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

Answer:

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.

Answer:

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

Answer:

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

Answer:

(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

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(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

Answer:

(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

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(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

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(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