Question 1: Compute the mean deviation from the median of the following distribution: 

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 10 20 5 10

Answer:  

Calculation of Mean Deviation from Median 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & \text{Cum. Freq.} &  |d_i| = |x_i - 25 |  &  f_i |d_i|  \\ \hline  0-10 &  5 & 5 & 5 & 20 & 100 \\   \hline  10-20 &  15 & 10 & 15 & 10 & 100 \\   \hline  20-30 & 25 & 20 & 35 & 0 & 0 \\  \hline  30-40 & 35 & 5 & 40 & 10 & 50 \\   \hline  40-50 & 45 & 10 & 50 & 20 & 200 \\   \hline   & & N = \sum f_i = 50 &  &  & \sum_{i=1}^{5} f_i |d_i|= 450 \\  \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} = 25 \text{ is } 35.

\displaystyle \text{The corresponding value of class } \text{ is } 20-30

\displaystyle \text{Therefore the median class } \text{ is } 20-30

\displaystyle \therefore l = 20, \hspace{0.5cm} f = 20,  \hspace{0.5cm} F = 15, \hspace{0.5cm} N = 50, \hspace{0.5cm} h = 10

\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h 

\displaystyle = 20 + \frac{\Big( \frac{50}{2} - 15 \Big) }{20} \times 10 

\displaystyle = 20 + \frac{\Big( 25 - 15 \Big) }{20} \times 10 

\displaystyle = 25 

\displaystyle \text{Mean deviation from the median } =  \frac{\sum_{i=1}^5 f_i|d_i|}{N} = \frac{450}{50} = 9

\\

Question 2: Find the mean deviation from the mean from the following data:

(i) 

Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Frequency 4 8 9 10 7 5 4 3

(ii) 

Classes 95-105 105-115 115-125 125-135 135-145 145-155
Frequency 9 13 16 26 30 12

(iii) 

Classes 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 6 8 14 16 4 2

Answer:

(i)

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & f_ix_i &  |x_i - \overline{X}|   &  f_i|x_i - \overline{X}|  \\ & & & &  = |x_i - 358 |&  \\ \hline  0-100 &  50 & 4 & 200 & 308 & 1232 \\   \hline  100-200 &  150 & 8 & 1200 & 208 & 1664 \\   \hline  200-300 & 250 & 9 & 2250 & 108 & 972 \\  \hline  300-400 & 350 & 10 & 3500 & 8 & 80 \\   \hline  400-500 & 450 & 7 & 3150 & 92 & 644 \\   \hline  500-600 & 550 & 5 & 2750 & 192 & 960 \\ \hline  600-700 & 650 & 4 & 2600 & 292 & 1168 \\   \hline  700-800 & 750 & 3 & 2250 & 392 & 1176 \\   \hline   & N=  &  & \sum_{i=1}^{8} f_ix_i  &  & \sum_{i=1}^{8} f_i |x_i - \overline{X}| \\   &  \sum f_i = 50 &  & = 17900 & & = 7896 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 17900

\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{17900}{50} = 358

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 7896 \text{ and } N = \sum f_i = 50

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{7896}{50} = 157.92

\\

(ii)

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & f_ix_i &  |x_i - \overline{X}|   &  f_i|x_i - \overline{X}|  \\ & & & & = |x_i - 128.58 | &  \\ \hline  95-105 &  100 & 9 & 900 & 28.58 & 257.22 \\   \hline  105-115 &  110 & 13 & 1430 & 18.58 & 241.54 \\   \hline  115-125 & 120 & 16 & 1920 & 8.58 & 137.28 \\ \hline  125-135 & 130 & 26 & 3380 & 1.42 & 36.92 \\   \hline  135-145 & 140 & 30 & 4200 & 11.42 & 342.6 \\   \hline  145-155 & 150 & 12 & 1800 & 21.42 & 257.04 \\ \hline   & N=  &  & \sum_{i=1}^{6} f_ix_i  &  & \sum_{i=1}^{6} f_i |x_i - \overline{X}|   \\   &  \sum f_i = 106 &  & = 13630 & & = 1272.6 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 106 \text{ and } \sum f_ix_i = 13630

\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{13630}{106} = 128.58

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 1272.6 \text{ and } N = \sum f_i = 106

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{1272.6}{106} = 12.005

\\

(iii)

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & f_ix_i &  |x_i - \overline{X}|   &  f_i|x_i - \overline{X}|  \\  & & & & = |x_i - 358 | &  \\ \hline  0-10 &  5 & 6 & 30 & 22 & 132\\   \hline  10-20 &  15 & 8 & 120 & 12 & 96 \\   \hline  20-30 & 25 & 14 & 350 & 2 & 28\\  \hline  30-40 & 35 & 16 & 560 & 8 & 128 \\   \hline  40-50 & 45 & 4 & 180 & 18 & 72 \\   \hline  50-60 & 55 & 2 & 110 & 28 & 56 \\  \hline   & N=  &  & \sum_{i=1}^{6} f_ix_i  &  & \sum_{i=1}^{6} f_i |x_i - \overline{X}|   \\   &  \sum f_i = 50 &  & = 1350 & & = 512 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 1350

\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{1350}{50} = 27

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 512 \text{ and } N = \sum f_i = 27

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{512}{50} = 10.24

\\

Question 3: Compute the mean deviation from the mean of the following distribution: 

Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of Students 8 10 15 25 20 18 9 5

Answer:

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & f_ix_i &  |x_i - \overline{X}| = |x_i - 49 |  &  f_i|x_i - \overline{X}|  \\  \hline  10-20 &  15 & 8 & 120 & 34 & 272\\   \hline  20-30 &  25 & 10 & 250 & 24 & 240 \\   \hline  30-40 & 35 & 15 & 525 & 14 & 210 \\   \hline  40-50 & 45 & 25 & 1125 & 4 & 100 \\  \hline  50-60 & 55 & 20 & 1100 & 6 & 120 \\ \hline  60-70 & 65 & 18 & 1170 & 16 & 288 \\  \hline  70-80 & 75 & 9 & 675 & 26 & 234 \\ \hline  80-90 & 85 & 5 & 425 & 36 & 180 \\ \hline   & N=  &  & \sum_{i=1}^{8} f_ix_i  &  & \sum_{i=1}^{8} f_i |x_i - \overline{X}|   \\   &  \sum f_i = 110 &  & = 5390 & & = 1644 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 110 \text{ and } \sum f_ix_i = 5390

\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{5390}{110} = 49

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 1644 \text{ and } N = \sum f_i = 110

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{1644}{110} = 14.95

\\

Question 4: The age distribution of 100 life insurance policy holders is as follows:

Age ( on nearest birthday) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of Persons 5 16 12 26 14 12 6 5

Calculate the mean deviation from the median age.

Answer:

To make the function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

Calculation of Mean Deviation from Median 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & \text{Cum. Freq.} &  |d_i|   &  f_i |d_i|  \\ & & & & = |x_i - 38.63 | &  \\ \hline  16.75 - 19.75 &  18.25 & 5 & 5 & 20.38 & 101.9 \\   \hline  19.75 - 25.75 &  22.75 & 16 & 21 & 15.88 & 254.08 \\   \hline  25.75 - 35.75 & 30.75 & 12 & 33 & 7.88 & 94.56 \\ \hline  35.75 - 40.75 & 38.25 & 26 & 59 & 0.38 & 9.88 \\   \hline  40.75-50.75 & 45.75 & 14 & 73 & 7.12 & 99.68 \\   \hline  50.75-55.75 & 53.25 & 12 & 85 & 14.62 & 175.44 \\   \hline  55.75-60.75 & 58.25 & 6 & 91 & 19.62 & 117.72 \\ \hline  60.75-70.75 & 65.75 & 5 & 96 & 27.12 & 135.6 \\   \hline   & & N = \sum f_i = 96 &  &  & \sum_{i=1}^{8} f_i |d_i|= 988.68 \\  \hline \end{array}

\displaystyle \text{Clearly,  } N = 96 \Rightarrow \frac{N}{2} = \frac{96}{2} = 48

\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 48 \text{ is } 59.

\displaystyle \text{The corresponding value of class } \text{ is } 35.75-40.75

\displaystyle \text{Therefore the median class } \text{ is } 35.75-40.75

\displaystyle \therefore l = 35.75, \hspace{0.5cm} f = 26,  \hspace{0.5cm} F = 33, \hspace{0.5cm} N = 96, \hspace{0.5cm} h = 5

\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h 

\displaystyle = 35.75 + \frac{\Big( \frac{96}{2} - 33 \Big) }{26} \times 5 

\displaystyle = 35.75 + \frac{\Big( 48 - 33 \Big) }{26} \times 5 

\displaystyle = 38.63 

\displaystyle \text{Mean deviation from the median } =  \frac{\sum_{i=1}^8 f_i|d_i|}{N} = \frac{988.86}{96} = 10.30

\\

Question 5: Find the mean deviation from the mean and from median of the following distribution:

Marks 0-10 10-20 20-30 30-40 40-50
No. of Students 5 8 15 16 6

Answer:

Calculation of Mean Deviation from Median 

\displaystyle \begin{array}  {|c | c| c| c| c | c | } \hline  \text{Class} & x_i &  f_i & \text{Cum. Freq.} &  |d_i| = |x_i - 28 |  &  f_i |d_i|  \\ \hline  0-10 &  5 & 5 & 5 & 23  & 115 \\   \hline  10-20 &  15 & 8 & 13 & 13 & 104 \\   \hline  20-30 & 25 & 15 & 28 & 3 & 45 \\\hline  30-40 & 35 & 16 & 44 & 7 & 112 \\   \hline  40-50 & 45 & 6 & 50 & 17 & 102 \\   \hline   & & N = \sum f_i = 50 &  &  & \sum_{i=1}^{5} f_i |d_i|= 478 \\  \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} = 25 \text{ is } 28.

\displaystyle \text{The corresponding value of class } \text{ is } 20-30

\displaystyle \text{Therefore the median class } \text{ is } 20-30

\displaystyle \therefore l = 20, \hspace{0.5cm} f = 15,  \hspace{0.5cm} F = 13, \hspace{0.5cm} N = 50, \hspace{0.5cm} h = 10

\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h 

\displaystyle = 20 + \frac{\Big( \frac{50}{2} - 13 \Big) }{15} \times 10 

\displaystyle = 20 + \frac{\Big( 25 - 13 \Big) }{15} \times 10 

\displaystyle = 28 

\displaystyle \text{Mean deviation from the median } =  \frac{\sum_{i=1}^5 f_i|d_i|}{N} = \frac{478}{50} = 9.56

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c | c| c| c| c | c | }  \hline  \text{Class} & x_i &  f_i & f_ix_i &  |x_i - \overline{X}| = |x_i - 27 |  &  f_i|x_i - \overline{X}|  \\  \hline  0-10 &  5 & 5 & 25 & 22 & 110\\  \hline  10-20 &  15 & 8 & 120 & 12 & 96\\ \hline  20-30 &  25 & 15 & 375 & 2 & 30 \\   \hline  30-40 & 35 & 16 & 560 & 8 & 128 \\   \hline  40-50 & 45 & 6 & 270 & 18 & 108 \\  \hline   & N=  &  & \sum_{i=1}^{5} f_ix_i  &  & \sum_{i=1}^{5} f_i |x_i - \overline{X}|   \\   &  \sum f_i = 50 &  & = 1350 & & = 472 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 50 \text{ and } \sum f_ix_i = 1350

\displaystyle \therefore \overline{X} = \frac{\sum f_ix_i}{N} = \frac{1350}{50} = 27

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 472 \text{ and } N = \sum f_i = 50

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{472}{50} = 9.44

\\

Question 6: Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
No. of Persons 5 6 12 14 26 12 16 9

Answer:

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains the same, while the function becomes continuous.

Calculation of Mean Deviation from Median 

\displaystyle \begin{array}  {|c | c| c| c| c | c | }  \hline  \text{Class} & x_i &  f_i & \text{Cum. Freq.} &  |d_i| = |x_i - 38 |  &  f_i |d_i|  \\    \hline  15.5-20.5 &  18 & 5 & 5 & 20  & 100 \\  \hline  20.5-25.5 &  23 & 6 & 11 & 15 & 90 \\  \hline  25.5-30.5 & 28 & 12 & 23 & 10 & 120 \\  \hline  30.5-35.5 & 33 & 14 & 37 & 5 & 70 \\  \hline  35.5-40.5 & 38 & 26 & 63 & 0 & 0 \\  \hline  40.5-45.5 & 43 & 12 & 75 & 5 & 60 \\  \hline  45.5-50.5 & 48 & 16 & 91 & 10 & 160 \\  \hline  50.5-55.5 & 53 & 9 & 100 & 15 & 135 \\  \hline   & & N = \sum f_i = 100 &  &  & \sum_{i=1}^{8} f_i |d_i|= 735 \\  \hline \end{array}

\displaystyle \text{Clearly,  } N = 100 \Rightarrow \frac{N}{2} = \frac{100}{2} =50

\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 50 \text{ is } 63.

\displaystyle \text{The corresponding value of class } \text{ is } 35.5-40.5

\displaystyle \text{Therefore the median class } \text{ is } 35.5-40.5

\displaystyle \therefore l = 35.5, \hspace{0.5cm} f = 26,  \hspace{0.5cm} F = 37, \hspace{0.5cm} N = 100, \hspace{0.5cm} h = 5

\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h 

\displaystyle = 35.5 + \frac{\Big( \frac{100}{2} - 37 \Big) }{26} \times 5 

\displaystyle = 35.5 + \frac{\Big( 50 - 37 \Big) }{26} \times 5 

\displaystyle = 38 

\displaystyle \text{Mean deviation from the median } =  \frac{\sum_{i=1}^8 f_i|d_i|}{N} = \frac{735}{100} = 7.35

Thus the mean deviation from the median age is 7.35 years

\\

Question 7: Calculate the mean deviation about the mean for the following frequency distribution:

Class Interval 0-4 4-8 8-12 12-16 16-20
Frequency 4 6 8 5 2

Answer:

In order to avoid tedious calculations of computing mean \overline{X}, let us compute \overline{X} by step-deviation method. The formula for the same is :

\displaystyle \overline{X} = a + h \Bigg( \frac{1}{N} \sum_{i=1}^{n} f_id_i \Bigg), \text{ where } d_i = \frac{x_i - a}{h}, a = \text{ assumed mean and } \\ \\ h = \text{ common factor. }

Let us take the assumed mean a = 10 and h = 4 and form the following table.

Calculation of Mean Deviation from Mean 

\displaystyle \begin{array}  {|c |c | c| c| c| c | c | }  \hline  \text{Class} & x_i &  f_i & d_i = \frac{x_i - 10}{4} & f_ix_i &  |x_i - \overline{X}| &  f_i|x_i - \overline{X}|  \\ & & & & & = |x_i - 9.2 |  &  \\ \hline  0-4 &  2 & 4 & -2 & -8 & 7.2 & 28.8\\  \hline  4-8 &  6 & 6 & -1  & -6 & 3.2 & 19.2\\   \hline  8-12 &  10 & 8 & 0  & 0 & 0.8 & 6.4 \\   \hline  12-16 & 14 & 5 & 1 & 5 & 4.8 & 24 \\   \hline  16-20 & 18 & 2 &2  & 4 & 8.8 & 17.6 \\  \hline   & & N=  &  & \sum_{i=1}^{5} f_ix_i   & & \sum_{i=1}^{5} f_i |x_i - \overline{X}|   \\   & &  \sum f_i = 25 &  & = -5 & & = 96 \\ \hline \end{array}

\displaystyle \text{Clearly, } N = 25 \text{ and } \sum f_ix_i = -5

\displaystyle \therefore \overline{X} = a + h \Bigg( \frac{1}{N} \sum_{i=1}^{n} f_id_i \Bigg)

\displaystyle  = 10 + 4 \Bigg( \frac{1}{25} \times ( - 5)  \Bigg)  = 10 - \frac{20}{25} = 10 - 0.8 = 9.2

From the above table we get

\displaystyle \sum f_i|x_i- \overline{X}| = 96 \text{ and } N = \sum f_i = 25

\displaystyle \therefore M.D. = \frac{1}{N} \sum f_i |x_1 - \overline{X}| = \frac{96}{25} = 3.84

\\

Question 8: Calculate the mean deviation from the median of the following data:

Class Interval 0-6 6-12 12-18 18-24 24-30
Frequency 4 5 3 6 2

Answer:

Calculation of Mean Deviation from Median 

\displaystyle \begin{array}  {|c | c| c| c| c | c | }  \hline  \text{Class} & x_i &  f_i & \text{Cum. Freq.} &  |d_i| = |x_i - 14 |  &  f_i |d_i|  \\   \hline  0-6 &  3 & 4 & 4 & 11  & 44 \\  \hline  6-12 &  9 & 5 & 9 & 5 & 25 \\  \hline  12-18 & 15 & 3 & 12 & 1 & 3 \\  \hline  18-24 & 21 & 6 & 18 & 7 & 42 \\  \hline  24-30 & 27 & 2 & 20 & 13 & 26 \\  \hline   & & N = \sum f_i = 20 &  &  & \sum_{i=1}^{5} f_i |d_i|= 140 \\  \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} = 10 \text{ is } 12.

\displaystyle \text{The corresponding value of class } \text{ is } 12-18

\displaystyle \text{Therefore the median class } \text{ is } 12-18

\displaystyle \therefore l = 12, \hspace{0.5cm} f = 3,  \hspace{0.5cm} F = 9, \hspace{0.5cm} N = 20, \hspace{0.5cm} h = 6

\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h 

\displaystyle = 12 + \frac{\Big( \frac{20}{2} - 9 \Big) }{3} \times 6 

\displaystyle = 12 + \frac{\Big( 10 - 9 \Big) }{3} \times 6 

\displaystyle = 14 

\displaystyle \text{Mean deviation from the median } =  \frac{\sum_{i=1}^5 f_i|d_i|}{N} = \frac{140}{20} = 7

\\