Question 1: Calculate the mean deviation from the median of the following frequency distribution:

 Heights in inches 58 59 60 61 62 63 64 65 66 No of students 15 20 32 35 53 22 20 10 8

We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array} {| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 61 | & f_i |d_i| \\ \hline 58 & 15 & 15 & 3 & 45 \\ \hline 59 & 20 & 35 & 2 & 40 \\ \hline 60 & 32 & 67 & 1 & 32 \\ \hline 61 & 35 & 102 & 0 & 0 \\ \hline 62 & 35 & 137 & 1 & 35 \\ \hline 63 & 22 & 159 & 2 & 44 \\ \hline 64 & 20 & 179 & 3 & 60 \\ \hline 65 & 10 & 189 & 4 & 40 \\ \hline 66 & 8 & 197 & 5 & 40 \\ \hline & N = \sum f_i = 197 & & & \sum_{i=1}^{n} f_i |d_i|= 336 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 197 \Rightarrow \frac{N}{2} = \frac{197}{2} = 98.5$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 98.5 \text{ is } 102.$

$\displaystyle \text{The corresponding value of } x \text{ is } 61.$

$\displaystyle \text{Therefore, the median } = 61$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{197} \times 336 = 1.7055$

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Question 2: The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:

 Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12

Compute the mean deviation about median.

We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 4 | & f_i |d_i| \\ \hline 0 & 14 & 14 & 4 & 56 \\ \hline 1 & 21 & 35 & 3 & 63 \\ \hline 2 & 25 & 60 & 2 & 50 \\ \hline 3 & 43 & 103 & 1 & 43 \\ \hline 4 & 51 & 154 & 0 & 0 \\ \hline 5 & 40 & 194 & 1 & 40 \\ \hline 6 & 39 & 233 & 2 & 78 \\ \hline 7 & 12 & 245 & 3 & 36 \\ \hline & N = \sum f_i = 245 & & & \sum_{i=1}^{n} f_i |d_i| = 366 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 245 \Rightarrow \frac{N}{2} = \frac{245}{2} = 122.5$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 122.5 \text{ is } 154.$

$\displaystyle \text{The corresponding value of } x \text{ is } 4.$

$\displaystyle \text{Therefore, the median } = 4$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{245} \times 366 = 1.493$

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Question 3: Calculate the mean deviation from the median of the following frequency distribution:

 $\displaystyle x_i$ 5 7 9 11 13 15 17 $\displaystyle f_i$ 2 4 6 8 10 12 8

We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 13 | & f_i |d_i| \\ \hline 5 & 2 & 2 & 8 & 16 \\ \hline 7 & 4 & 6 & 6 & 24 \\ \hline 9 & 6 & 12 & 4 & 24 \\ \hline 11 & 8 & 20 & 2 & 16 \\ \hline 13 & 10 & 30 & 0 & 0 \\ \hline 15 & 12 & 42 & 2 & 24 \\ \hline 17 & 8 & 50 & 4 & 32 \\ \hline & N = \sum f_i = 50 & & & \sum_{i=1}^{n} f_i |d_i| = 136 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \Rightarrow \frac{N}{2} = \frac{50}{2} = 25$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 25 \text{ is } 30.$

$\displaystyle \text{The corresponding value of } x \text{ is } 13.$

$\displaystyle \text{Therefore, the median } = 13$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{50} \times 136 = 2.72$

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Question 4: Find the mean deviation from the mean for the data:

(i)

 $\displaystyle x_i$ 5 7 9 10 12 15 $\displaystyle f_i$ 8 6 2 2 2 6

(ii)

 $\displaystyle x_i$ 5 10 15 20 25 $\displaystyle f_i$ 7 4 6 3 5

(iii)

 $\displaystyle x_i$ 10 30 50 70 90 $\displaystyle f_i$ 4 24 28 16 8

(iv)

 $\displaystyle x_i$ 20 21 22 23 24 $\displaystyle f_i$ 6 4 5 1 4

(v)

 $\displaystyle x_i$ 1 3 5 7 9 11 13 15 $\displaystyle f_i$ 3 3 4 1 47 4 3 4

(i) Calculation of mean deviation about mean.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & f_i x_i & |x_i - \overline{x} | & f_i |x_i - 9| \\ \hline 5 & 8 & 40 & 4 & 32 \\ \hline 7 & 6 & 42 & 2 & 12 \\ \hline 9 & 2 & 18 & 0 & 0 \\ \hline 10 & 2 & 20 & 1 & 2 \\ \hline 12 & 2 & 24 & 3 & 6 \\ \hline 15 & 6 & 90 & 6 & 36 \\ \hline & N = \sum f_i = 26 & \sum_{i=1}^{n} f_i x_i = 234 & & \sum_{i=1}^{n} f_i |x_i - 9| = 88 \\ \hline \end{array}$

$\displaystyle \text{Mean } = \overline{X} = \frac{1}{N} \Bigg( \sum_{i=1}^{n} f_ix_i \Bigg) = \frac{234}{26} = 9$

$\displaystyle \text{Mean Deviation } = M.D. = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - 9| = \frac{1}{26} \times 88 = 3.39$

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(ii) Calculation of mean deviation about mean.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & f_i x_i & |x_i - \overline{x} | & f_i |x_i - 14| \\ \hline 5 & 7 & 35 & 9 & 63 \\ \hline 10 & 4 & 40 & 4 & 16 \\ \hline 15 & 6 & 90 & 1 & 6 \\ \hline 20 & 3 & 60 & 6 & 18 \\ \hline 25 & 5 & 125 & 11 & 55 \\ \hline & N = \sum f_i = 25 & \sum_{i=1}^{n} f_i x_i = 350 & & \sum_{i=1}^{n} f_i |x_i - 14| = 158 \\ \hline \end{array}$

$\displaystyle \text{Mean } = \overline{X} = \frac{1}{N} \Bigg( \sum_{i=1}^{n} f_ix_i \Bigg) = \frac{350}{25} = 14$

$\displaystyle \text{Mean Deviation } = M.D. = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - 14| = \frac{1}{25} \times 158 = 6.32$

$\\$

(iii) Calculation of mean deviation about mean.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & f_i x_i & |x_i - \overline{x} | & f_i |x_i - 50| \\ \hline 10 & 4 & 40 & 40 & 160 \\ \hline 30 & 24 & 720 & 20 & 480 \\ \hline 50 & 28 & 1400 & 0 & 0 \\ \hline 70 & 16 & 1120 & 20 & 320 \\ \hline 90 & 8 & 720 & 40 & 320 \\ \hline & N = \sum f_i = 80 & \sum_{i=1}^{n} f_i x_i = 4000 & & \sum_{i=1}^{n} f_i |x_i - 50| = 1280 \\ \hline \end{array}$

$\displaystyle \text{Mean } = \overline{X} = \frac{1}{N} \Bigg( \sum_{i=1}^{n} f_ix_i \Bigg) = \frac{4000}{80} = 50$

$\displaystyle \text{Mean Deviation } = M.D. = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - 50| = \frac{1}{80} \times 1280 = 16$

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(iv) Calculation of mean deviation about mean.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & f_i x_i & |x_i - \overline{x} | & f_i |x_i - 21.65| \\ \hline 20 & 6 & 120 & 1.65 & 9.9 \\ \hline 21 & 4 & 84 & 0.65 & 2.6 \\ \hline 22 & 5 & 110 & 0.35 & 1.75 \\ \hline 23 & 1 & 23 & 1.35 & 1.35 \\ \hline 24 & 4 & 96 & 2.35 & 9.4 \\ \hline & N = \sum f_i = 20 & \sum_{i=1}^{n} f_i x_i = 433 & & \sum_{i=1}^{n} f_i |x_i - 21.65| = 25 \\ \hline \end{array}$

$\displaystyle \text{Mean } = \overline{X} = \frac{1}{N} \Bigg( \sum_{i=1}^{n} f_ix_i \Bigg) = \frac{433}{20} = 21.65$

$\displaystyle \text{Mean Deviation } = M.D. = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - 21.65| = \frac{1}{20} \times 25 = 1.25$

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(v) Calculation of mean deviation about mean.

$\displaystyle \begin{array}{| c| c| c| c | c | } \hline x_i & f_i & f_i x_i & |x_i - \overline{x} | & f_i |x_i - 8| \\ \hline 1 & 3 & 3 & 7 & 21 \\ \hline 3 & 3 & 9 & 5 & 15 \\ \hline 5 & 4 & 20 & 3 & 12 \\ \hline 7 & 14 & 98 & 1 & 14 \\ \hline 9 & 7 & 63 & 1 & 7 \\ \hline 11 & 4 & 44 & 3 & 12 \\ \hline 13 & 3 & 39 & 5 & 15 \\ \hline 15 & 4 & 60 & 7 & 28 \\ \hline & N = \sum f_i = 42 & \sum_{i=1}^{n} f_i x_i = 336 & & \sum_{i=1}^{n} f_i |x_i - 8| = 124 \\ \hline \end{array}$

$\displaystyle \text{Mean } = \overline{X} = \frac{1}{N} \Bigg( \sum_{i=1}^{n} f_ix_i \Bigg) = \frac{336}{42} = 8$

$\displaystyle \text{Mean Deviation } = M.D. = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - 8| = \frac{1}{42} \times 124 = 2.95$

Question 5: Find the mean deviation from the median for the data:

(i)

 $\displaystyle x_i$ 15 21 27 30 35 $\displaystyle f_i$ 3 5 6 7 8

(ii)

 $\displaystyle x_i$ 74 89 42 54 91 94 35 $\displaystyle f_i$ 20 12 2 4 5 3 4

(iii)

 $\displaystyle x_i$ 10 11 12 14 15 $\displaystyle f_i$ 2 3 8 3 4

(i) We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array} {| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 30 | & f_i |d_i| \\ \hline 15 & 3 & 3 & 15 & 45 \\ \hline 21 & 5 & 8 & 9 & 45 \\ \hline 27 & 6 & 14 & 3 & 18 \\ \hline 30 & 7 & 21 & 0 & 0 \\ \hline 35 & 8 & 29 & 5 & 40 \\ \hline & N = \sum f_i = 29 & & & \sum_{i=1}^{n} f_i |d_i|= 148 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 29 \Rightarrow \frac{N}{2} = \frac{29}{2} = 14.5$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 14.5 \text{ is } 30.$

$\displaystyle \text{The corresponding value of } x \text{ is } 30.$

$\displaystyle \text{Therefore, the median } = 30$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{29} \times 148 = 5.10$

$\\$

(ii) We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array} {| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 74 | & f_i |d_i| \\ \hline 35 & 4 & 4& 39 & 156 \\ \hline 42 & 2 & 6 & 32 & 64 \\ \hline 54 & 4 & 10 & 20 & 80 \\ \hline 74 & 20 & 30 & 0 & 0 \\ \hline 89 & 12 & 42 & 15 & 180 \\ \hline 91 & 5 & 47 & 17 & 85 \\ \hline 94 & 3 & 50 & 20 & 60 \\ \hline & N = \sum f_i = 50 & & & \sum_{i=1}^{n} f_i |d_i|= 625 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 50 \Rightarrow \frac{N}{2} = \frac{50}{2} = 25$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 25 \text{ is } 30.$

$\displaystyle \text{The corresponding value of } x \text{ is } 74.$

$\displaystyle \text{Therefore, the median } = 74$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{50} \times 775 = 12.5$

$\\$

(iii) We have to calculate mean deviation about median. Hence, first we calculate median.

$\displaystyle \begin{array} {| c| c| c| c | c | } \hline x_i & f_i & \text{Cum. Frequency} & |d_i| = |x_i - 12 | & f_i |d_i| \\ \hline 10 & 2 & 2 & 2 & 4 \\ \hline 11 & 3 & 5 & 1 & 3 \\ \hline 12 & 8 & 13 & 0 & 0 \\ \hline 14 & 3 & 16 & 2 & 6 \\ \hline 15 & 4 & 20 & 3 & 12 \\ \hline & N = \sum f_i = 20 & & & \sum_{i=1}^{n} f_i |d_i|= 25 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } N = 20 \Rightarrow \frac{N}{2} = \frac{20}{2} = 10$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} \text{ is } 10 \text{ is } 13.$

$\displaystyle \text{The corresponding value of } x \text{ is } 12.$

$\displaystyle \text{Therefore, the median } = 12$

$\displaystyle \text{Therefore Mean Deviation }= \frac{1}{N} \sum_{i = 1}^{n} f_i |d_i| = \frac{1}{20} \times 25 = 1.25$