Notes On Mean Deviation about Median for Grouped Data - CBSE Class 11 Maths
Data can be grouped in two ways: Discrete frequency distribution Continuous frequency distribution   Mean deviation about the median for a discrete frequency distribution.     Height in cms (xi)    Number of Students (fi)           147             8           148            12           150            15           152            10           155              5   Total number of Students (N) = 50    Mean deviation about the median of the given data:   Median height of the data –     Height in cms (xi)    Number of Students (fi)  Cumulative Frequency            147             8              8           148            12              20           150            15              35           152            10              45           155              5              50   Total number of Students (N) = 50    The number of students is 50, which is an even number.   Median of the given data (M) = Average of the 25th and 26th observations Median of the given data (M) = Average of the 25th and 26th observations = $\frac{{\text{25}}^{\text{th}}\text{observation +}{\text{26}}^{\text{th}}\text{observation}}{\text{2}}$ = $\frac{\text{150 + 150}}{\text{2}}$ = 150 cm   Find the absolute values of the deviations of the heights from the median 150 and the products of the deviations and their frequencies.     Height in cms (xi)    Number of Students (fi)  Cumulative Frequency  $\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{| = |}{\text{x}}_{\text{i}}\text{- 150|}$           147             8              8          3     24           148            12              20          2     24           150            15              35          0       0           152            10              45          2     20           155              5              50          5     25   Total number of Students (N) = 50   $\sum _{\text{i = 1}}^{\text{5}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M|= 93   The sum of the products of the deviations and their frequencies is equal to 93.   Mean deviation about the median = MD (M) = $\frac{\text{1}}{\text{n}}$ $\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M | = $\frac{\text{93}}{\text{50}}$ = 1.86 cm.   Therefore, the mean deviation about the median is equal to 1.86 cm.   Mean deviation about the median of a continuous frequency distribution:   Consider the frequency distribution table which shows the grouped data of a tree plantation programme conducted in 60 schools.     Number of Trees Planted (Class interval)    Number of Schools (fi)           5 - 25                12          25 - 45                 8          45 - 65                14          65 - 85                20          85 - 105                  6   Total number of Schools (N) = 60     Median of the given data:     Number of Trees Planted (Class interval)    Number of Schools (fi)  Cumulative Frequency (cf)           5 - 25                12         12          25 - 45                 8         20          45 - 65                                               14                            34                                 65 - 85                20         54          85 - 105                  6         60   Total number of Schools (N) = 60     $\frac{\text{n}}{\text{2}}\text{=}\frac{\text{60}}{\text{2}}$ = 30 Hence, 45 – 65 is the median class. Median = $\mathrm{\text{l + (}}\frac{\frac{\text{n}}{\text{2}}\text{- cf}}{\text{f}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}×\mathrm{\text{h}}$ Median (M) =  = = $\mathrm{\text{45 + (}}\frac{\text{10}}{\text{14}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}×\mathrm{\text{20}}$ = $\mathrm{\text{45 + (}}\frac{\text{200}}{\text{14}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}$ = $\mathrm{\text{45 + 14,28}}$ = $\mathrm{\text{59.28}}$   Mid-values of the class intervals and absolute values of the deviations of the mid-values from the median $\mathrm{\text{59.28}}$ are calculated.     Number of Trees Planted (Class interval)    Number of Schools (fi)  Cumulative Frequency (cf)  $\text{|}{\text{x}}_{\text{i}}\text{- M| = |}{\text{x}}_{\text{i}}\text{- 59.28|}$  ${\text{f}}_{\text{i}}\text{|}{\text{x}}_{\text{i}}\text{- M|}$           5 - 25                12         12          44.28  531.36          25 - 45                 8         20          24.28  194.24          45 - 65                                               14                            34                                 4.28  59.96          65 - 85                20         54          15.28  314.4          85 - 105                  6         60          35.28  214.32   Total number of Schools (N) = 60    ∑ ${\text{f}}_{\text{i}}\text{|}{\text{x}}_{\text{i}}\text{- M|}$ = 1314.24   Products of the deviations and their frequencies, the sum of the products of the deviations and their frequencies is equal to 1314.24 Mean deviation about the median = MD (M) = = $\frac{\text{1}}{\text{N}}$ $\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M| =   $\frac{\text{1314.24}}{\text{60}}$   = 21.904 Therefore, the mean deviation about the median is 21.904

#### Summary

Data can be grouped in two ways: Discrete frequency distribution Continuous frequency distribution   Mean deviation about the median for a discrete frequency distribution.     Height in cms (xi)    Number of Students (fi)           147             8           148            12           150            15           152            10           155              5   Total number of Students (N) = 50    Mean deviation about the median of the given data:   Median height of the data –     Height in cms (xi)    Number of Students (fi)  Cumulative Frequency            147             8              8           148            12              20           150            15              35           152            10              45           155              5              50   Total number of Students (N) = 50    The number of students is 50, which is an even number.   Median of the given data (M) = Average of the 25th and 26th observations Median of the given data (M) = Average of the 25th and 26th observations = $\frac{{\text{25}}^{\text{th}}\text{observation +}{\text{26}}^{\text{th}}\text{observation}}{\text{2}}$ = $\frac{\text{150 + 150}}{\text{2}}$ = 150 cm   Find the absolute values of the deviations of the heights from the median 150 and the products of the deviations and their frequencies.     Height in cms (xi)    Number of Students (fi)  Cumulative Frequency  $\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{| = |}{\text{x}}_{\text{i}}\text{- 150|}$           147             8              8          3     24           148            12              20          2     24           150            15              35          0       0           152            10              45          2     20           155              5              50          5     25   Total number of Students (N) = 50   $\sum _{\text{i = 1}}^{\text{5}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M|= 93   The sum of the products of the deviations and their frequencies is equal to 93.   Mean deviation about the median = MD (M) = $\frac{\text{1}}{\text{n}}$ $\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M | = $\frac{\text{93}}{\text{50}}$ = 1.86 cm.   Therefore, the mean deviation about the median is equal to 1.86 cm.   Mean deviation about the median of a continuous frequency distribution:   Consider the frequency distribution table which shows the grouped data of a tree plantation programme conducted in 60 schools.     Number of Trees Planted (Class interval)    Number of Schools (fi)           5 - 25                12          25 - 45                 8          45 - 65                14          65 - 85                20          85 - 105                  6   Total number of Schools (N) = 60     Median of the given data:     Number of Trees Planted (Class interval)    Number of Schools (fi)  Cumulative Frequency (cf)           5 - 25                12         12          25 - 45                 8         20          45 - 65                                               14                            34                                 65 - 85                20         54          85 - 105                  6         60   Total number of Schools (N) = 60     $\frac{\text{n}}{\text{2}}\text{=}\frac{\text{60}}{\text{2}}$ = 30 Hence, 45 – 65 is the median class. Median = $\mathrm{\text{l + (}}\frac{\frac{\text{n}}{\text{2}}\text{- cf}}{\text{f}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}×\mathrm{\text{h}}$ Median (M) =  = = $\mathrm{\text{45 + (}}\frac{\text{10}}{\text{14}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}×\mathrm{\text{20}}$ = $\mathrm{\text{45 + (}}\frac{\text{200}}{\text{14}}\mathrm{\text{)}}\mathrm{\text{}}\mathrm{\text{}}$ = $\mathrm{\text{45 + 14,28}}$ = $\mathrm{\text{59.28}}$   Mid-values of the class intervals and absolute values of the deviations of the mid-values from the median $\mathrm{\text{59.28}}$ are calculated.     Number of Trees Planted (Class interval)    Number of Schools (fi)  Cumulative Frequency (cf)  $\text{|}{\text{x}}_{\text{i}}\text{- M| = |}{\text{x}}_{\text{i}}\text{- 59.28|}$  ${\text{f}}_{\text{i}}\text{|}{\text{x}}_{\text{i}}\text{- M|}$           5 - 25                12         12          44.28  531.36          25 - 45                 8         20          24.28  194.24          45 - 65                                               14                            34                                 4.28  59.96          65 - 85                20         54          15.28  314.4          85 - 105                  6         60          35.28  214.32   Total number of Schools (N) = 60    ∑ ${\text{f}}_{\text{i}}\text{|}{\text{x}}_{\text{i}}\text{- M|}$ = 1314.24   Products of the deviations and their frequencies, the sum of the products of the deviations and their frequencies is equal to 1314.24 Mean deviation about the median = MD (M) = = $\frac{\text{1}}{\text{N}}$ $\sum _{\text{i = 1}}^{\text{n}}{{\text{f}}_{\text{i}}\text{| x}}_{\text{i}}$  - M| =   $\frac{\text{1314.24}}{\text{60}}$   = 21.904 Therefore, the mean deviation about the median is 21.904

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