Subject: Compulsory Maths
When the frequency of each class is added in succession, the total frequency is called cumulative frequency and is denoted by c.f. The average value of data is called mean. data is grouped in individual, discrete and continuous series. When a given distribution arranged either ascending or in descending order then the value that lies in the middle of the distribution is called median.In the given data the most repeated value or data is called mode.
A collection of facts, such as numbers, words, measurements, observations or even just descriptions of things is known as data.
The data which is collected in initial phase which may not be in an order is known as Raw data. This data can't be used to make a decision.
The raw data which wellarranged properly for further analysis after collecting is known as Arrayed data.
An observation is an individual fact or information in the form of numerical figure.
The frequency of a particular data value is the number of times the data value occurs. For example, if four students have a score of 70 in mathematics, and then the score of 70 is said to have a frequency of 3. The frequency of a data value is represented by f.
The tabular representation of given data with their frequencies is known as frequency table.
Cumulative Frequency Table
Cumulative Frequency is used to determine the number of observations that lie above (or below) a particular value in a data set. The is calculated using a Cumulative Frequency distribution table, which can be constructed from stem and leaf plots or directly from the data.
Constructing Cumulative Frequency Table:
(i) Arrange the given data in increasing order.
(ii) Go on adding frequency of each class one after another.
For example:
62 63 63 60 55 66 65 66 67 62 68 69 73
63 62 60 59 63 70 65 67 62 66 68 65 64
The cumulative frequency of a particular class is the sum of the frequency of that class and the frequencies of all the classes above it.
Cumulative Frequency Table
Class weight (in kg) (X)  Frequency(f)  Cumulative Frequency (c.f) 
55  59  1  1 
59  63  7  1 + 7 = 8 
63  67  11  8 + 11 = 19 
67  71  6  19 + 6 = 25 
71  75  1  25 + 1 = 26 
When the frequency of each class is added in succession, the total frequency is called cumulative frequency and is denoted by c.f.
Line graph
A linegraph especially useful in the field of statistic and science. Line graph are more popular than other graphs because the visual characteristics in them show the trends of data clearly and it is easy to draw. A linegraph is a visual comparison of two variablesshown on xaxes and yaxes.
Pie Chart
A pie chart (or a circle chart) is a circular statistical graph, which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice (and consequently its central angle and area are proportional to the quantity it represents.
Study of pie chart
1. At first, look at the chart to find the number of sectors on the different title.
2. Add all the quantities of each title.
3. Calculate the value of quantity represented by 10 on the basis of 3600 equivalent to total values (quantity) of the sectors.
4. Find the value of each title (sector) separately according to the angles indicated in the pie chart.
Introduction of the Histogram
A histogram is a graphical representation of the distribution of numerical data.
Let us see the frequency table and the histogram given below. 110 employee of a finance company pays the following income tax to the government of their monthly income.
Monthly Income Tax 
No.of taxpayer (frequency) 
100  150 150  200 200  250 250  300 300  350 350  400 400  450 450  500 
8 16 20 12 24 15 10 5 
Here, the first bar represents 8 taxpayers paying more than Rs.100 but less than Rs. 150. Similarly, the second bar represents 16 tax payers 16 taxpayers paying more than Rs. 150 but less than Rs. 200. The highest bar represents 24 workers paying more than Rs. 300 but less than Rs. 350. In this way, the above figure obtained by drawing a set of bars having the equal base on the xaxis and the height equivalent to the frequency of the given data or the collected data after representing it in the continuous frequency distribution table with various class interval known as the histogram.
Introduction and Construction of an Ogive (cumulative frequency curve)
The numbers of periods per week assigned to 50 teachers in a boarding school is as follows:
No. of Periods  No. of Teachers 
0  6 6  12 12  18 18  24 24  30 30  36 
3 10 20 10 5 2 
This can be represented as in the following table:
No. of periods  Frequency  Cumulative Frequency 
0  6  3  3 
6  12  10  3 + 10 = 13 
12  18  20  13 = 20 = 33 
18  24  10  33 + 10 = 43 
24  30  5  43 + 5 = 48 
30  36  2  48 + 2 = 50 
The above cumulative frequency table can be represented in the 'less than type' simple frequency table as in the following:
Periods  Frequency 
Less than 6  3 
Less than 12  13 
Less than 18  33 
Less than 24  43 
Less than 30  48 
Less than 36  50 
Representing this data in graph:
The line so formed is called ogive or cumulative frequency curve.
To draw the ogive of any data,
1 . First of all, find the cumulative frequency of each data or class.
2 . Convert the cumulative frequency table into 'less than type' frequency table.
3 . Plot each data of ' less than type ' frequency table and its frequency on the graph and then join the plotted points freely.
The average value of data is called mean or arithmetic mean. Data is grouped into three categories:
1 . Individual Series
2 . Discrete Series
3 . Continuous Series.
**This includes the process of calculating the mean of the individual mean of individual series and discrete series only. **
Data expressed without mentioning frequency is called individual series.
Data having frequency but no class interval is called discrete series.
When a given data arranged either in ascending or in descending order then the value that lies in the middle of the distribution is called median.
If there is N number, then the value at the ( \(\frac{N+1}{2}\) ) place is the median. But if N is even, then the average value of the two terms is the median.
Quartiles are those values of a distribution which divide the given data into four equal parts. A quartile is a type of quantile. The first quartile (Q_{1}) is defined as the middle number between the smallest number and the median of the data set. The second quartile (Q_{2}) is the median of the data. The third quartile (Q_{3}) is the middle value between the median and the highest value of the data set. Quartile divides the whole data into 25%, 50%, and 75%.
For individual and discrete series
First Quartile (Q1) = the value of (\(\frac{N+1}{4}\))^{th} item
Second Quartile (Q2) = the value of 2 \(\times\) (\(\frac{N+1}{4}\))^{th} item = (\(\frac{N+1}{2}\))^{th} item
Third Quartile (Q3) = the value of 3 \(\times\) (\(\frac{N+1}{4}\))^{th} item
The frequency distribution table for 231 people of a village classified according to their ages is as follows:
Age Group (year)  No.of people 
0  5  18 
5  10  15 
10  15  14 
15  20  31 
20  25  33 
25  30  60 
30  35  23 
35  40  20 
40  45  12 
45  50  5 
Representing the above table in a cumulative frequency:
Age group (in years)  Cumulative Frequency 
0  5  18 
5  10  33 
10  15  47 
15  20  111 
20  25  171 
25  30  194 
30  35  214 
35  40  226 
45  50  231 
The total materials expenditure of Rs. 9800 for a student to study in higher education in Kathmandu is given in the pie chart below. Answer the following questions on the basis of the pie chart.
(i) how much is spent for books?
(ii). How much is the expenditure for house rent and food more than that of education?
(i) . Expenditure on books = 7% of 9800
= \(\frac{7}{100}\) \(\times\) 9800
= Rs. 686
(ii) Percentage of expenditure on house rent and food = 100%  (7% + 22% + 7% + 16%)
= 100%  52%
= 48%.
Hence ,
expendituren on house rent and food = 48% of Rs. 9800
= Rs. 9800 \(\times\) \(\frac{48}{100}\)
= Rs. 4.704
Expenditure on education = 22% of Rs. 9800
= Rs. 2156
\(\therefore\) (rs.4704  Rs. 2156) = Rs. 2548 is more expenditure on house rent and food than education Ans.
The different materials in a bundle of 100 meters of cloths weighing 25 kg were found as shown in th pie chart given below :
Answers the following questions on the basis of the pie chart :
(i). What is the weight of cotton in the bundle of cloth ?
(ii). Which of two raw materials have equal weight ? What is the weight of those items ?
(iii). How many times is the weight of Nylon heavier than cotton ?What is the weight of Nylom ?
(iv). What is the percentage of polyster ?
Here, total weight = 360
i.e. 360 = 25kg
or 1\(^{o}\) = \(\frac{25}{360}\) kg
(i). Here , the angle subtended at centre for cotton
= 360 (198 + 36 + 36 +18 ) = 72
But , 1 weight = \(\frac{25}{360}\)k.g
\(\therefore\) 72 weight = \(\frac{25}{360}\) \(\times\) 72 =5 k.g
Hence , weight of cotton = 5 k.g Ans.
The blood groups of 900 patients who came to Bir Hospital , Ktahmandu being injured in accidents throughout as follows.
Draw a pie chart on the basis of the given data. (There are 4 types of human blood i.e , A , B , AB and O.)
Types Of blood  A  B  AB  O 
Percentage  40%  10%  5%  45% 
Here , total percentage = 100%
Now , let 100% = total angle formed at the centre.
i.e. 100% = 360
or , 1% = \(\frac{360}{100}\) = 3.6
Now , to find out the central angle formed at centre of each for the blood group are as follows :
(i). 40% blood of group A = 40 \(\times\) 3.6 = 144
(ii). 10% blood of group B = 10 \(\times\) 3.6 = 36
(iii). 5% blood of group AB = 5 \(\times\) 3.6 = 18
(iv). 45% blood of group O = 45 \(\times\) 3.6 = 162
The percentage distribution of different elements in the body of a man of 60 kg is given below. Express the weight in percentage ond represent them in a pie.
Element  Muscles  Bones  Blood  Fats  Others 
Percentage  45%  18%  12%  20%  5% 
Here, total percentage = 100%
Here , let 100% = 360
\(\therefore\) 1% \(\frac{360}{100}\) = 3.6
Now , to find angle for each part of the body formed at the centre of pie chart are as follows :
(i). 45% of muscles = 45 \(\times\) 3.6 = 162
(ii). 18% of bones = 18 \(\times\) 3.6 = 64.8
(iii). 12% of blood = 12 \(\times\) 3.6 = 43.2
(iv). 20% of fats = 20 \(\times\) 3.6 = 72
(v). 5% of others = 5 \(\times\) 3.6 = 18
Here ,
360\(^o\) = 60 k.g
\(\therefore\) 1 = \(\frac{60}{360}\) = \(\frac{1}{6}\) k.g.
\(\therefore\) 162 muscles = \(\frac{1}{6}\) \(\times\) 162 = 27 Kg.
64. 8 bones = \(\frac{1}{6}\) \(\times\) 64.8 = 10.8 k.g
43.2 blood = \(\frac{1}{6}\) \(\times\) 43.2 =7.2k.g.
72 fats = \(\frac{1}{6}\) \(\times\) 72 = 12 k.g.
18 others = \(\frac{1}{6}\) \(\times\) 18 = 3 k.g .
The daily schedule of class 9 students is as follows : Convert these data into percentage and then represent them in a pie chart :
Work  School  Sports  Homework  Food  Misc  Sleeping 
Time (hour)  7  2  3  1 \(\frac{1}{2}\)  2 \(\frac{1}{2}\)  8 
Here , total time = 24 hrs
wE Covert given data into percentage as follows :
School's book = \(\frac{7}{24}\) \(\times\) 100% = \(\frac{175}{6}\)%
Sleeping work = \(\frac{8}{24}\) \(\times\) 100% = \(\frac{100}{3}\)%
Sport's work = \(\frac{2}{24}\) \(\times\) 100% =\(\frac{25}{3}\)%
Home work = \(\frac{3}{24}\) \(\times\) 100% = \(\frac{25}{2}\)%
Food = 1 \(\frac{1}{2}\) \(\times\) 100% = \(\frac{25}{4}\)%
Misc. work = 2 \(\frac{1}{2}\) \(\times\) 100% = \(\frac{125}{12}\)%
Total angle formed at the centre of each pie chart = 360
Here , let 100% = 360
1% = \(\frac{360}{100}\) = 3.6
Now , we find the angles formed at the centre for each of work as follows :
School \(\frac{175}{6}\) =\(\frac{175}{6}\) \(\times\) 3.6 = 105
Sleeping \(\frac{100}{3}\) =\(\frac{100}{3}\) \(\times\) 3.6 = 120
Sports \(\frac{25}{3}\) =\(\frac{25}{3}\) \(\times\) 3.6 = 30
Home work \(\frac{25}{2}\) =\(\frac{25}{2}\) \(\times\) 3.6 = 45
Fooding \(\frac{25}{4}\) = \(\frac{25}{4}\) \(\times\) 3.6 = 22.5
Misc. \(\frac{125}{12}\) = \(\frac{125}{12}\) \(\times\) 3.6 = 37.5
A survey of 360 household of a municipality was conducted regarding the use of fuel for cooking. The data were found as follows :
Fuel Type  wood  electricity  gas  Kerosene 
Percentage  12.5%  22.5%  25%  40% 
Represent this data in a pie chart .
Fuel  Percent 
Wood  12.5% 
Electricity  22.5% 
Gas  25% 
Kerosene  40% 
Total  100% 
Total percentage of consumed fuel = 100%
Angle formed at the centre of pie chart = 360
Here , 100% = 360
Now , we convert fuel percet into degree as follows :
Wood (12.5%) = 12.5 \(\times\) 3.6 = 45
Electricity (22.5%) = 22.5 \(\times\) 3.6 = 81
Gas (25%) = 25 \(\times\) 3.6 = 90
Kerosene (40%) = 40 \(\times\) = 144
Construct the histogram for each of the following data with equal size of class intervals and make a less than type cumulative frequency table for each of them. Then draw the ogive :
Class  Frequency 
10 15 15 20 20  25 25  30 30  35 
2 4 3 2 3 
As the scale on X  axis start from 10 , a break is drawn near the ogive and the graph is drawn using scale beginning from 10 but not the origin itself. In order to draw the histogram , take 1cm = 5 units on X axis representing the class intervals and 1 cm = 1 unit on Yaxis representing the frequencies of given data , and draw rectangles as shown given below :
Now , convert the above table in cumulative frequency table.
Class  Frequency  Cumulative Frequency 
10  15 15  20 20  25 25  30 30  35  2 4 3 2 3  2 6 9 11 14 
Now , to draw the CF curve , convert the above table in less than type CF table by including imagined class interval 5  10 woth frequency as O show below :
Class  Cumulative Frequency 
Less than `10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35  0 2 6 9 11 14 
In order to draw the cumulative frequency curve of the data from above 'less than' frequency table let us take point 1cm = 5 units on X axis and 1 cm = 2 unit on y axis and plot the points and joined smothly.
Construct the histogram for the data with equal size of class intervals and make a less than type cumulative frequency table for each of them. Then draw the ogive :
Class  Frequency 
12 15 15  18 18  21 21  24 24  27 27  30 
2 4 3 6 4 1 
To draw the histogram from the above table , let us mark the points by taking the scale 2cm = 3 units and 1 cm = 1 unit in Y axis and draw the rectangle as shown in the graph given below :
Now , converting the above table into cumulative frequency table , we have
Class  Frequency  Cumulative Frequency 
12  15 15  18 18  21 21  24 24  27 27  30  2 4 3 6 4 1  2 6 9 15 19 20 
Again for drawing the ogive , we convert the above table into 'less than' type cumulative frequency table by including imagined class interval 9  12 with frequency 0 as follows :
Class  Cumulative Frequency 
Less than 12 Less than 15 Less than 18 Less than 21 Less than 24 Less than 27 Less than 30  0 2 6 9 15 19 20 
In order to draw the ogive of the data from above 'less than' frequency table , let us plot the points by taking the scale 1cm = 3 units on X  axis representing less than class interval and 1 cm = 2 units on Y axis representing the cumulative frequencies and join freely as shown in the given below ,
The pulse rate per minute of 34 students of class IX was measured and found to be as follows:
61 56 72 65 67 59 68 82 60 66 52 59 73 68
66 61 51 59 84 68 79 75 68 58 71 62 57 53
74 70 66 71 54
Represent the above data by frequency distribution table and cumulative frequency table by taking the class intervals 55  60, 60  65,.....
Class interval 
Tally Marks 
Frequency 
50 – 55 
//// 
4 
55 – 60 
//// / 
6 
60 – 65 
//// 
5 
65 – 70 
//// //// 
9 
70 – 75 
//// / 
6 
75 – 80 
/ / 
2 
80 – 85 
// 
2 


Total = 34 
Cumulative frequency Table:
Class  Interval  Cumulative Frequency 
50  55  4 
55  60  10 
60  65  15 
65  70  24 
70  75  30 
75  80  32 
80  85  34 
The marks obtained in a unit test of 25 full marks by 23 students of class IX are as follows :
17 15 12 5 18 24 20 10 14
20 23 17 22 21 14 9 13
21 24 20 25 16
Represent the above data in a frequency distribution table by classifying them into the class 0  5 , 5  10. . . ,
Numbers  Tally Marks  Frequency 
0 – 5  /  1 
5 – 10  //  2 
10 – 15  ////  5 
15 – 20  ////  5 
20 – 25  //// ////  9 
25  30  /  1 
Total  23 
The votes secured by three candidates A , B and C from the total polling of 12000 for one seat of the parliament member in the general parliamentary election in the year 2056 are presented in the following pie chart :
Study the pie chart and answer the following questions :
(i). How many votes did B get ?
(ii). Who secured the highest vote ?
Votes secured by A = 90
(i). Central angle for votes secured by B = 360  120  90 = 150
We have given
360 =12000 votes
1 = \(\frac{12000}{360}\) votes
150 = \(\frac{12000}{360}\) \(\times\) 150 votes
= 5000 votes.
(ii) . Candidate who secured the highest vote
By question , votes secured by A = 90
90 = \(\frac{12000}{360}\) \(\times\) 90 votes = 3000 votes.
Similarly , votes secured by C = 120
120 = \(\frac{12000}{360}\) \(\times\) 120 votes = 4000 votes.
Hence , B gets highest vote.
Find the arithmetic mean of given discrete data.
Wages(rs.)  120  150  175  200  225  300 
No. of workers  12  17  11  13  4  3 
Wages(in rs.)  Frequency(f)  fx 
120 150 175 200 225 300  12 17 11 13 4 3  1440 2250 1925 2600 900 900 
N = \(\sum\)f = 60  \(\sum\)fx = 10315 
By formula ,
Mean = \(\overline {x}\) = \(\frac{∑fx}{N}\) = \(\frac{10315}{60}\) = 179.12 Ans.
Find the arithmetic mean.
Marks Obtained  32  40  45  50  55  65  82 
No. of students  9  12  17  13  11  5  4 
Marks obtained(x)  Frequency(f)  fx 
32 40 45 50 55 65 82  9 12 17 13 11 5 4  288 480 765 650 605 325 328 
By formula ,
Mean = \(\overline {x}\) = \(\frac{∑fx}{N}\) = \(\frac{3441}{71}\) = 48.46 Ans.
The mean of the given data is 17 , what is the value of x ?
Marks Obtained  5  10  15  20  `25  30 
No. of students  2  5  10  x  4  2 
Marks Obtained(m)  Frequency(f)  fm 
5 10 15 20 25 30  2 5 10 x 4 2  10 50 150 20x 100 60 
N = \(\sum\)f = 23 + x  \(\sum\)fm = 370 + 20x 
By formula
Mean = \(\overline {m}\) = \(\frac{\sum fx}{N}\) = \(\frac{370 + 20x}{23 + x}\)
According to question ,
Mean = \(\overline{m}\) = 17
17 = \(\frac{370 + 20x}{23 + x}\)
or , 391 + 17x = 370 + 20x
or , 391  370 = 20x  17x
or , 21 = 3x
x = \(\frac{21}{3}\) = 7 Ans.
If the mean of the given data below is 5 . 6, find the value of p:
x  2  4  6  8  10 
f  7  4  p  5  4 
x  f  fx 
2 4 6 8 10  7 4 p 5 4  14 16 6p 40 40 
N = \(\sum\)f = 20 + p  \(\sum\)fx = 110 + 6p 
By formula ,
Mean \(\overline {x}\) = \(\frac{\sum fx}{N}\)
or , 5.6 = \(\frac{110 + 6P}{20 + P}\)
OR , 112 + 5.6p = 110 + 6p
or , 112 110 = 6p  5.6p
or, 2 = 0.4 \(\times\) p
\(\therefore\) p = 5 Ans.
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