Statistics

Subject: Compulsory Maths

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Overview

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.

Statistics

Data

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 well-arranged 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.

Frequency

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.

Frequency table

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.

Graphical representation of data

Line graph

A line-graph 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 line-graph is a visual comparison of two variables-shown on x-axes and y-axes.

line-graph

 

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.

Histogram and Cumulative Frequency Curve

.

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 x-axis 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.

Measure of Central Tendency

Arithmetic Mean

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. **

1 . Individual Series

Data expressed without mentioning frequency is called individual series.

2 . Discrete Series

Data having frequency but no class interval is called discrete series.

Median

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

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 (Q1) is defined as the middle number between the smallest number and the median of the data set. The second quartile (Q2) is the median of the data. The third quartile (Q3) 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

Things to remember
  • Median = [\(\frac{N+1}{2}\)]thposition
  • First quartile(Q1) = [\(\frac{N+1}{4}\)]th position
  • Second quartile(Q2) = [\(\frac{N+1}{2}\)]th position
  • Third quartile(Q3) = 3[\(\frac{N+1}{4}\)]th position
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Statistics
Statistics - Mean, Median, Mode
Statistics intro: Mean, median, and mode | Data and statistics
Statistics: The average | Descriptive statistics | Probability and Statistics | Khan Academy
Questions and Answers

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

(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.

figure
figure

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

figure
figure

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 .

figure
figure

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

figure
figure

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

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 Y-axis 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.

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 ,

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.

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.

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.

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.

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.

Quiz

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