Properties of Arithmetic Mean

In this class, We discuss Properties of Arithmetic Mean.

For Complete YouTube Video: Click Here

The reader should have a prior understanding of arithmetic mean. Click Here.

1) if X’ is the arithmetic mean of n observations x1,x2, . . . xn. then (x1 – x’)+(x2 – x’) + . . .+(xn – x’) =0

Sum of deviations of the set of values from their arithmetic mean.

Understand the property with an example

1, 2, 3, 4, 5

X’ = mean = (1 + 2+ 3 +4 +5)/5

X’ = 3

(1-3)+(2-3) + (3-3) + (4-3) + (5-3) = 0

(1-3) gives how much the data value 1 deviates from the mean value 3.

The sum of all the deviations of the data values from the mean is zero.

Proof:

X’ = (X1+X2+. . +Xn)/N

(X1+X2+. . +Xn) = X’N – Important to understand.

The sum of all the data values is equal to X’N.

(x1 – x’)+(x2 – x’) + . . .+(xn – x’) = (X1+X2+. . +Xn) – NX’

NX’ – NX’ = 0

2) Mean of composite series

Mean of heights of N1 girls is X1′

Mean of heights of N2 boys is X2′

X1′ = (xg1 + xg2 + . . . +xgn1)/N1

X2′ = (xb1 + xb2 + . . . +xbn2)/N2

Total Mean X’ is as follows.

X’ = ((xg1 + xg2 + . . . +xgn1) + (xb1 + xb2 + . . . +xbn2)) / (N1 + N2)

X’ = (N1X1′ + N2X2′)/(N1 + N2)

3) If each of the data values is increased or decreased by a fixed value then the mean of the new data is increased or decreased by the fixed value.

Example:

1, 2, 3, 4, 5

X ‘ = ( 1+ 2+ 3+ 4+ 5)/5 = 3

Each data value is increased by 5.

6,7,8,9,10

X’new = ( 6+7+8+9+10)/ 5 = 8

X’new = X’ + 5

4) Similar to property 3 with multiplication or division

Example:

1, 2, 3, 4, 5

X ‘ = ( 1+ 2+ 3+ 4+ 5)/5 = 3

Each data value is multiplied by 5

5,10,15,20,25

X’new = (5+10+15+20+25)/5 = 15

Example:

Each data value multiplied by b and added with a.

X’new = a + bX’