Notes On Variance - CBSE Class 11 Maths
Limitation of mean deviation: If the degree of the variability in a data is high, then the median cannot be considered as a central tendency. The sum of the deviations from the mean is more than the sum of the deviations from the median. Thus, mean deviation about the mean is not more scientific. For calculating mean deviation, we consider the absolute values of the deviations. Thus, we cannot carry out further algebraic operations. Therefore, mean deviation about the mean or median is not sufficient to measure the dispersion. So, we need to extend the measure of dispersion to an appropriate measure of dispersion. Variance is such a measure of dispersion.   To eliminate the negative signs of the deviations, we take their absolute values. An alternative method to eliminate the negative sign of the deviations is to take the square of the deviations. Let n observations be x1, x2, x3,...., xn. Mean ($\stackrel{_}{\text{x}}$) = (x11+ x22+ x33+....+ xnn)/n Deviations of the observations from the mean: (x1 - $\stackrel{_}{\text{x}}$), (x2 - $\stackrel{_}{\text{x}}$), (x3 - $\stackrel{_}{\text{x}}$),...., (xn - $\stackrel{_}{\text{x}}$) (x1 - $\stackrel{_}{\text{x}}$)2, (x2 - $\stackrel{_}{\text{x}}$)2, (x3 - $\stackrel{_}{\text{x}}$)2,...., (xn - $\stackrel{_}{\text{x}}$)2 (x1 - $\stackrel{_}{\text{x}}$)2, (x2 - $\stackrel{_}{\text{x}}$)2, (x3 - $\stackrel{_}{\text{x}}$)2,...., (xn - $\stackrel{_}{\text{x}}$)2 = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$      The sum of the squares of the deviations is equal to 0, which implies that each deviation is equal to 0. There is no dispersion since all the observations are equal to the mean. The sum of the squares of the deviations is small, which implies that the observations are closer to the mean. Hence, the degree of dispersion is low. If the sum of the squares of the deviations is large, then the degree of dispersion is high. Mean of the squares of the deviations = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$      The variance of a data is the mean of the squares of the deviations of the observations from the mean. Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$       This is the formula used to calculate the variance of an ungrouped data.   Variance of a discrete frequency distribution  Observations (xi)  x1  x2  x3  ...   xn  Frequency (fi)  f1  f2  f3  ...   fn Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$     Sum of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}$ fi   Variance of a continuous frequency distribution  Mid-values (xi)  x1  x2  x3  ...   xn  Frequency (fi)  f1  f2  f3  ...   fn Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$     Sum of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}$ fi   Consider the given frequency distribution table of trees planted in 20 schools in a tree plantation programme.  Trees Planted  9  11  15  18  22  25  30  No. of Schools  5  2  4  4  3  1  1   Variance of the given data:   Mean of the given distribution         Trees Planted (xi)            No. of Schools (fi)    fixi                 9                  5     45                11                  2     22                15                  4     60                18                  4     72                 22                  3     66                25                  1     25                30                  1     30      N = $âˆ‘{\text{f}}_{\text{i}}\text{= 20}$  $âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}\text{= 320}$ Mean ($\stackrel{_}{\text{x}}$) = $\frac{âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}}{\text{N}}$              = $\frac{\text{320}}{\text{20}}$ = 16   Deviations of the numbers of trees planted from the mean and the squares of the deviations are calculated.   The product of the squares of the deviations and their frequencies are obtained         Trees Planted (xi)            No. of Schools (fi)    fixi  $\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{) =}{\text{x}}_{\text{i}}\text{- 16}$  ${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$  fi${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$                 9                  5     45         â€“ 7     49       245                11                  2     22         â€“ 5     25        50                15                  4     60         â€“ 1      1          4                18                  4     72            2      4         16                 22                  3     66            6      36        108                25                  1     25           9      81          81                30                  1     30          14    196        196      N = $âˆ‘{\text{f}}_{\text{i}}\text{= 20}$  $âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}\text{= 320}$  âˆ‘ fi${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$  = 700    The sum of the products of the squares of the deviations and the frequencies is equal to 700.   Variance ( Ïƒ2 )  = $\frac{\text{1}}{\text{N}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$                             = $\frac{\text{700}}{\text{20}}$                        = 35 âˆ´ The variance of the given frequency distribution = 35.   If each item is increased (or decreased) by a fixed constant, then the variance does not alter. In other words, variance is independent of change of origin. For a series in the form a, a+d, a+2d, a+3d,...., a + (n - 1)d, the variance is given by: Ïƒ2  $\mathrm{\text{}}$$\mathrm{\text{=}}{\text{(c.d.)}}^{2}\text{(}\frac{{\text{n}}^{2}\mathrm{\text{- 1}}}{12}\text{)}$, where c.d. is the common difference of the series and n is the number of terms in the series.

#### Summary

Limitation of mean deviation: If the degree of the variability in a data is high, then the median cannot be considered as a central tendency. The sum of the deviations from the mean is more than the sum of the deviations from the median. Thus, mean deviation about the mean is not more scientific. For calculating mean deviation, we consider the absolute values of the deviations. Thus, we cannot carry out further algebraic operations. Therefore, mean deviation about the mean or median is not sufficient to measure the dispersion. So, we need to extend the measure of dispersion to an appropriate measure of dispersion. Variance is such a measure of dispersion.   To eliminate the negative signs of the deviations, we take their absolute values. An alternative method to eliminate the negative sign of the deviations is to take the square of the deviations. Let n observations be x1, x2, x3,...., xn. Mean ($\stackrel{_}{\text{x}}$) = (x11+ x22+ x33+....+ xnn)/n Deviations of the observations from the mean: (x1 - $\stackrel{_}{\text{x}}$), (x2 - $\stackrel{_}{\text{x}}$), (x3 - $\stackrel{_}{\text{x}}$),...., (xn - $\stackrel{_}{\text{x}}$) (x1 - $\stackrel{_}{\text{x}}$)2, (x2 - $\stackrel{_}{\text{x}}$)2, (x3 - $\stackrel{_}{\text{x}}$)2,...., (xn - $\stackrel{_}{\text{x}}$)2 (x1 - $\stackrel{_}{\text{x}}$)2, (x2 - $\stackrel{_}{\text{x}}$)2, (x3 - $\stackrel{_}{\text{x}}$)2,...., (xn - $\stackrel{_}{\text{x}}$)2 = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$      The sum of the squares of the deviations is equal to 0, which implies that each deviation is equal to 0. There is no dispersion since all the observations are equal to the mean. The sum of the squares of the deviations is small, which implies that the observations are closer to the mean. Hence, the degree of dispersion is low. If the sum of the squares of the deviations is large, then the degree of dispersion is high. Mean of the squares of the deviations = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$      The variance of a data is the mean of the squares of the deviations of the observations from the mean. Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$       This is the formula used to calculate the variance of an ungrouped data.   Variance of a discrete frequency distribution  Observations (xi)  x1  x2  x3  ...   xn  Frequency (fi)  f1  f2  f3  ...   fn Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$     Sum of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}$ fi   Variance of a continuous frequency distribution  Mid-values (xi)  x1  x2  x3  ...   xn  Frequency (fi)  f1  f2  f3  ...   fn Ïƒ2 = $\frac{\text{1}}{\text{n}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$     Sum of the frequencies: N = $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}$ fi   Consider the given frequency distribution table of trees planted in 20 schools in a tree plantation programme.  Trees Planted  9  11  15  18  22  25  30  No. of Schools  5  2  4  4  3  1  1   Variance of the given data:   Mean of the given distribution         Trees Planted (xi)            No. of Schools (fi)    fixi                 9                  5     45                11                  2     22                15                  4     60                18                  4     72                 22                  3     66                25                  1     25                30                  1     30      N = $âˆ‘{\text{f}}_{\text{i}}\text{= 20}$  $âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}\text{= 320}$ Mean ($\stackrel{_}{\text{x}}$) = $\frac{âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}}{\text{N}}$              = $\frac{\text{320}}{\text{20}}$ = 16   Deviations of the numbers of trees planted from the mean and the squares of the deviations are calculated.   The product of the squares of the deviations and their frequencies are obtained         Trees Planted (xi)            No. of Schools (fi)    fixi  $\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{) =}{\text{x}}_{\text{i}}\text{- 16}$  ${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$  fi${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$                 9                  5     45         â€“ 7     49       245                11                  2     22         â€“ 5     25        50                15                  4     60         â€“ 1      1          4                18                  4     72            2      4         16                 22                  3     66            6      36        108                25                  1     25           9      81          81                30                  1     30          14    196        196      N = $âˆ‘{\text{f}}_{\text{i}}\text{= 20}$  $âˆ‘{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}\text{= 320}$  âˆ‘ fi${\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$  = 700    The sum of the products of the squares of the deviations and the frequencies is equal to 700.   Variance ( Ïƒ2 )  = $\frac{\text{1}}{\text{N}}$ $\underset{\text{i = 1}}{\overset{\text{n}}{âˆ‘}}{{\text{f}}_{\text{i}}\text{(}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{)}}^{\text{2}}$                             = $\frac{\text{700}}{\text{20}}$                        = 35 âˆ´ The variance of the given frequency distribution = 35.   If each item is increased (or decreased) by a fixed constant, then the variance does not alter. In other words, variance is independent of change of origin. For a series in the form a, a+d, a+2d, a+3d,...., a + (n - 1)d, the variance is given by: Ïƒ2  $\mathrm{\text{}}$$\mathrm{\text{=}}{\text{(c.d.)}}^{2}\text{(}\frac{{\text{n}}^{2}\mathrm{\text{- 1}}}{12}\text{)}$, where c.d. is the common difference of the series and n is the number of terms in the series.

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