Class 10 – Statistics
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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 
Frequency 

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 will also be large. We can’t do anything with the frequencies but we can change each xito 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.
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.
x_{i} 
73 
72 
71 
70 
69 
68 
67 
66 
65 
f_{i} 
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.
x_{i} 
f_{i} 
di = xi – a 
fidi 
73 
2 
7369 = 4 
8 
72 
4 
7269 = 3 
12 
71 
6 
7169 = 2 
12 
70 
10 
7069 = 1 
10 
69 
11 
6969 = 0 
0 
68 
7 
6869 = 1 
7 
67 
5 
6769 = 2 
10 
66 
4 
6669 = 3 
12 
65 
1 
6569 = 4 
4 

From above table, the mean of the deviations,
Now let us find the relation between .
Since, in obtaining di we subtract ‘a’ from each xi
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 
20 
26 
30 
34 
40 
48 
50 
53 
60 
70 
75 
80 
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., 2030 and not in 1020.
Class interval 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
8090 
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 midvalue.
∴ Midvalue (or) class mark
Class interval 
Number of students (f_{i}) 
Class marks (x_{i}) 
f_{i}x_{i} 
1020 
2 
15 
30 
2030 
7 
25 
175 
3040 
7 
35 
245 
4050 
13 
45 
585 
5060 
7 
55 
385 
6070 
1 
65 
65 
7080 
13 
75 
975 
8090 
3 
85 
255 

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) 
80100 
100120 
120140 
140160 
160180 
Number of workers 
20 
30 
20 
40 
90 
Solution:
Daily wages 
Number of workers 
Class mark xi 
fixi 
di =xi – a 
fidi 
80100 
20 
90 
1800 
 40 
 800 
100120 
30 
110 
3300 
 20 
 600 
120140 
20 
130 (a) 
2600 
0 
0 
140160 
40 
150 
6000 
20 
800 
160180 
90 
170 
15300 
40 
3600 

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) Stepdeviation 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 multiply with f_{i}. 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 
di = xi – a 
fidi 

fiui 
80100 
20 
90 
1800 
 40 
 800 
2 
40 
100120 
30 
110 
3300 
 20 
 600 
1 
30 
120140 
20 
130 (a) 
2600 
0 
0 
0 
0 
140160 
40 
150 
6000 
20 
800 
1 
40 
160180 
90 
170 
15300 
40 
3600 
2 
180 

Now, we calculate ui
Let
Here, again the let us find the relation between
We have
So,
Therefore,
Stepdeviation 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 stepdeviation method.
Class interval 
019 
2039 
4059 
6079 
8099 
100119 
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 x_{i} 

f_{i}u_{i} 
019 
9 
9.5 
 3 
 27 
2039 
16 
29.5 
 2 
 32 
4059 
24 
49.5 
 1 
 24 
6079 
15 
69.5 
0 
0 
8099 
4 
89.5 
1 
4 
100119 
2 
109.5 
2 
4 

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.