Statistics

Subject: Optional Maths

Overview

Statistics is the branch of mathematics which deals with data.This note gives information about the Statistics, Mean/average, and Median .

Statistics

Statistics is the branch of mathematics which deals with data collection tabulation of data, analysis of data and drawing conclusion for such purpose

Eg ;The marks obtained by 10 students in a unit test in mathematics are given below.The total marks were 20 and the pass marks were 10.

6 , 11, 14, 14, 15, 17, 17, 17, 18, 20 .

From the data , we can classify the data as follows

 Marks Obtained Telly marks No of Students 6 / 1 11 / 1 14 // 2 15 / 1 17 /// 3 18 / 1 20 / 1 Total 10

From the collected in a class of 10 students, we have classified the data as shown above .Marks which are here are variables so it is called variate . The number which is repeated is called frequency .The total number of frequency is the total number of students .

Here frequency is denoted by f and its sum by Σƒ is also denoted by N .The variate is usually denoted by X .

Usually, data can separate into three parts :

1. Individual series .
2. Discrete series .
3. Continuous series .

Individual series ;It is usually listed in dates which are collected in small scale .They are also known as raw data . eg;The amount of money brought by 6students is

eg;The amount of money brought by 6students of school for their tiffin.

Rs 4, Rs 5, Rs 7, Rs 10, Rs 11, These kind of data are less in number and need not be arranged in a table .

Here, the number of data is denoted by n .

In the above example , n = 5 .

Discrete series ;Tose series which are formed by discrete values are called discrete series .

The marks obtained by 10 students in a class test is given below;

 Marks 6 10 12 14 15 18 20 No of students 1 2 1 1 3 1 1

Here, marks denote the variate values and the number of students denotes the frequencies .

Continuous Series ;Those series which can be represented by a continuous variable is called continuous series . we mostly use this kind of data to insert large scale of data.
eg;

The marks obtained by 100 students in an examination is given below :

 Marks 0 -10 10 - 20 20 - 30 30 - 40 40 - 50 Numbers of students 10 30 20 15 25

In this series, the data lie within the group .Its Interval or groups are called class intervals.The end points of a class interval are called limits .

eg; 0 - 10 0 is called lower limit.

Mean \ Average

Here,the average of data is very important .Mean or average of data which single represents the whole set of data This represents the central data .Here, a percentage of students is the average marks of all the marks he or she gets in their exam.

Mean of data is calculated by adding all the data togeather and dividing them by the total number of data .

Mathematically,

Mean =$\frac{Sum\ of \ the\ data}{Total\ no\ of \ data}$

=$\frac{∑χ}{N}$

Where x is the variate values

∑χ is the sum of all variates

n is a total number of dates

Note :

∑ is read as sigma or summation .

In discrete series

In case of discrete series or ungrouped repeated data , we find mean by

Mean ($\overline{X}$) = $\frac{∑ƒχ}{N}$

Where,X is the variate

ƒ is the frequency

ƒx is the product of ƒ and x

∑ƒχ is the sum of ƒχ.

N is the total number of frequency.

Median

Those data which divides the given set of data into two equal halves is called median . It is denoted by Md . Here we need to arrange our data in either ascending or descending order for calculating the median .

Median ( Md ) = value of $\frac{(n+1)^{th}}{2}$

Where , n = total no of items .

Things to remember
• The data which divides the given set of data into two equal halves is called  median.
• Median ( Md ) = value of  $\frac{(n+1)^{th}}{2}$                                                                                                                                                                       Where, n = total no of items .
• 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.
Statistics - Find the median

Statistics is the branch of mathematics which deals with datas.

Those data which divides the given set of data into two equal halves is called median .

The formula to calculate median in individual series;

Median  ( Md ) = value of $\frac{(n+1)^{th}}{2}$item

The data median must be arranged in either ascending or descending order .

Median divides the set of data into  two parts .

Solution:

Calculation of missing frequency

 Marks ( x ) No of stdents ( ƒ ) ƒx 10 2 20 20 4 80 30 7 210 40 4 160 50 3 150 ∑ƒ= N = 20 ∑ƒx=620

Now ,

Mean  $\overline{X}$ = $\frac{ ∑ƒχ}{N}$

=  $\frac{620}{20}$=31

Hence , mean marks = 31 .

Solution :

Calculation of missing frequency

 X ƒ Fx 5 2 10 10 5 50 15 a 15a 20 7 140 25 4 100 30 2 60 ∑ƒ=20 + a ∑ƒx = 360 + 15a

We know ,

Mean $\overline{X}$=$\frac{∑ƒx}{N}$

According to  the question ,

Mean = 17

17=$\frac{360 + 15a}{20 + a }$

or17 ( 20 + a ) =360 + 15a

or340 + 17a = 360 + 15a

or17a - 15a  = 360 - 340

or,               2a = 20

∴           a = 10

Solution :

Here, the data are 3 ,7 , 9 , 10 , 15

Which are already in ascending  order .

So , number of data ( n ) = 5

Now , Median ( Md ) = value of $\frac{( n + 1 )^{th}}{2}$item

=value of $\frac{( 5 + 1 )^{th}}{2}$item

=value of $\frac{6}{2}^{th}$item

∴ Median ( Md ) = 9 .

Solution :

Here , The given data are

20, 16, 12, 10 , 24, 28, 30, 32 .

Since the data are not arranged , arranging them in ascending  order we get 10, 12, 16, 20, 24, 28, 30, 32

Now , number of data (n) = 8

So, Median (Md ) = value of $\frac{(8+1)^{th}}{2}$item

= value of  $\frac{9}{2}^{th}$ item

= value of  4.5thitem

Now , 4.5thitem  is not positive so, 4.5thitem is the mean of 4.5thitem and 5thitem

So, median (Md) =$\frac{ value of 4^{th}item + value of 5^{th}item }{2}$

= $\frac{20\ +\ 24}{2}$

=$\frac{44}{2}$

=22

∴ Md = 22.

Solution :

The dates are 18, 11, 14, p, 17,

Number of dates (n) = 5

∑x                              = 18 + 11 + 14 + p + 17

=60 + p

Now, we know , mean $\overline{X}$= 14

So, mean $\overline{X}$ = $\frac{∑X}{n}$

or, 14 = $\frac{60 + p}{5}$

or, 70=60 + p

or, 70-60 = p

or, 10= p

∴ The value of p is 10 .

Solution ,

Here the datas are  1, 2, 3, 4, 5.

number of datas ( n ) = 5

∑x = 1 + 2+3+4+5

=15

Now, mean $\overline{(X)}$=$\frac{∑X}{n}$

=$\frac{15}{5}$

=3