data presentation: 
 Circle or Pie Graph 
-commonly used to represents the relationship of different components of a 
particular data. It is a circular chart divided into sectors, illustrating numerical proportion. 
Pie chart represents the total sample or population. The sectors should be proportional to 
the percentage components of the data in which the total area was 100%. 
Example 
 Line Graph 
– It shows the relationships between two sets of quantities. It was done by plotting 
the points of x along the horizontal axes and y in the vertical axes. Broken lines connect the 
plotted points, at last line segment was finally formed. 
Example 
This line graph shows the midday temperature over a period of 7 days.
7 
You can see at a glance that the temperature was at its highest on Monday and that it 
started to fall in the middle of the week before rising again at the end of the week. 
 Bar Graph 
It is the comparisons of numerical values of a given item over a period. It was composed of 
bars, rectangle or rectangular prism of equal width. It can be drawn vertically or 
horizontally in single or paired bar graphs and it begins with zero 
For example
8 
CHAPTER III FREQUENCY DISTRIBUTION 
Frequency Distribution or Frequency Table is a tabular summary of a set of data that 
shows the total number of times a specific data item occured or fell under one of the 
distinct classes named 
Two Types of data in Frequency Distribution 
1. Grouped Data 
2. Ungrouped Data 
UNGROUPED DATA 
 It names all the data in tabular form and tells how many times the value occured. 
Example 1 : 
Consider the following data obtained when a dice tossed 40 times. 
2 5 4 6 3 6 1 6 
1 2 5 4 6 3 2 1 
5 6 6 1 2 4 4 4 
4 2 1 3 5 6 4 6 
5 2 3 6 3 2 3 1 
Construct a frequency table 
Solution : 
Create a three column table. In the first row, you will put the numbers in the dice. In the 
second row, the tally of each numbers in the dice appeared. In the last column, the 
frequency by which the values occured. 
Number on the Dice Tally Frequency 
1 |||||| 6 
2 ||||||| 7 
3 |||||| 6 
4 ||||||| 7 
5 ||||| 5 
6 ||||||||| 9 
SUM 40
9 
To get the range of the data we have the highest value and the lowest value of the dice, 6 
and 1 respectively. Thus, 
R = Highest Value – Lowest Value 
R = 6 – 1 
R = 5 
GROUPED DATA 
 Data that are organized and arranged in the form of frequency distribution 
Example 2: 
Ms. Hautea teaches math to 70 students. She has a list of the number of days each 
student was absent in her class for the entire year. 
13 8 5 6 3 4 1 1 2 9 
2 9 10 11 18 20 16 15 17 9 
4 2 1 8 7 7 2 19 11 10 
5 8 6 4 7 19 22 5 10 13 
14 11 14 14 2 3 8 9 9 11 
26 5 15 12 1 6 5 9 6 15 
2 5 6 8 7 7 6 6 1 1 
( Construct a Frequency Table) 
Solution: 
STEP 1: Look for the range of the data. 
Range = Highest Value – Lowest Value 
R = 26 – 1 
R = 25 
STEP 2: Look for the classes 
Approximate No. Of Classes = [(2)(Total number of observation)]0.3333 
No. Of Classes = [(2)(70)]0.3333 
= (140)0.3333 
= 5.19 ≈ 5 
STEP 3: Determine the class interval and class limits 
Approximate Class Interval = _ _Rannge_ _____ 
Number of Classes 
Class Interval = 25_ 
5 
= 5
10 
Class Limits is the smallest and the largest values that would be placed in a given. 
The lower limit is the lowest possible value in the class while the upper limit is the 
highest possible value in the class. Where in 1, 6, 11, 16, 21, and 26 are the lower limits 
while 5, 10, 15, 20, 25, and 30 are the upper limit. 
CLASS 
1 – 5 
6 – 10 
11 – 15 
16 – 20 
21 – 25 
26 – 30 
STEP 4: Find the number of observations (frequency) in each class 
Number of Absenses ( CLASS ) Frequency 
1 – 5 23 
6 – 10 25 
11 – 15 14 
16 – 20 6 
21 – 25 1 
26 – 30 1 
SUM 70 
STEP 5: Determine the class boundaries and the class mark 
Upper True Class Boundaries ( UTCB ) = Upper Limit + 0.5 
Lower True Class Boundaries (LTCB ) = Lower Limit – 0.5 
Class Mark = Upper Limit + Lower Limit 
2 
Number of 
Absenses 
( CLASS ) 
Frequency 
Class Boundaries 
LTCB - UTCB 
Class Mark 
( x ) 
1 – 5 23 0.5 5.5 3 
6 – 10 25 5.5 10.5 8 
11 – 15 14 10.5 15.5 13 
16 – 20 6 15.5 20.5 18 
21 – 25 1 20.5 25.5 23 
26 – 30 1 25.5 30.5 28 
SUM 70
11 
STEP 6: Cumulative Frequency Distribution 
Relative Cummulative Frequency = Frequency ( number of observations/occurences 
Total Number of Frequency 
Number of 
Absenses 
( CLASS ) 
Frequency 
Less than 
Cummulative 
Frequency 
( <CF ) 
Relative 
Cummulative 
Frequency 
Percentage 
Cummulative 
Frequency 
1 – 5 23 23 0.33 33% 
6 – 10 25 48 0.36 36% 
11 – 15 14 62 0.2 20% 
16 – 20 6 68 0.09 9% 
21 – 25 1 69 0.01 1% 
26 – 30 1 70 0.01 1% 
SUM 70 100%
12 
CHAPTER IV MEASURES OF CENTRAL TENDENCY 
Measures of Central Tendency 
 Is a single number that describes a set of data 
 It is also called measure of central location 
 The most commonly used measures of central tendency are mean, median, and 
mode. It can be computed depends if it is ungrouped and grouped data. 
Ungrouped Data 
-data that are not yet organized 
Grouped Data 
-data that are organized and arranged in the form of frequency distribution 
Summation Notation 
The following are the weights of 5 people (in kilograms): 
40, 45, 30, 50, 55 
We can use symbols such as X1=40, X2=45, and so on, where x stands for the weight and 
the subscript for person it represents. We can write it as: 
X1+X2+X3+X4+X5 
We can use summation notation symbol Σ to write the sum. 
5 
Σ 푋푖 
푖= 1 
Where i indicates the subscript of the first term in the summation and the number on top, 
5, indicates the subscript of the last term of summation. 
Example 1: 
If X1=4, X2=5, X3=9, X4=6, X5=7, find 
5 
Σ 푋푖 
푖= 1 
Solution: 
5 
Σ 푋푖 
푖= 1 
=X1+X2+X3+X4+X5 
=4+5+9+6+7 
=31
13 
Example 2: 
If X1=13, X2=15, X3=19, find 
3 
Σ(푋푖 − 1) 
푖=1 
Solution: 
3 
Σ(푋푖 − 1) 
푖= 1 
=(X1-1) + (X2-1) + (X3-1) 
= (13-1) + (15-1) + (19-1) 
=12+14+18 
=44 
Rules on Summation Notation 
Rule No.1: The summation notation is distributive over addition. 
푛 
Σ(푋푖 + 푌푖) = 
푖 =1 
푛 
Σ 푋푖 + Σ 푌푖 
푖=1 
푛 
푖=1 
Rule No.2: If c is a constant, then 
푛 
Σ 퐶푋푖 = 푐 Σ 푋푖 
푖 =1 
푛 
푖=1 
Rule No. 3: If c is constant then 
푛 
Σ 푐 = 푛푐 
푖 =1 
Examples: 
Use the rules on summation to write out the expansion of the given expression. 
1. 
3 
Σ(2푋푖 + 3) 
푖= 1 
= 
3 
Σ 2푋푖 + Σ 3 
푖=1 
3 
푖= 1 
Rule No.1 
= 
3 
Σ 2푋푖 + 3(3) 
푖= 1 
Rule No.3 
=
14 
3 
Σ 2푋푖 + 9 
푖= 1 
Rule No.2 
=2(X1+X2+X3) +9 
2. 
5 
Σ 4푋푖 
푖= 1 
5 
= 4 Σ 푋푖 
푖= 1 
= 4(푋1 + 푋2 + 푋3 + 푋4 + 푋5) 
Three Measures of Central Tendency 
A. Arithmetic Mean or Mean 
-referred as the average 
1. Ungrouped Data 
The mean is the sum of the values of variable divided by the number of 
observations. 
Formula: 
푛 
푥 = Σ 푋푖 
푖=1 
n 
where: 
x=sample mean 
n=total number of items in the sample 
Xi=the ith observed value 
Σ=summation notation 
Example 1: 
Find the mean score of 7 students whose quiz scores are 85, 90, 87, 83, 80, 75, and 85. 
Solution: 
x=85+90+87+83+80+75+85 = 585 = 83.57 
7 7 
Example 2: 
What is the mean age of a group of 5 students whose ages are 12, 15, 19, 20, and 18? 
Solution: 
x=12+15+19+20+18 = 84 = 16.8 
5 5
15 
2. Grouped Data 
Formula: 
푥 = Σ 푓푖푋푖 
n 
where: n=total number of observations 
Xi=the class mark of the ith class interval 
fi=the frequency of the ith interval 
Example 1: 
Find the mean score of 50 students in periodic exam in Math. 
Class Interval Frequency of 
Scores(f) 
Class Mark (x) fx 
95-99 3 97 291 
90-94 10 92 920 
85-89 15 87 1305 
80-84 10 82 820 
75-79 5 77 385 
70-74 5 72 360 
65-69 2 67 134 
n=50 Σfx=4215 
Solution: 
We can find the class mark by adding the class interval divided by two. The fx is simply 
multiplying the frequency by the class mark. We can now find the summation of fx by 
adding all of the values. Then, substitute to the formula: 
푥 = Σ 푓푖푋푖 =4215 =84.3 
n 50 
Thus, the mean score of 50 students in periodic exam is 84.3. 
B. Median 
-middlemost number that divides the values of observations into halves 
-it is referred as counting average 
1. Ungrouped Data 
To find the median, we must arrange the data in decreasing or increasing of 
magnitude. Then, find the middle value if it is odd and arithmetic average if it is even. 
Example 1: 
Find the median of the following test scores: 25, 16, 12, 23, and 18 
Solution: 
Arrange the data: 12, 16, 18, 23,25 
Then, obviously it has 5 scores so the middle of 5 is 3. Thus, the median of the set of 
scores is 18.
16 
Example 2 
Find the median of the following set of scores in Math: 
4, 6, 9, 7, 3, 10 
Solution: 
Arrange the data: 3, 4, 6, 7, 9, 10 
Then, the middle of the scores are 6 and 7 so we can use the arithmetic average to find 
the median. Thus, 6+7 = 6.5. The median is 6.5. 
2 
2. Grouped Data 
Formula: 
Md=Ll + (푛 
2 − 푐푓푏) ci 
f 
where: Ll=lower limit of assumed median determine by expression n/2 
cfb=cumulative frequency just below the assumed median 
f= corresponding frequency 
ci= class interval 
Example 1: 
Compute for the median using grouped data 
Class interval f 
95-99 1 
90-94 2 
85-89 5 
80-84 8 
75-79 22 
70-74 12 
65-69 6 
60-64 4 
n=60 
Solution: 
Class interval F Class Boundary <cf 
95-99 1 94.5-99.5 60 
90-94 2 89.5-94.5 59 
85-89 5 84.5-89.5 57 
80-84 8 79.5-84.5 52 
75-79 22 74.5-79.5 44 
70-74 12 69.5-74.5 22 
65-69 6 64.5-69.5 10 
60-64 4 59.5-64.5 4
17 
n=60 
Solution: 
To obtain the median class: 
First, solve for n = 60 = 30th 
2 2 
Then, we can now locate where 30th item is equal or nearest but not greater than the value 
in the less than cumulative frequency (<cf) distribution. 
The median class is the 75-79 class interval. We can now substitute to the formula. 
Md=Ll+ (푛 
2 − 푐푓푏) ci 
f 
=74.5+ (60 
2 − 22) 5 
22 
=74.5+(30 − 22) 5 
22 
=74.5+1.82 
Md =76.32 
Thus, the median is 76.32. 
C. Mode 
-is the number that occurs most frequently 
 It is possible to have more than one mode. If there are two modes, it is 
called bimodal. If three, it is called trimodal. 
1. Ungrouped Data 
Example 1: 
Find the mode of the following set of items. 3, 7, 9, 8, 7, and 2 
Solution: 
The number that occurs most frequent is 7. In the 5 numbers, 7 occur twice so it is 
the mode. It is uni-modal. 
Example No.2 
Determine the mode of the following: 15, 20, 10, 20, 10, and 25 
Solution: 
The numbers that occur most frequent are 10 and 20. It is bimodal.
18 
2. Grouped Data 
Mode = x = Xlb+ 푑1 c 
d1+d2 
where: Xlb=lower class boundary 
d1=difference between the frequency of the modal class and the frequency of the 
class next lower in value 
d2= difference between the frequency of the modal class and the frequency of the 
class next higher in value 
c=class interval 
Example 1: 
Classes Frequency(f) Less than cumulative 
frequency (<cf) 
1-8 4 4 
9-16 5 9 
17-24 12 21 
25-32 15 36 
33-40 9 45 
41-48 10 55 
n=55 
Solution: 
The modal class is the 25-32, since it is the class interval with highest frequency. We can 
substitute to the formula: 
x^=Xlb+ 푑1 푐 
d1+d2 
=24.5+ 15-12 8 
(15-12)+(15-9) 
=24.5+ 3 8 
3+6 
=24.5+2.67 
=27.17 
Thus, the mode is 27.17.
19 
CHAPTER V Measures of Location or Fractiles 
Measures of Location or fractiles are values below which a specified fraction or 
percentage of observations in given set must fall. 
A. Percentiles 
-are values that divide a set of observations into 100 equal parts 
P1, read as first percentile, is the value which 1% of the values fall 
P99, is the value below which 99% of the values fall 
To compute for the ith percentile: 
Pi=the value of the i(n+1) th observation in the array 
100 
Example: 
The following were the scores of 8 students in a quiz: 
4, 3, 5, 6, 8, 6, 7, 9 
Solution: 
First, we need to arrange the data from lowest to highest. That is, 
3, 4, 5, 6, 6, 7, 8, 9 
Then, substitute 
P70= 70(8+1) th observation=6.3 or the 7th observation 
100 
Therefore, the 70th percentile is 8, which interpreted as: 70% of the scores are below 8. 
Approximating the ith Percentile from a Frequency Distribution 
Formula: 
Pi=LCBpi+c ( 푖푛 
100 −< 푐푓푝푖 − 1) 
fpi 
where: 
The Pith class where the <cf is equal to, or exceeds for the first time, in 
100 
LCBpi=the lower class boundary of the pith class 
c =class size of the Pith class 
fpi =frequency of the Pith class 
<cfpi-1=less than cumulative frequency of the class preceding the Pith class
20 
Example: (Refer to the example on the scores of 100 students in an achievement test). 
Find the 40th percentile. 
Score Frequency(f) <cf 
60-64 10 10 
65-69 8 18 
70-74 12 30 
75-79 25 55 
80-84 20 75 
85-89 25 100 
Total n=100 
Solution: 
P40 
th class is the class containing 40x100 th observation which is the 40th observation. 
100 
Thus, from the above FDT it can be found at the 75-79 class interval. We now substitute 
to the formula. 
P40=74.5+5 40-30 
25 
=74.5+5(0.4) 
=74.5+2 
=76.5 
Therefore, forty percent of the scores in the achievement test is below 76.5. 
B. Deciles 
- are values that divide the array into 10 equal parts. 
D1, read as first decile, is the value below which 10% of the values fall 
D9, read as ninth decile, is the value below which 20% of the values fall 
To compute for the ith decile: 
Di= the value of the i(n+1) th observation 
100
21 
Example: 
From the given set scores in a quiz find the 5th Decile or D5. 
5, 3, 7, 6, 4, 8, 9 
Solution: 
First, arrange the data from lowest to highest. 
D5= 5(7+1) th = 40 observation= 4th observation 
10 10 
Therefore, the 5th decile is 6, this implies that: 50% of the scores in the quiz are below 6. 
Approximating the ith decile from a Frequency Distribution 
Formula to be used: 
D1= LCBdi+c ( 푖푛 
100 −< 푐푓푑푖 − 1) 
fDi 
where: 
The Dith class is the class where the <cf is equal to, or exceeds for the first time, in . 
10 
LCBDi=the lower class boundary of the Dith class 
c =class size of the Dith class 
fDi =frequency of the Dith class 
<cfDi-1=less than cumulative frequency of the class preceding the Dith class 
Example: (Refer to the example on the scores of 100 students in an achievement test) 
Find the 8th Decile. 
Score Frequency(f) <cf 
60-64 10 10 
65-69 8 18 
70-74 12 30 
75-79 25 55 
80-84 20 75 
85-89 25 100 
Total n=100 
Solution: 
D8th class is the class containing 8x100 th observation which is the 80th observation 
10 
The said observation can be found at the 85-89 class interval. 
D8=84.5+5 80-75 
25 
=84.5+5(0.2) 
=84.5+1 
=85.5
22 
Thus, 80% of the scores are below 85.5. 
C. Quartiles 
-are values that divide the array into 4 equal parts. 
Q1, read as first quartile, is the value below which 25% of the values fall 
Q2, is the value below which 50% of the values fall 
Q3, is the value below which 75% of the value fall 
Example: From the given set scores in a quiz. Find the 2nd percentile. 
2, 5, 7, 8, 9 
Solution: 
We need to use the formula: i(n+1) th .That is, 
100 
Q2= 2(5+1) th 12 = observation = 3rd observation 
4 4 
Therefore, the 2nd quartile is 7, this implies that: 50% of the scores in the quiz are below 
7. 
Approximating the ith Quartile from a Frequency Distribution 
Formula to be used: 
Qi= LCBQi + c (푖푛 
4 −< 푐푓푄푖 − 1) 
fQi 
where: 
The Qith class is the class where the <cf is equal to, or exceeds for the first time, in 
4 
LCBQi=the lower class boundary of the Qith class 
c =class size of the Qith class 
fQi =frequency of the Qith class 
<cfQi-1=less than cumulative frequency of the class preceding the Qith class 
Example: (Refer to the example on the scores of 100 students in an achievement test) 
Find the 3rd Quartile. 
Score Frequency(f) <cf 
60-64 10 10 
65-69 8 18 
70-74 12 30 
75-79 25 55 
80-84 20 75 
85-89 25 100 
Total n=100 
Solution: 
The third quartile class is the class containing 3x100 th observation which is the 75th 
observation. 4
23 
The said observation can be found at the 80-84 class interval. 
We will use the formula, that is, 
Q3=79.5+5 75-55 
20 
=79.5+5(1) 
=79.5+5 
=84.5 
Thus, 75% of the scores are below 84.5.
24 
ASSESSMENT 
TEST I 
A. Write the following expressions in summation notation. 
1. X7+X8+X9+X10 
2. (X3-2) + (X4-2) + (X5-2) + (X6-2) 
3. 2X1+2X2+2X3 
4. (X1+4) + (X2+4) + (X3+4) 
5. 12+22+32+42 
B. Write the following without summation signs and simplify if possible. 
5 
1. Σ 6 
푖=3 
10 
2. Σ 푖 
푖=4 
3 
3. Σ 2푋푖 
푖=1 
4 
4. Σ 푋푖푌푖 
푖=1 
3 
5. Σ 푋푖 + 
푖=1 
2 
Σ(푌푖 − 2) 
푖=1 
C. Solve for the mean of the given ungrouped data. 
1. The scores in the long quiz of the 5 students are as follows: 88, 90, 85, 93 and 87. 
2. The weight of the ten students (in kilogram) is: 40, 45, 49, 38, 52, 60, 70, 59, 47, and 
48. 
3. The grades of Maria in 5 subjects are as follows: 89, 90, 93, 87, and 92. 
4. The 4 students have a grade of 95, 87, 74, and 83 in their Math subject. 
5. The age of the 6 college students are: 21, 24, 19, 18, 17 and 25. 
D. Solve for the mean of grouped data. 
Class Interval Frequency (f) 
48-50 12 
45-47 9 
42-44 5 
39-41 6 
36-38 2 
33-35 1
25 
n=35 
E. Find the median of the following test scores. 
1. 15, 13, 19, 14, 20 
2. 5, 6, 11, 19, 20, 25, 30, 40 
3. 38, 28, 54, 44, 17, 20 
4. 8, 11, 17, 19, 21, 27, 30 
5. 2, 7, 6, 4, 5 
F. Compute for the median using grouped data. 
Class interval f 
46-50 7 
41-45 18 
36-40 6 
31-35 12 
26-30 5 
21-25 2 
n=50 
G. Find for the mode of the given ungrouped data. 
1.6, 8, 6, 9, 7, 10, 11, 6 
2. 4, 7, 21, 10, 9, 7, 10 
3. 3, 9, 9, 4, 3, 7, 4, 8, 10 
4. 25, 27, 18, 40, 27, 27 
5. 2, 1, 9, 7, 9, 3 
H. Solve for the mode of the given grouped data. 
Classes Frequency(f) Less than cumulative 
frequency (<cf) 
6-10 8 8 
11-15 11 19 
16-20 14 33 
21-25 7 40 
26-30 5 45 
31-35 5 50 
n=50
26 
I. Solve the following. 
The following were the scores of 6 students in a quiz: 
6, 5, 10, 8, 3, 4 
1. Find the 40th percentile 
2. Find the 6th decile 
3. Find the 2nd quartile 
4. Find the 60th percentile 
5. Find the 3rd quartile 
J. Solve the following. 
Find the 2nd Quartile, 6th Decile and 80th Percentile of the average weight of 30 students. 
Weight (in kilograms) No. of students 
76-80 2 
71-75 1 
66-70 5 
61-65 10 
56-60 6 
51-55 6 
n=30 
TEST II 
Direction: Solve the following measure of dispersion given below. 
1.) Find the range of each ungrouped set of data. 
a. 21, 35, 45, 25, 31, 54, 47, 39, 40, 28 ____________ 
b. Jenny took 10 Abstract Algebra tests for the first semester. What is the range of her 
test scores if her scores given are 85, 80, 75, 90, 95, 79, 87, 73, 83, and 92 as the results?
27 
2.) Find the range of grouped data in the given set below. 
Class Intervals Frequency Class Boundaries Class Mark 
5 – 9 
10 – 14 
15 – 19 
20 – 24 
25 – 29 
30 – 34 
35 – 39 
2 
4 
6 
8 
6 
4 
2 
4.5 - 9.5 
9.5 - 14.5 
14.5 - 19.5 
19.5 - 24.5 
24.5 - 29.5 
29.5 - 34.5 
34.5 - 39.5 
7 
12 
17 
22 
27 
32 
37 
3.) Fill in the table below. Solve for the Quartile Deviation, Average and Standard 
Deviation of grouped data. 
Classes 
5 – 9 
10 – 14 
15 – 19 
20 – 24 
25 – 29 
30 – 34 
35 – 39 
40 – 44 
45 – 49 
50 – 54 
Frequency 
3 
8 
12 
11 
10 
6 
15 
8 
9 
18 
n = 100 
X 
7 
12 
17 
22 
27 
32 
37 
42 
47 
52 
fx 
x - ẋ 
|x - ẋ| x - ẋ2 
f|x - ẋ| <cf

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Elementary Statistics

  • 1. 1 PART I INTRODUCTION TO STATISTICS
  • 2. 2 CHAPTER I INTRODUCTION TO STATISTICS Nature Statistics is derived from the Latin word status which means “state”. In the word statistics, it refers to the actual numbers derived from data and a method of analysis. Definition Statistics is a branch of mathematics which concerned with the methods for collecting, organizing, presenting, analyzing and interpreting quantitative data to aid in drawing valid conclusions and in decision making. TYPES OF STATISTICS Descriptive statistics  A type of statistics, which focuses on the collecting, summarizing, and presenting a mass of data so as to yield meaningful information. Examples: 1. A math teacher wants to determine the percentage of students who passed the examination. 2. Lula is a bowler wants to find her bowling average for the past 10 games. Inferential statistic  A type of statistics, that deals with making generalizations and analyzing sample data to draw conclusions about a population. This is a process of obtaining information about a large group from the study of a smaller group. Examples: 1. Kim is a basketball player wants to estimate his chance of winning the MVP award based on his current season averages and averages of his opponents. 2. A manager would like to predict base on previous year’s sales, the sales performance of company for the next six years. BASIC STATISTICAL TERMS 1. Population is consisting of all elements such as events, objects, and individuals whose characteristic is being studied. Example: The researcher would like to determine the number of male BSE students in CvSU-Imus campus. 2. Variable is a characteristic of an item or individual that will be analyzed using statistics. Variables are usually denoted by any capital letter. 3. Sample- is a portion of population selected for study.
  • 3. 3 4. Parameter- a numerical measure that describes a characteristic of a population. 5. Statistics- a numerical measure that describes a characteristic of a sample. TYPES OF VARIABLE Qualitative variable- a variable, that cannot be measured numerically but can be classified into different categories. Examples: Names, Gender, Hair color, subjects enrolled in a semester, species. Quantitative variable- consists of variable that can be measured numerically. Examples: price of a house, height, gross sales, numbers of cars own. CLASSIFICATION OF QUANTITATIVE VARIABLE Discrete Quantitative Variable – a variable results from either a finite number of possible values or a countable number of possible values. In other words, a discrete variable can assume only certain values no intermediate values. Examples: number of patient, number of sold cars, number of book Continuous Quantitative Variable- a variable results from infinitely many possible values that can be associated with points on a continuous scale in such a way that there are no gaps of interruptions. Examples: time, height of a person, weight LEVELS OF MEASUREMENT The level of measurement of data determines the algebraic operations that can be performed and th statistical tools that can be applied to the data set. LEVEL 1. Nominal is characterized by data that consist of names, labels, or categories only. Examples: gender, marital status, employment, religion, address, degree program LEVEL 2. Ordinal it involves data that may be arranged in some order, but differences between data values either cannot be determined or are meaningless. The data measured can be ordered or rank. Examples: grades of the students, military rank, job position, year level LEVEL 3. Interval In this level has precise differences between measures but there is no true zero. Examples: temperature, IQ score LEVEL 4. Ratio is the interval level modified to include the inherent zero starting point. For values at this level, differences and ratios are meaningful. Examples: weekly allowance, area, and volume
  • 4. 4 Some Commonly Used Symbols in Statistics Σ capital letter sigma denotes the sum of, and summation of f small letter f denotes as frequencies F capital letter F denotes cummulative frequencies n small letter n denotes sample size i small lettre i denotes interval N capital letter N denotes population size X capital letter X denotes independent variable Y capital letter Y denotes dependent variable 푥̅ mean of the sample μ capiltal letter m denotes population mean
  • 5. 5 CHAPTER II PRESENTATION AND GRAPHS  Mass or bulk of data that is collected from population sample based on the observations, from primary or secondary source is still considered a raw data.  Raw data is the data recorded in the sequence in which they are collected and before they are processed or ranked. This decision cannot be derive easily if the results are not yet organized because it does not give a clear idea or presentation of what has gathered Three different forms of presentation  Textual form of presentation - These forms of presentation make use of words, sentences and paragraph in presenting data collected. For example  Attitudes to women with children working vary depending on the stage of the youngest child, with support for working outside the home increasing, as the children get older.  When the youngest child is pre-school age the majority of respondents, (57%) thought the woman should stay at home. About a third thought she should work part-time, and there was little agreement with her working full-time outside the home (2%).  Once the youngest child is at school, most people think women should work part-time (71%) and there is little support for women not working outside the home at all (6%).  Once the children have left home, the majority of people think women should be in paid work full-time (64%).  Tabular form of presentation - It shows the data in a more concise and systematic manner. A way of classifying related numerical facts in horizontal arrays called lines or rows and vertical arrays called columns. The space common to a particular raw and column called cell. Example
  • 6. 6  Graphical presentation -it is the most interesting and most effective means of organizing and presenting statistical data. Graphs tell a story with visuals rather than in words or numbers and can help readers or students to understand and interpret the findings of the data collected. Kinds of Graphs that commonly used in data presentation:  Circle or Pie Graph -commonly used to represents the relationship of different components of a particular data. It is a circular chart divided into sectors, illustrating numerical proportion. Pie chart represents the total sample or population. The sectors should be proportional to the percentage components of the data in which the total area was 100%. Example  Line Graph – It shows the relationships between two sets of quantities. It was done by plotting the points of x along the horizontal axes and y in the vertical axes. Broken lines connect the plotted points, at last line segment was finally formed. Example This line graph shows the midday temperature over a period of 7 days.
  • 7. 7 You can see at a glance that the temperature was at its highest on Monday and that it started to fall in the middle of the week before rising again at the end of the week.  Bar Graph It is the comparisons of numerical values of a given item over a period. It was composed of bars, rectangle or rectangular prism of equal width. It can be drawn vertically or horizontally in single or paired bar graphs and it begins with zero For example
  • 8. 8 CHAPTER III FREQUENCY DISTRIBUTION Frequency Distribution or Frequency Table is a tabular summary of a set of data that shows the total number of times a specific data item occured or fell under one of the distinct classes named Two Types of data in Frequency Distribution 1. Grouped Data 2. Ungrouped Data UNGROUPED DATA  It names all the data in tabular form and tells how many times the value occured. Example 1 : Consider the following data obtained when a dice tossed 40 times. 2 5 4 6 3 6 1 6 1 2 5 4 6 3 2 1 5 6 6 1 2 4 4 4 4 2 1 3 5 6 4 6 5 2 3 6 3 2 3 1 Construct a frequency table Solution : Create a three column table. In the first row, you will put the numbers in the dice. In the second row, the tally of each numbers in the dice appeared. In the last column, the frequency by which the values occured. Number on the Dice Tally Frequency 1 |||||| 6 2 ||||||| 7 3 |||||| 6 4 ||||||| 7 5 ||||| 5 6 ||||||||| 9 SUM 40
  • 9. 9 To get the range of the data we have the highest value and the lowest value of the dice, 6 and 1 respectively. Thus, R = Highest Value – Lowest Value R = 6 – 1 R = 5 GROUPED DATA  Data that are organized and arranged in the form of frequency distribution Example 2: Ms. Hautea teaches math to 70 students. She has a list of the number of days each student was absent in her class for the entire year. 13 8 5 6 3 4 1 1 2 9 2 9 10 11 18 20 16 15 17 9 4 2 1 8 7 7 2 19 11 10 5 8 6 4 7 19 22 5 10 13 14 11 14 14 2 3 8 9 9 11 26 5 15 12 1 6 5 9 6 15 2 5 6 8 7 7 6 6 1 1 ( Construct a Frequency Table) Solution: STEP 1: Look for the range of the data. Range = Highest Value – Lowest Value R = 26 – 1 R = 25 STEP 2: Look for the classes Approximate No. Of Classes = [(2)(Total number of observation)]0.3333 No. Of Classes = [(2)(70)]0.3333 = (140)0.3333 = 5.19 ≈ 5 STEP 3: Determine the class interval and class limits Approximate Class Interval = _ _Rannge_ _____ Number of Classes Class Interval = 25_ 5 = 5
  • 10. 10 Class Limits is the smallest and the largest values that would be placed in a given. The lower limit is the lowest possible value in the class while the upper limit is the highest possible value in the class. Where in 1, 6, 11, 16, 21, and 26 are the lower limits while 5, 10, 15, 20, 25, and 30 are the upper limit. CLASS 1 – 5 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 STEP 4: Find the number of observations (frequency) in each class Number of Absenses ( CLASS ) Frequency 1 – 5 23 6 – 10 25 11 – 15 14 16 – 20 6 21 – 25 1 26 – 30 1 SUM 70 STEP 5: Determine the class boundaries and the class mark Upper True Class Boundaries ( UTCB ) = Upper Limit + 0.5 Lower True Class Boundaries (LTCB ) = Lower Limit – 0.5 Class Mark = Upper Limit + Lower Limit 2 Number of Absenses ( CLASS ) Frequency Class Boundaries LTCB - UTCB Class Mark ( x ) 1 – 5 23 0.5 5.5 3 6 – 10 25 5.5 10.5 8 11 – 15 14 10.5 15.5 13 16 – 20 6 15.5 20.5 18 21 – 25 1 20.5 25.5 23 26 – 30 1 25.5 30.5 28 SUM 70
  • 11. 11 STEP 6: Cumulative Frequency Distribution Relative Cummulative Frequency = Frequency ( number of observations/occurences Total Number of Frequency Number of Absenses ( CLASS ) Frequency Less than Cummulative Frequency ( <CF ) Relative Cummulative Frequency Percentage Cummulative Frequency 1 – 5 23 23 0.33 33% 6 – 10 25 48 0.36 36% 11 – 15 14 62 0.2 20% 16 – 20 6 68 0.09 9% 21 – 25 1 69 0.01 1% 26 – 30 1 70 0.01 1% SUM 70 100%
  • 12. 12 CHAPTER IV MEASURES OF CENTRAL TENDENCY Measures of Central Tendency  Is a single number that describes a set of data  It is also called measure of central location  The most commonly used measures of central tendency are mean, median, and mode. It can be computed depends if it is ungrouped and grouped data. Ungrouped Data -data that are not yet organized Grouped Data -data that are organized and arranged in the form of frequency distribution Summation Notation The following are the weights of 5 people (in kilograms): 40, 45, 30, 50, 55 We can use symbols such as X1=40, X2=45, and so on, where x stands for the weight and the subscript for person it represents. We can write it as: X1+X2+X3+X4+X5 We can use summation notation symbol Σ to write the sum. 5 Σ 푋푖 푖= 1 Where i indicates the subscript of the first term in the summation and the number on top, 5, indicates the subscript of the last term of summation. Example 1: If X1=4, X2=5, X3=9, X4=6, X5=7, find 5 Σ 푋푖 푖= 1 Solution: 5 Σ 푋푖 푖= 1 =X1+X2+X3+X4+X5 =4+5+9+6+7 =31
  • 13. 13 Example 2: If X1=13, X2=15, X3=19, find 3 Σ(푋푖 − 1) 푖=1 Solution: 3 Σ(푋푖 − 1) 푖= 1 =(X1-1) + (X2-1) + (X3-1) = (13-1) + (15-1) + (19-1) =12+14+18 =44 Rules on Summation Notation Rule No.1: The summation notation is distributive over addition. 푛 Σ(푋푖 + 푌푖) = 푖 =1 푛 Σ 푋푖 + Σ 푌푖 푖=1 푛 푖=1 Rule No.2: If c is a constant, then 푛 Σ 퐶푋푖 = 푐 Σ 푋푖 푖 =1 푛 푖=1 Rule No. 3: If c is constant then 푛 Σ 푐 = 푛푐 푖 =1 Examples: Use the rules on summation to write out the expansion of the given expression. 1. 3 Σ(2푋푖 + 3) 푖= 1 = 3 Σ 2푋푖 + Σ 3 푖=1 3 푖= 1 Rule No.1 = 3 Σ 2푋푖 + 3(3) 푖= 1 Rule No.3 =
  • 14. 14 3 Σ 2푋푖 + 9 푖= 1 Rule No.2 =2(X1+X2+X3) +9 2. 5 Σ 4푋푖 푖= 1 5 = 4 Σ 푋푖 푖= 1 = 4(푋1 + 푋2 + 푋3 + 푋4 + 푋5) Three Measures of Central Tendency A. Arithmetic Mean or Mean -referred as the average 1. Ungrouped Data The mean is the sum of the values of variable divided by the number of observations. Formula: 푛 푥 = Σ 푋푖 푖=1 n where: x=sample mean n=total number of items in the sample Xi=the ith observed value Σ=summation notation Example 1: Find the mean score of 7 students whose quiz scores are 85, 90, 87, 83, 80, 75, and 85. Solution: x=85+90+87+83+80+75+85 = 585 = 83.57 7 7 Example 2: What is the mean age of a group of 5 students whose ages are 12, 15, 19, 20, and 18? Solution: x=12+15+19+20+18 = 84 = 16.8 5 5
  • 15. 15 2. Grouped Data Formula: 푥 = Σ 푓푖푋푖 n where: n=total number of observations Xi=the class mark of the ith class interval fi=the frequency of the ith interval Example 1: Find the mean score of 50 students in periodic exam in Math. Class Interval Frequency of Scores(f) Class Mark (x) fx 95-99 3 97 291 90-94 10 92 920 85-89 15 87 1305 80-84 10 82 820 75-79 5 77 385 70-74 5 72 360 65-69 2 67 134 n=50 Σfx=4215 Solution: We can find the class mark by adding the class interval divided by two. The fx is simply multiplying the frequency by the class mark. We can now find the summation of fx by adding all of the values. Then, substitute to the formula: 푥 = Σ 푓푖푋푖 =4215 =84.3 n 50 Thus, the mean score of 50 students in periodic exam is 84.3. B. Median -middlemost number that divides the values of observations into halves -it is referred as counting average 1. Ungrouped Data To find the median, we must arrange the data in decreasing or increasing of magnitude. Then, find the middle value if it is odd and arithmetic average if it is even. Example 1: Find the median of the following test scores: 25, 16, 12, 23, and 18 Solution: Arrange the data: 12, 16, 18, 23,25 Then, obviously it has 5 scores so the middle of 5 is 3. Thus, the median of the set of scores is 18.
  • 16. 16 Example 2 Find the median of the following set of scores in Math: 4, 6, 9, 7, 3, 10 Solution: Arrange the data: 3, 4, 6, 7, 9, 10 Then, the middle of the scores are 6 and 7 so we can use the arithmetic average to find the median. Thus, 6+7 = 6.5. The median is 6.5. 2 2. Grouped Data Formula: Md=Ll + (푛 2 − 푐푓푏) ci f where: Ll=lower limit of assumed median determine by expression n/2 cfb=cumulative frequency just below the assumed median f= corresponding frequency ci= class interval Example 1: Compute for the median using grouped data Class interval f 95-99 1 90-94 2 85-89 5 80-84 8 75-79 22 70-74 12 65-69 6 60-64 4 n=60 Solution: Class interval F Class Boundary <cf 95-99 1 94.5-99.5 60 90-94 2 89.5-94.5 59 85-89 5 84.5-89.5 57 80-84 8 79.5-84.5 52 75-79 22 74.5-79.5 44 70-74 12 69.5-74.5 22 65-69 6 64.5-69.5 10 60-64 4 59.5-64.5 4
  • 17. 17 n=60 Solution: To obtain the median class: First, solve for n = 60 = 30th 2 2 Then, we can now locate where 30th item is equal or nearest but not greater than the value in the less than cumulative frequency (<cf) distribution. The median class is the 75-79 class interval. We can now substitute to the formula. Md=Ll+ (푛 2 − 푐푓푏) ci f =74.5+ (60 2 − 22) 5 22 =74.5+(30 − 22) 5 22 =74.5+1.82 Md =76.32 Thus, the median is 76.32. C. Mode -is the number that occurs most frequently  It is possible to have more than one mode. If there are two modes, it is called bimodal. If three, it is called trimodal. 1. Ungrouped Data Example 1: Find the mode of the following set of items. 3, 7, 9, 8, 7, and 2 Solution: The number that occurs most frequent is 7. In the 5 numbers, 7 occur twice so it is the mode. It is uni-modal. Example No.2 Determine the mode of the following: 15, 20, 10, 20, 10, and 25 Solution: The numbers that occur most frequent are 10 and 20. It is bimodal.
  • 18. 18 2. Grouped Data Mode = x = Xlb+ 푑1 c d1+d2 where: Xlb=lower class boundary d1=difference between the frequency of the modal class and the frequency of the class next lower in value d2= difference between the frequency of the modal class and the frequency of the class next higher in value c=class interval Example 1: Classes Frequency(f) Less than cumulative frequency (<cf) 1-8 4 4 9-16 5 9 17-24 12 21 25-32 15 36 33-40 9 45 41-48 10 55 n=55 Solution: The modal class is the 25-32, since it is the class interval with highest frequency. We can substitute to the formula: x^=Xlb+ 푑1 푐 d1+d2 =24.5+ 15-12 8 (15-12)+(15-9) =24.5+ 3 8 3+6 =24.5+2.67 =27.17 Thus, the mode is 27.17.
  • 19. 19 CHAPTER V Measures of Location or Fractiles Measures of Location or fractiles are values below which a specified fraction or percentage of observations in given set must fall. A. Percentiles -are values that divide a set of observations into 100 equal parts P1, read as first percentile, is the value which 1% of the values fall P99, is the value below which 99% of the values fall To compute for the ith percentile: Pi=the value of the i(n+1) th observation in the array 100 Example: The following were the scores of 8 students in a quiz: 4, 3, 5, 6, 8, 6, 7, 9 Solution: First, we need to arrange the data from lowest to highest. That is, 3, 4, 5, 6, 6, 7, 8, 9 Then, substitute P70= 70(8+1) th observation=6.3 or the 7th observation 100 Therefore, the 70th percentile is 8, which interpreted as: 70% of the scores are below 8. Approximating the ith Percentile from a Frequency Distribution Formula: Pi=LCBpi+c ( 푖푛 100 −< 푐푓푝푖 − 1) fpi where: The Pith class where the <cf is equal to, or exceeds for the first time, in 100 LCBpi=the lower class boundary of the pith class c =class size of the Pith class fpi =frequency of the Pith class <cfpi-1=less than cumulative frequency of the class preceding the Pith class
  • 20. 20 Example: (Refer to the example on the scores of 100 students in an achievement test). Find the 40th percentile. Score Frequency(f) <cf 60-64 10 10 65-69 8 18 70-74 12 30 75-79 25 55 80-84 20 75 85-89 25 100 Total n=100 Solution: P40 th class is the class containing 40x100 th observation which is the 40th observation. 100 Thus, from the above FDT it can be found at the 75-79 class interval. We now substitute to the formula. P40=74.5+5 40-30 25 =74.5+5(0.4) =74.5+2 =76.5 Therefore, forty percent of the scores in the achievement test is below 76.5. B. Deciles - are values that divide the array into 10 equal parts. D1, read as first decile, is the value below which 10% of the values fall D9, read as ninth decile, is the value below which 20% of the values fall To compute for the ith decile: Di= the value of the i(n+1) th observation 100
  • 21. 21 Example: From the given set scores in a quiz find the 5th Decile or D5. 5, 3, 7, 6, 4, 8, 9 Solution: First, arrange the data from lowest to highest. D5= 5(7+1) th = 40 observation= 4th observation 10 10 Therefore, the 5th decile is 6, this implies that: 50% of the scores in the quiz are below 6. Approximating the ith decile from a Frequency Distribution Formula to be used: D1= LCBdi+c ( 푖푛 100 −< 푐푓푑푖 − 1) fDi where: The Dith class is the class where the <cf is equal to, or exceeds for the first time, in . 10 LCBDi=the lower class boundary of the Dith class c =class size of the Dith class fDi =frequency of the Dith class <cfDi-1=less than cumulative frequency of the class preceding the Dith class Example: (Refer to the example on the scores of 100 students in an achievement test) Find the 8th Decile. Score Frequency(f) <cf 60-64 10 10 65-69 8 18 70-74 12 30 75-79 25 55 80-84 20 75 85-89 25 100 Total n=100 Solution: D8th class is the class containing 8x100 th observation which is the 80th observation 10 The said observation can be found at the 85-89 class interval. D8=84.5+5 80-75 25 =84.5+5(0.2) =84.5+1 =85.5
  • 22. 22 Thus, 80% of the scores are below 85.5. C. Quartiles -are values that divide the array into 4 equal parts. Q1, read as first quartile, is the value below which 25% of the values fall Q2, is the value below which 50% of the values fall Q3, is the value below which 75% of the value fall Example: From the given set scores in a quiz. Find the 2nd percentile. 2, 5, 7, 8, 9 Solution: We need to use the formula: i(n+1) th .That is, 100 Q2= 2(5+1) th 12 = observation = 3rd observation 4 4 Therefore, the 2nd quartile is 7, this implies that: 50% of the scores in the quiz are below 7. Approximating the ith Quartile from a Frequency Distribution Formula to be used: Qi= LCBQi + c (푖푛 4 −< 푐푓푄푖 − 1) fQi where: The Qith class is the class where the <cf is equal to, or exceeds for the first time, in 4 LCBQi=the lower class boundary of the Qith class c =class size of the Qith class fQi =frequency of the Qith class <cfQi-1=less than cumulative frequency of the class preceding the Qith class Example: (Refer to the example on the scores of 100 students in an achievement test) Find the 3rd Quartile. Score Frequency(f) <cf 60-64 10 10 65-69 8 18 70-74 12 30 75-79 25 55 80-84 20 75 85-89 25 100 Total n=100 Solution: The third quartile class is the class containing 3x100 th observation which is the 75th observation. 4
  • 23. 23 The said observation can be found at the 80-84 class interval. We will use the formula, that is, Q3=79.5+5 75-55 20 =79.5+5(1) =79.5+5 =84.5 Thus, 75% of the scores are below 84.5.
  • 24. 24 ASSESSMENT TEST I A. Write the following expressions in summation notation. 1. X7+X8+X9+X10 2. (X3-2) + (X4-2) + (X5-2) + (X6-2) 3. 2X1+2X2+2X3 4. (X1+4) + (X2+4) + (X3+4) 5. 12+22+32+42 B. Write the following without summation signs and simplify if possible. 5 1. Σ 6 푖=3 10 2. Σ 푖 푖=4 3 3. Σ 2푋푖 푖=1 4 4. Σ 푋푖푌푖 푖=1 3 5. Σ 푋푖 + 푖=1 2 Σ(푌푖 − 2) 푖=1 C. Solve for the mean of the given ungrouped data. 1. The scores in the long quiz of the 5 students are as follows: 88, 90, 85, 93 and 87. 2. The weight of the ten students (in kilogram) is: 40, 45, 49, 38, 52, 60, 70, 59, 47, and 48. 3. The grades of Maria in 5 subjects are as follows: 89, 90, 93, 87, and 92. 4. The 4 students have a grade of 95, 87, 74, and 83 in their Math subject. 5. The age of the 6 college students are: 21, 24, 19, 18, 17 and 25. D. Solve for the mean of grouped data. Class Interval Frequency (f) 48-50 12 45-47 9 42-44 5 39-41 6 36-38 2 33-35 1
  • 25. 25 n=35 E. Find the median of the following test scores. 1. 15, 13, 19, 14, 20 2. 5, 6, 11, 19, 20, 25, 30, 40 3. 38, 28, 54, 44, 17, 20 4. 8, 11, 17, 19, 21, 27, 30 5. 2, 7, 6, 4, 5 F. Compute for the median using grouped data. Class interval f 46-50 7 41-45 18 36-40 6 31-35 12 26-30 5 21-25 2 n=50 G. Find for the mode of the given ungrouped data. 1.6, 8, 6, 9, 7, 10, 11, 6 2. 4, 7, 21, 10, 9, 7, 10 3. 3, 9, 9, 4, 3, 7, 4, 8, 10 4. 25, 27, 18, 40, 27, 27 5. 2, 1, 9, 7, 9, 3 H. Solve for the mode of the given grouped data. Classes Frequency(f) Less than cumulative frequency (<cf) 6-10 8 8 11-15 11 19 16-20 14 33 21-25 7 40 26-30 5 45 31-35 5 50 n=50
  • 26. 26 I. Solve the following. The following were the scores of 6 students in a quiz: 6, 5, 10, 8, 3, 4 1. Find the 40th percentile 2. Find the 6th decile 3. Find the 2nd quartile 4. Find the 60th percentile 5. Find the 3rd quartile J. Solve the following. Find the 2nd Quartile, 6th Decile and 80th Percentile of the average weight of 30 students. Weight (in kilograms) No. of students 76-80 2 71-75 1 66-70 5 61-65 10 56-60 6 51-55 6 n=30 TEST II Direction: Solve the following measure of dispersion given below. 1.) Find the range of each ungrouped set of data. a. 21, 35, 45, 25, 31, 54, 47, 39, 40, 28 ____________ b. Jenny took 10 Abstract Algebra tests for the first semester. What is the range of her test scores if her scores given are 85, 80, 75, 90, 95, 79, 87, 73, 83, and 92 as the results?
  • 27. 27 2.) Find the range of grouped data in the given set below. Class Intervals Frequency Class Boundaries Class Mark 5 – 9 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 35 – 39 2 4 6 8 6 4 2 4.5 - 9.5 9.5 - 14.5 14.5 - 19.5 19.5 - 24.5 24.5 - 29.5 29.5 - 34.5 34.5 - 39.5 7 12 17 22 27 32 37 3.) Fill in the table below. Solve for the Quartile Deviation, Average and Standard Deviation of grouped data. Classes 5 – 9 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 Frequency 3 8 12 11 10 6 15 8 9 18 n = 100 X 7 12 17 22 27 32 37 42 47 52 fx x - ẋ |x - ẋ| x - ẋ2 f|x - ẋ| <cf