## Note on Statistics

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### 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.

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 IncomeTax No.of taxpayer (frequency) 100 - 150150 - 200200 - 250250 - 300300 - 350350 - 400400 - 450450 - 500 81620122415105

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 - 66 - 1212 - 1818 - 2424 - 3030 - 36 310201052

This can be represented as in the following table:

 No. of periods Frequency CumulativeFrequency 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

• 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
.

### Very Short Questions

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.

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.

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

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 .

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

 Numbers Tally Marks Frequency 0 – 5 / 1 5 – 10 // 2 10 – 15 //// 5 15 – 20 //// 5 20 – 25 //// //// 9 25 - 30 / 1 Total 23

Votes secured by A = 90
(i). Central angle for votes secured by B = 360 - 120 - 90 = 150
We have given
1 = $$\frac{12000}{360}$$ votes
150 = $$\frac{12000}{360}$$ $$\times$$ 150 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 120150175200225300 1217111343 1440225019252600900900 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 32404550556582 91217131154 288480765650605325328

By formula ,
Mean = $$\overline {x}$$ = $$\frac{∑fx}{N}$$ = $$\frac{3441}{71}$$ = 48.46 Ans.

 Marks Obtained(m) Frequency(f) fm 51015202530 2510x42 105015020x10060 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 246810 74p54 14166p4040 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.

0%

95

50

80

20

45.25

30.33

35.85

20.30

11

15

10

9

1620

1440

1330

1220

20

30

45

25

15

35

20

12

25

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31

22

6

5

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6

2

8

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29

69

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38

52

10

65

51

34

13

22

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19

38

58

85

22

4

23

16

12

9.5

11.3

2.4

14.5

31,40

65,65

55,55

60,65

25,25

35,35

61,25

22,21

2,4

5,5

7,8

5,6

22,21

40,51

39,43

51,39

23

15

25

11

16

45

28

11

18

8

10

13

19

6

10

3

28

49

72

21

28

21

33

27

66

61

16

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42

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35

23

3

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35

45

55

25

• ## You scored /31

Forum Time Replies Report
##### Kirann

the mean of 5 different numbers is 6 and the mean 4 different numbers is 15. find the mean of 9 numbers.

##### Keshav

Find the missing value of a, if the mean is 12.58.X 5 8 10 12 a 20 25f 2 5 8 22 7 4 2