# 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 x1, x2,… xn, are observations with respective frequencies f1, f2,, . . ., fn then this means observation x1, occurs f1 times, x2, occurs f2, times, and so on. Now, the sum of the values of all the observations = f1 x1 + f2 x2 + . . . + fn xn, and the number of observations = f1 + f2 + . . . + fn.

 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