Notes On Mean Deviation for Ungrouped Data - CBSE Class 11 Maths
The deviation of an observation x from a fixed value k is equal to x - k. Commonly, we take the central tendency as the value of k. To find the dispersion, we need to find the deviations about the central tendency, like the mean or the median. Absolute measure is the mean of the deviations about a central tendency. Half of the deviations are negative and the others are positive. The sum of the deviations about the mean is zero. Proof: Let the observations be x1,x2,x3,.....,xn. Mean = (x1,x2,x3,.....,xn)/n Sum of the deviations = (x1 - (x1,x2,x3,.....,xn)/n) + (x2 - (x1,x2,x3,.....,xn)/n) +.... + (xn - (x1,x2,x3,.....,xn)/n) Let x1,x2,x3,.....,xn = k Sum of the deviations = (x1 - k/n + (x2 - k/n) +.... + (xn - k/n). = [(nx1 - k) + (nx2 - k) +.... + (nxn - k)]/n. = [n(x1+x2+x3+.....+xn) - nk]/n = (nk - nk)/n = 0 Hence, the sum of the deviations about the mean is zero. Mean of the deviations = Sum of the deviations/ Number of deviations = 0/n = 0 Mean deviation about the mean of an ungrouped data: Observations in the ungrouped data: x1,x2,x3,.....,xn To find the mean deviation about the mean: Step 1: Find the mean of the given observations. Let it be k. Step2: Find the deviation of each observation from the mean. x1-k,x2-k,x3-k,.....,xn-k Step 3: Find the absolute values of the deviations. |x1-k|,|x2-k|,|x3-k|,.....,|xn-k| Step 4: Find the mean of the absolute values of the deviations. |x1-k|,|x2-k|,|x3-k|,.....,|xn-k|/n Mean deviation about the mean = |x1-k|+|x2-k|+|x3-k|+.....+|xn-k|/n Mean deviation about the mean = MD ($\stackrel{_}{\text{x}}$) = $\frac{\text{1}}{\text{n}}$$\sum _{\text{i=1}}^{\text{n}}\text{}$$\left|{\text{x}}_{i}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}\stackrel{_}{\text{x}}\right|$, where $\stackrel{_}{\text{x}}$ = mean. Mean deviation about the median = MD (M) = $\frac{\text{1}}{\text{n}}$$\sum _{\text{i=1}}^{\text{n}}$$\left|{\text{x}}_{i}\mathrm{\text{}}\mathrm{\text{-M}}\right|$, where M = median. Find the mean deviation about the mean for the following data: 56, 65, 41, 63, 51, 59, 37, 46, 68 Mean of the given data ($\stackrel{_}{\text{x}}$) = (56+65+41+63+51+59+37+46+68)/9 = 486/9 = 54 The sum of the absolute values of the deviations of the observations from the mean $\stackrel{_}{\text{x}}$, i.e. xi - $\stackrel{_}{\text{x}}$ = |56 - 54| + |65 - 54| + |41 - 54| + |63 - 54| + |51 - 54| + |59 - 54| + |37 - 54| + |46 - 54| + |68 - 54| = 2 + 11 + 13 + 9 + 3 + 5 + 17 + 8 + 14 = 82 Mean deviation about mean =$\frac{\sum _{\text{i = 1}}^{\text{9}}\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{|}}{\text{9}}$  = 82/9 = 9.11 The marks scored by 9 students in a mathematics test are: 50, 69, 20, 33, 53, 39, 40, 65, 59 Ascending order of observations: 20, 33, 39, 40, 50, 53, 59, 65, 69 Median (M) = ((n+1)/2)th observation = ((9+1)/2)th term observation = 5th observation = 50 Sum of the absolute values of the deviations of the observations from median = |20 - 50| + |33 - 50| + |39 - 50| + |40 - 50| + |50 - 50| + |53 - 50| + |59 - 50| + |65 - 50| + |69 - 50| = 30 + 17 + 11 + 10 + 0 + 3 +9 + 15 + 19 = 114 Mean Deviation about Median  =$\frac{\sum _{\text{i = 1}}^{\text{9}}\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{|}}{\text{9}}$   = 114/9 = 12.66 Hence, the mean deviation about the median is 12.66

#### Summary

The deviation of an observation x from a fixed value k is equal to x - k. Commonly, we take the central tendency as the value of k. To find the dispersion, we need to find the deviations about the central tendency, like the mean or the median. Absolute measure is the mean of the deviations about a central tendency. Half of the deviations are negative and the others are positive. The sum of the deviations about the mean is zero. Proof: Let the observations be x1,x2,x3,.....,xn. Mean = (x1,x2,x3,.....,xn)/n Sum of the deviations = (x1 - (x1,x2,x3,.....,xn)/n) + (x2 - (x1,x2,x3,.....,xn)/n) +.... + (xn - (x1,x2,x3,.....,xn)/n) Let x1,x2,x3,.....,xn = k Sum of the deviations = (x1 - k/n + (x2 - k/n) +.... + (xn - k/n). = [(nx1 - k) + (nx2 - k) +.... + (nxn - k)]/n. = [n(x1+x2+x3+.....+xn) - nk]/n = (nk - nk)/n = 0 Hence, the sum of the deviations about the mean is zero. Mean of the deviations = Sum of the deviations/ Number of deviations = 0/n = 0 Mean deviation about the mean of an ungrouped data: Observations in the ungrouped data: x1,x2,x3,.....,xn To find the mean deviation about the mean: Step 1: Find the mean of the given observations. Let it be k. Step2: Find the deviation of each observation from the mean. x1-k,x2-k,x3-k,.....,xn-k Step 3: Find the absolute values of the deviations. |x1-k|,|x2-k|,|x3-k|,.....,|xn-k| Step 4: Find the mean of the absolute values of the deviations. |x1-k|,|x2-k|,|x3-k|,.....,|xn-k|/n Mean deviation about the mean = |x1-k|+|x2-k|+|x3-k|+.....+|xn-k|/n Mean deviation about the mean = MD ($\stackrel{_}{\text{x}}$) = $\frac{\text{1}}{\text{n}}$$\sum _{\text{i=1}}^{\text{n}}\text{}$$\left|{\text{x}}_{i}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}\stackrel{_}{\text{x}}\right|$, where $\stackrel{_}{\text{x}}$ = mean. Mean deviation about the median = MD (M) = $\frac{\text{1}}{\text{n}}$$\sum _{\text{i=1}}^{\text{n}}$$\left|{\text{x}}_{i}\mathrm{\text{}}\mathrm{\text{-M}}\right|$, where M = median. Find the mean deviation about the mean for the following data: 56, 65, 41, 63, 51, 59, 37, 46, 68 Mean of the given data ($\stackrel{_}{\text{x}}$) = (56+65+41+63+51+59+37+46+68)/9 = 486/9 = 54 The sum of the absolute values of the deviations of the observations from the mean $\stackrel{_}{\text{x}}$, i.e. xi - $\stackrel{_}{\text{x}}$ = |56 - 54| + |65 - 54| + |41 - 54| + |63 - 54| + |51 - 54| + |59 - 54| + |37 - 54| + |46 - 54| + |68 - 54| = 2 + 11 + 13 + 9 + 3 + 5 + 17 + 8 + 14 = 82 Mean deviation about mean =$\frac{\sum _{\text{i = 1}}^{\text{9}}\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{|}}{\text{9}}$  = 82/9 = 9.11 The marks scored by 9 students in a mathematics test are: 50, 69, 20, 33, 53, 39, 40, 65, 59 Ascending order of observations: 20, 33, 39, 40, 50, 53, 59, 65, 69 Median (M) = ((n+1)/2)th observation = ((9+1)/2)th term observation = 5th observation = 50 Sum of the absolute values of the deviations of the observations from median = |20 - 50| + |33 - 50| + |39 - 50| + |40 - 50| + |50 - 50| + |53 - 50| + |59 - 50| + |65 - 50| + |69 - 50| = 30 + 17 + 11 + 10 + 0 + 3 +9 + 15 + 19 = 114 Mean Deviation about Median  =$\frac{\sum _{\text{i = 1}}^{\text{9}}\text{|}{\text{x}}_{\text{i}}\text{-}\stackrel{_}{\text{x}}\text{|}}{\text{9}}$   = 114/9 = 12.66 Hence, the mean deviation about the median is 12.66

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