Mean ( Direct method)

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 xix = ------ 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 843x = ------- = ------- = 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 xix = ------- 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 xix = -------        Σ fi
Mean = (1310 / 50)
= 26.2

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

From median to measures of central tendency

Statistics