Mean ( Direct method)

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Mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations.
If x
1 , x 2 ,… x n , are observations with respective frequencies f 1 , f 2 ,, . . ., f n then this means observation x 1 , occurs f 1 times, x 2 , occurs f 2 , times, and so on. Now, the sum of the values of all the observations = f 1 x 1 + f 2 x 2 + . . . + f n x n , and the number of observations = f 1 + f 2 + . . . + f n .

f1 x1 + f2 x2 + . . . + fn xn
x = ---------------------------------
           f1 + f2 + . . . + fn

This can be written in short way using Greek letter Σ
       Σfi xi
x = ------
Σ fi

Example 1 :
Weight (in Kgs) 67 70 72 73 75
Number of students 4 3 2 2 1

Find the average weight.
Solution :
Weight (in kg)xi
Frequency (fi)
fixi
67
4
268
70
3
210
72
2
144
73
2
146
75
1
75
N=Σ fi = 12
Σ fixi = 843

       Σfi xi 843
x = ------- = ------- = 70.25 kg
Σ fi 12

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Example 2 :
If the average of the following data is 20.2, find the value of p:
x
10
15
20
25
30
f
6
8
P
10
6
Solution :
x
f
f x
10
6
60
15
8
120
20
P
20P
25
10
250
30
6
180
30 + P
610 + 20P

       Σfi xi
x = -------
Σ fi

20.2 = ( 610 + 20P) / ( 30 + P)
20.2 ( 30 + P ) = 610 + 20P
606 + 20.2P = 610 + 20P
20.2P – 20P = 610 – 606
0.2 P = 4
∴ P = 4 / 0.2
P = 20
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Example 3: Find the average of the following data :
Marks
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
No.of students
4
10
18
12
6

Solution: First find the class mark,
Class Mark = ( upper class limit + Lower class limit ) /2
Example : Class Mark = ( 10 + 0 ) / 2 = 10 / 2 = 5
Prepare the frequency table :
Class interval
fi
Class Mark (xi)
fi xi
0 - 10
4
5
20
10 - 20
10
15
150
20 -30
18
25
450
30 - 40
12
35
420
40 - 50
6
45
270
Σ fi = 50
Σ fi xi = 1310

       Σfi xi
x = -------
        Σ fi
Mean = (1310 / 50)
= 26.2


• Direct method.
Short cut method.
Step - Deviation method.

From median to measures of central tendency

Statistics

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