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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.
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