Subject: Optional Maths
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 :
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.
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.
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 .
What do you mean by statistics ?
Statistics is the branch of mathematics which deals with datas.
What do you mean by median ?
Those data which divides the given set of data into two equal halves is called median .
What is the formula to calculate median in individual series ?
The formula to calculate median in individual series;
Median ( Md ) = value of \(\frac{(n+1)^{th}}{2}\)item
In which order the data median must be arranged ?
The data median must be arranged in either ascending or descending order .
Into how many parts median divides the set of data ?
Median divides the set of data into two parts .
1.Calculate the mean from the data given below .
Marks  10  20  30  40  50 
No. of students  2  4  7  4  3 
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 .
Find the value of a from the data given below whose mean is 17 .
X  5  10  15  20  25  30 
ƒ  2  5  a  7  4  2 
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
Find the median from the set of data given below .
3 ,7, 9,10,15 .
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 .
Find the median from the given data :20 , 16 , 12 , 10 , 24 , 28 , 30 , 32 ,
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 4.5^{th}item
Now , 4.5^{th}item is not positive so, 4.5^{th}item is the mean of 4.5^{th}item and 5^{th}item
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.
The average of 18 , 11, 14, p, 17 is 14 .Find the value of p.
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, 7060 = p
or, 10= p
∴ The value of p is 10 .
Find the mean of 1, 2, 3, 4, 5.
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
© 201920 Kullabs. All Rights Reserved.