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Variance and Standard Deviation


To overcome the problem of negative signs involved in mean deviations about the mean, we take the absolute values of the deviations, ignoring the negative signs. We also saw how this may cause inaccurate and unscientific results. So instead of taking absolute deviation, we consider squared deviations or standard deviation.

Standard deviation is defined as the positive square root of the sum of the squares of all deviations of the observations from their Arithmetic Mean.

Let be n observations and be their mean.
Then is the sum of all the squares of the deviations from

If this sum is zero, then each of the quantity has to be zero, which means that there is no dispersion at all from the mean.

If is a small quantity, then there there is low degree of dispersion and if is large quantity, then there is a high degree of dispersion from the mean.

Value measures the variance and the square root of variance is the standard deviation, denoted by (Sigma).

 So

Standard Derivation for discrete distribution


Let the set of values be
  

If is easily calculated and is a round figure then .
If is in decimals, then it is better to use assumed mean for calculating S.D.

Let A be the assumed mean of
n readings
Calculate
Let
then

Standard Derivation for a Continuous Distribution


If are the frequencies for the class intervals given in the data and if is the mean of the distribution then the standard deviation where are the deviations from the actual mean . But this is a trouble some process if turns out to be a decimal. So we can do the calculations by assuming a mean close to the centre value. Then



Where ( A being assumed mean)
As in the case of determination of mean by step deviation method, if there is a common factor for the deviations, we can divide by that factor (say h) each of the deviations and then use the formula where , h is the common factor and A is the assumed mean

Merits of S.D
  1. It is well defined and suited for algebraic treatment
  2. It is based on all the observations.
  3. It is the most commonly used measure of dispersion.
  4. It is useful for comparing the variability of two or more distributions.
Example 1:
Find the S.D of the set of values: 45, 36, 40, 37, 39, 42, 45, 35, 40, 39

Solution:
The total of the set of values is 398 and hence = 39.8

It is not very easy to compute the S.D so let us use the formula

 

45

36

40

37

39

42

45

35

40

39

5

4

0

3

1

2

5

5

0

1

25

16

0

9

1

4

25

25

0

1

398

2

106


= 3.25

Example 2:
The weekly salaries of a group of employees are given in the following table. Find the mean and S. D of the salaries.

Salary (Rs.)

75

80

85

90

95

100

No.of persons

3

7

18

12

6

4


Solution:

75

80

85

90

95

100

3

7

18

12

6

4

2

1

0

1

2

3

6

7

0

12

12

12

12

7

0

12

24

36

 

50

 

23

91

 

 

​Example 3:
Find the S.D from the following data:
Weight in kg

35 - 45

45 - 55

55 - 65

65 - 75

75 - 85

No. of persons

18

22

30

6

4


Solution:
Assumed mean

Class

35 - 45

45 - 55

55 - 65

65 - 75

75 - 85

40

50

60

70

80

18

22

30

6

4

2

1

0

1

2

36

22

0

6

8

72

22

0

6

16

   

80

 

44

116

S.D =

Note: If you are asked to find the variance in any question, stop doing before taking the square root. That will be the answer.

eg. Variance in example (1) is 10.56
in example (2) it is Rs 1.61 × 25 = Rs. 40 .25 ,
in Example (3) it is 1.15 × 100 kg = 115kg.

Example 4:
The mean and S.D of 20 items is found to be 10 and 2 respectively. At the time of checking, it was found that one item 8 was incorrect. Calculate the mean and S.D if the wrong item is replaced by 12 give your answer correct to 2 decimal places.


Solution:

Wrong = (20 x 10) = 200
Corrected = 200 - 8 + 12 = 204.
 Corrected mean = = 10.2
                     
Wrong
Corrected .
 Corrected
  Corrected mean = 10.2
Corrected SD = 1.99




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