As originally posted on Edvie.com  

STATISTICS 

 

Class 10 Notes 

 Introduction: Statistics is the study of

collection, analysis, interpretation, and

organization of data. In applying statistics

to, e.g., a scientific, industrial, or societal

problem, it is conventional to begin with a

statistical population or a statistical model

process to be studied. Populations can be

diverse topics such as “all persons living

in a country” or “every atom composing a

crystal”. Statistics deals with all aspects of

data including the planning of data

collection in terms of the design of

surveys and experiments.  

 Mean of ungrouped data: The arithmetic

mean of a statistical data is defined as the

quotient of the sum of all the terms or

entries divided by the number of items. 

If are the items given, then 

 

This is usually denoted by . 

  i) As some values of x will be less and

some more than , in some sense lies

at the centre of all the values justifying

that it is the measure of central tendency. 

 ii) If to each item x is added a number

k, then the new mean = . 

iii) If each item is multiplied by a

number k, then the new mean = k x .  

 Example: Find the arithmetic mean of

numbers 7, 6, 10, 2, 5, 8, 9. 

Solution:

. 

Example: The sum of 15 observations of

a data is 420. Find the mean. 

Solution: Here n = 15,  

 

Properties of arithmetic mean: 

1. Algebraic sum of

deviations of a set of values from

their arithmetic mean is zero.

2. The sum of the squares of

the deviations of a set of values is

minimum when taken about the

mean.

Example: Using the formula ,  

prove that .  

Solution:  

 

 

 i.e., the algebraic sum of deviations of a set of values from their arithmetic mean is zero. 

Example: The mean of a data is 9. If each observation is multiplied by 3 and then 1 is added to each

result, find the mean of the new observations. 

Solution: Let the observations be . 

Given  

 

Each observation is multiplied by 3 and then 1 is added i.e., .

Then the mean is given by 

Now 

 

 ∴ Mean of new observations = 28.    Mean of an ungrouped frequency distribution: 

i) Direct method: If the entries occurs times respectively then the arithmetic mean is

The above formula can be stated in simpler form as

 

Where is the total frequency. Example: The frequency distribution of the number of heads obtained in tossing five coins 100

times is given below. Find the mean of the data. 

Number of heads  0  1  2  3  4  5 

Frequency  1  7  20  64  5  3 

 Solution: To find the mean from the given frequency distribution we prepare the table as below. 

Number of heads

( x i ) Frequency

( f i) f ix i 

0  1  0 

1  7  7 

2  20  40 

3  64  192 

4  5  20 

5  3  15 

Total  N = 100  

 

ii) Assumed mean method: The frequencies and the values of the variable are quite large

numbers. The product fi x i will also be large. We can’t do anything with the frequencies but we

can change each xi to a smaller number, so that our calculations become easy.

In this method assume one of the observations which is convenient as assumed mean ( a ). Then

find the deviation ( d ) of other observations from the assumed mean.

i.e., d i = x i – a

Any number can be taken as assumed mean ( a ) but for accurate values, it is generally taken as that

value of the variable which has the greatest frequency in the frequency distribution or which is

near about the middle of the frequency distribution.  

 Consider the following data to calculate the mean for the frequency distribution by assumed mean

method. 

xi   73  72  71  70  69  68  67  66  65 

fi   2  4  6  10  11  7  5  4  1 

 

 

From the above table. It is clear that, the variable which has greatest frequency is 69 and also it is in

the middle of the frequency distribution. 

∴ The assumed mean ( a ) = 69.  Solution: Let the assumed mean a = 70. 

xi  fi  d i = x i – a  f id i  73  2  73-69 = 4  8 72  4  72-69 = 3  12 71  6  71-69 = 2  12 70  10  70-69 = 1  10 69  11  69-69 = 0  0 68  7  68-69 = -1  -7 67  5  67-69 = -2  -10 66  4  66-69 = -3  -12 65  1  65-69 = -4  -4  

  

 

From above table, the mean of the deviations,  

Now let us find the relation between . 

Since, in obtaining d i we subtract ‘ a ’ from each x i  

So, in order to get the mean we need to add ‘ a ’ to d .  

This can be explained mathematically as: 

Mean of deviations,  

So,  

 

 

Therefore  

Now substituting the value of a , and from the table, we get 

 

 

Mean of the grouped data: 

i) Direct method: Sometimes, data is so large that it is difficult to study the data and also

difficult to find measures of central tendency (Mean, Mode, Median). In this situation we make

group of the data with suitable class intervals called as grouped data.

Example: 

Marks  10  2

0 26  3

0 34  4

0 48  5

0 53  6

0 70  75  8

0 

Number of students  2  4  3  2  5  6  7  2  5  1  6  7  3 

 Now, convert the data of above table into grouped data by forming class intervals of width say 10.  

While allocating frequencies to each class interval, students whose score is equal to any upper class

boundary would be considered in the next class. e.g., 4 students who have obtained 20 marks

would be considered in the next class. i.e., 20-30 and not in 10-20.  

 Class interval  10-2

0 20-3

0 30-4

0 40-5

0 50-6

0 60-7

0 70-8

0 80-9

0 

Number of students  2  7  7  13  7  1  13  3 

 In a grouped data, it is assumed that the frequency of each class is concentrated at its mid-value. 

∴ Mid-value (or) class mark  

Class interval  Number of students ( fi )  Class marks (x i ) fix i  

10-20  2  15  30 

20-30  7  25  175 

30-40  7  35  245 

40-50  13  45  585 

50-60  7  55  385 

60-70  1  65  65 

70-80  13  75  975 

80-90  3  85  255 

  Σf i = 53    Σf ix i = 2715 

 

 This new method of finding mean is known as direct method.  

ii) Assumed mean method:

Example: Calculate the mean daily wage of the workers given below:  

Daily wages (in Rs)  80-10

0 100-120  120-140  140-160  160-180 

Number of

workers 20  30  20  40  90 

 Solution:  

Daily

wages Number of

workers Class mark x i  fix i  d =x i – a  fid i  

80-100  20  90  1800  - 40  - 800 

100-120  30  110  3300  - 20  - 600 

120-140  20  130 ( a )  2600  0  0 

140-160  40  150  6000  20  800 

160-180  90  170  15300  40  3600 

  Σf i = 200    Σf ix i = 29000 

  Σf idi = 3000 

 Let the assumed mean a = 130. 

Direct method:  

 

 

 

 Assumed mean method: We use the formula derived for finding the mean of ungrouped data using

assumed mean here. 

 

 

 

iii) Step-deviation method: According to the formula , we have to find

to calculate the mean. Some times when the frequencies are large in number, this becomes

cumbersome. This can be simplified if the class interval of each class of grouped data is same.

As in above example, if we divide all the values of deviations by 20, we would get smaller

numbers which we then multifly with fi . Here 20 is the class size of each class integral.

So, let where a is the assumed mean and h is the class size. 

Daily wages 

Number of

workers Class

mark xi  fixi  d i = xi – a  fidi  

 ui = h

x −ai   

 

fiu i  

80-100  20  90  1800  - 40  - 800  -2  -40 

100-120  30  110  3300  - 20  - 600  -1  -30 

120-140  20  130 ( a )  2600  0  0  0  0 

140-160  40  150  6000  20  800  1  40 

160-180  90  170  15300  40  3600  2  180 

  Σfi = 200    Σfixi =

29000   Σfidi

=

3000 

  Σfiui =

150 

 Now, we calculate u i  

Let  

Here, again the let us find the relation between  

We have  

So,  

 

 

 

Therefore,  Now, a = 130, h = 20, Σf iui = 150, Σfi = 200 

 

Step-deviation Method is a very short method and should always be used for grouped data where

class interval sizes are equal.  

 Example: Find the mean of the following frequency distribution by step-deviation method. 

Class interval  0-19  20-39  40-59  60-79  80-99  100-119 Frequency  9  16  24  15  4  2 

 Solution: Here class length are equal. 

Class length = h = 20. a = 69.5, h = 20. 

Class interval  Frequency f i  

Class mark xi   ui = hx −ai  

 fiui  

0-19  9  9.5  - 3  - 27 20-39  16  29.5  - 2  - 32 40-59  24  49.5  - 1  - 24 60-79  15  69.5  0  0 80-99  4  89.5  1  4 

100-119  2  109.5  2  4   Σfi = 70      Σfiui =

-85  

 

 Mean of the combined distributions: When two sets of scores have been combined into a single distribution, 

then the mean of the combined distribution is the weighted mean of the means of the components, the 

weight being the total frequencies in those components. 

In other words,   

Where   is the mean of the combined distribution. 

 

 are the means of the component distributions. 

 are the total frequencies of the component distributions.   

 

GET MORE FREE STUDY MATERIAL & PRACTICE TESTS ON 

Edvie.com 

 

Follow us on Facebook