NCERT Notes for Class 10 Maths Chapter 14 Statistics
Class 10 Maths Chapter 14 Statistics Notes
Chapter Name  Statistics Notes 
Class  CBSE Class 10 
Textbook Name  Mathematics Class 10 
Related Readings 

Statistics
Statistics is one of the parts of mathematics in which we study about the collecting, organizing, analyzing, interpreting and presenting data.
Statistics is very helpful in real life situations as it is easy to understand if we represent a data in a particular number which represents all numbers. This number is called the measure of central tendency.
Some of the central tendencies commonly in use are:
1. Mean
It is the average of “n” numbers, which is calculated by dividing the sum of all the numbers by n.
The mean of n values x_{1} , x_{2} , x_{3} , ...... x_{n} is given by
2. Median
If we arrange the numbers in an ascending or descending order then the middle number of the series will be median. If the number of series is even then the median will be the average of two middle numbers.
If n is odd then the median is observation.
If the n is even then the median is the average of observation.
3. Mode
The number which appears most frequently in the series then it is said to be the mode of n numbers.
Mean of Grouped Data (without Class Interval)
If the data is organized in such a way that there is no class interval then we can calculate the mean by
where, x_{1} , x_{2} , x_{3} , ..... x_{n} are the observations
f_{1} , f_{2} , f_{3} , ...... f_{n} are the respective frequencies of the given observations.
Example
Grouped Population Mean 

X 
f 
fx 
20 
40 
800 
40 
60 
2400 
60 
30 
1800 
80 
50 
4000 
100 
20 
2000 
200 
∑fx = 11000 
Here, x_{1} , x_{2} , x_{3} , x_{4} , x_{5} are 20, 40, 60, 80, 100 respectively and f_{1} , f_{2} , f_{3} , f_{4} , f_{5} are 40, 60, 30, 50, 20 respectively.
Mean of Grouped Data (with ClassInterval)
When the data is grouped in the form of class interval then the mean can be calculated by three methods.
1. Direct Method
In this method, we use a midpoint which represents the whole class. It is called the class mark. It is the average of the upper limit and the lower limit.
Example: A teacher marks the test result of the class of 55 students for mathematics. Find the mean for the given group.
Marks of Students 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Frequency 
27 
10 
7 
5 
4 
2 
To find the mean we need to find the midpoint or class mark for each class interval which will be the x and then by multiplying frequency and midpoint we get fx.
Marks of Students 
Frequency (f) 
Midpoint (x) 
fx 
0 – 10 
27 
5 
135 
10 – 20 
10 
15 
150 
20 – 30 
7 
25 
175 
30 – 40 
5 
35 
175 
40 – 50 
4 
45 
180 
50 – 60 
2 
55 
110 


∑f = 55 
∑fx = 925 
2. Deviation or Assumed Mean Method
If we have to calculate the large numbers then we can use this method to make our calculations easy. In this method, we choose one of the x’s as assumed mean and let it as “a”. Then we find the deviation which is the difference of assumed mean and each of the x. The rest of the method is the same as the direct method.
Example: If we have the table of the expenditure of the company's workers in the household, then what will be the mean of their expenses?
Expense (Rs.) 
100 – 150 
150 – 200 
200 – 250 
250 – 300 
300 – 350 
350 – 400 
Frequency 
24 
40 
33 
28 
30 
22 
Solution
As we can see that there are big values of x to calculate so we will use the assumed mean method.
Here we take 275 as the assumed mean.
Expenses (Rs.) 
Frequency(f) 
Mid value(x) 
d= x – 275 
fd 
100 – 150 
24 
125 
–150 
–3600 
150 – 200 
40 
175 
–100 
–4000 
200 – 250 
36 
225 
–50 
–1650 
250 – 300 
28 
275 
0 
0 
300 – 350 
30 
325 
50 
1500 
350 – 400 
22 
375 
100 
2200 

∑f = 180 


∑fd = –5550 
3. Step Deviation Method
In this method, we divide the values of d with a number "h" to make our calculations easier.
Example: The wages of the workers are given in the table. Find the mean by step deviation method.
Wages 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
No. of workers 
8 
9 
12 
11 
6 
Solution
Wages 
No. of workers(f) 
Mid – Point(x) 
Assume mean(a) = 35 d = x – a 
H = 10, u = (x – a)/h 
fu 
10 – 20 
8 
15 
–20 
–2 
–16 
20 – 30 
9 
25 
–10 
–1 
–9 
30 – 40 
12 
35 
0 
0 
0 
40 – 50 
11 
45 
10 
1 
11 
50 – 60 
6 
55 
20 
2 
12 

∑f = 46 



∑fu = –2 
Mode of Grouped Data
In the ungrouped data the most frequently occurring no. is the mode of the sequence, but in the grouped data we can find the class interval only which has the maximum frequency number i.e. the modal class.
The value of mode in that modal class is calculated by
l = lower class limit of the modal class
h = class interval size
f_{1} =frequency of the modal class
f_{0} =frequency of the preceding class
f_{2} = frequency of the succeeding class
Example: The table of the marks of the students of a class is given. Find the modal class and the mode.
Marks 
0 – 20 
20 – 40 
40 – 60 
60 – 80 
80 – 100 
No. of Students 
4 
8 
6 
7 
5 
Solution
Here we can see that the class interval with the highest frequency 8 is 20 – 40.
So, this is our modal class.
Modal class = 20 – 40
Lower limit of modal class (l) = 20
Class interval size (h) = 20
Frequency of the modal class(f_{1}) = 8
Frequency of the preceding class(f_{0}) = 4
Frequency of the succeeding class (f_{2}) = 6
Median of Grouped Data
To find the median of a grouped data, we need to find the cumulative frequency and n/2
Then we have to find the median class, which is the class of the cumulative frequency near or greater than the value of n/2.
Cumulative Frequency is calculated by adding the frequencies of all the classes preceding the given class.
Then substitute the values in the formula
where,
l = lower limit of median class
n = no. of observations
cf = cumulative frequency of the class preceding to the median class
f = frequency of the median class
h = size of class
Example: Find the median of the given table.
Class Interval 
Frequency 
Cumulative Frequency(fc) 

1 – 5 
4 
4 
4 
6 – 10 
3 
7 
4 +3 = 7 
11 – 15 
6 
13 
7+6 = 13 
16 – 20 
5 
18 
13+ 5 = 18 
21 – 25 
2 
20 
18 + 2 = 20 

N = 20 


Solution
Let’s find the n/2.
n = 20, so n/2 = 20/2 = 10
The median class is 11  15 as its cumulative frequency is 13 which is greater than 10.
Remark: The empirical relation between the three measures of central tendency is
3 Median = Mode + 2 Mean
Graphical Representation of Cumulative Frequency Distribution
The graph makes the data easy to understand. So to make the graph of the cumulative frequency distribution, we need to find the cumulative frequency of the given table. Then we can plot the points on the graph.
The cumulative frequency distribution can be of two types:
1. Less than ogive
To draw the graph of less than ogive we take the lower limits of the class interval and mark the respective less than frequency. Then join the dots by a smooth curve.
2. More than ogive
To draw the graph of more than ogive we take the upper limits of the class interval on the xaxis and mark the respective more than frequency. Then join the dots.
Example: Draw the cumulative frequency distribution curve for the following table.
Marks of Students 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
No. of students 
7 
10 
14 
20 
6 
3 
Solution
To draw the less than and more than ogive, we need to find the less than cumulative frequency and more than cumulative frequency.
Now, we plot all the points on the graph and we get two curves.
Remark
 The class interval should be continuous to make the ogive curve.
 The xcoordinate at the intersection of the less than and more than ogive is the median of the given data.