Question 1: Calculate the mean deviation about the median of the following observations:

(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

(v) 38, 79, 48, 34, 63, 42, 55, 44, 53, 47

(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 2354, 2780, 3011, 3020, 3541, 4150, 5000$

$\displaystyle \text{Here, Median } (M) = 3020 \text{ and } n = 7$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 3011 9 2780 240 3020 0 2354 666 3541 521 4150 1130 5000 1980 Total 4546

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{7} \times 4546 = 649.42$

$\\$

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 34, 38, 42, 44, 46, 48, 54, 55, 63, 70$

$\displaystyle \text{Here, Median } (M) = \frac{46+ 48}{2} = 47 \text{ and } n = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 38 9 70 23 48 1 34 13 42 5 55 8 63 16 46 1 54 7 44 3 Total 86

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 86 =8.6$

$\\$

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 30, 34, 38, 40, 42, 44, 50, 51, 60, 66$

$\displaystyle \text{Here, Median } (M) = \frac{42+ 44}{2} = 43 \text{ and } n = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 34 9 66 23 30 13 38 5 44 1 50 7 40 3 60 17 42 1 51 8 Total 87

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 87 =8.7$

$\\$

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 22, 24, 25, 27, 28, 29, 30, 31, 41, 42$

$\displaystyle \text{Here, Median } (M) = \frac{28+ 29}{2} = 28.5 \text{ and } n = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 22 6.5 24 4.5 30 1.5 27 1.5 29 0.5 31 2.5 25 3.5 28 0.5 41 12.5 42 13.5 Total 47

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 47 =4.7$

$\\$

(v) 38, 79, 48, 34, 63, 42, 55, 44, 53, 47

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 34, 38, 42, 44, 47, 48, 53, 55, 63, 70$

$\displaystyle \text{Here, Median } (M) = \frac{47+ 48}{2} = 47.5 \text{ and } n = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 38 9.5 70 22.5 48 0.5 34 13.5 63 15.5 42 5.5 55 7.5 44 3.5 53 5.5 47 0.5 Total 84

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 84 =8.4$

$\\$

Question 2: Calculate the mean deviation about the mean of the following observations:

(i) 4, 7, 8, 9, 10, 12, 13, 17

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

(i) 4, 7, 8, 9, 10, 12, 13, 17

$\displaystyle \text{Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - \overline{x} |$$\displaystyle |d_i| = | x_i - \overline{x} |$ 4 6 7 3 8 2 9 1 10 0 12 2 13 3 17 7 Total 24

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{8} \times 24 =3$

$\\$

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

$\displaystyle \text{Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{13+ 17+ 16+ 14+ 11+ 13+ 10+ 16+ 11+ 18+ 12+ 17}{12} = 14$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 13 1 17 3 16 2 14 0 11 3 13 1 10 4 16 2 11 3 18 4 12 2 17 3 Total 28

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{12} \times 28 =2.33$

$\\$

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

$\displaystyle \text{Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{38+ 70+ 48+ 40+ 42+ 55+ 63+ 46+ 54+ 44}{10} = 50$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 38 12 70 20 48 2 40 10 42 8 55 5 63 13 46 4 54 4 44 6 Total 84

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 84 =8.4$

$\\$

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

$\displaystyle \text{Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{36+ 72+ 46+ 42+ 60+ 45+ 53+ 46+ 51+ 49}{10} = 50$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 36 14 72 22 46 4 42 8 60 10 45 5 53 3 46 4 51 1 49 1 Total 72

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 72 =7.2$

$\\$

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

$\displaystyle \text{Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{57+ 64+ 43+ 67+ 49+ 59+ 44+ 47+ 61+ 59}{10} = 55$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 57 2 64 9 43 12 67 12 49 6 59 4 44 11 47 8 61 6 59 4 Total 74

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 74 =7.4$

$\\$

Question 3: Calculate the mean deviation of the following income groups of fine and seven members from their medians:

 I Income in Rs. II Income in Rs. 4000 300 4200 4000 4400 4200 4600 4400 4800 4600 4800 5800

1st Data Set

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 4000, 4200, 4400, 4600, 4800$

$\displaystyle \text{Here, Median } (M) = 4400 \text{ and } n = 5$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 4000 400 4200 200 4400 0 4600 200 4800 400 Total 1200

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{5} \times 1200 = 240$

2nd Data Set

$\displaystyle \text{Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 300, 4000, 4200, 4400, 4600, 4800, 5800$

$\displaystyle \text{Here, Median } (M) = 4400 \text{ and } n = 7$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 300 4100 4000 400 4200 200 4400 0 4600 200 4800 400 5800 1400 Total 6700

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{7} \times 6700 = 957.14$

$\\$

Question 4: The lengths (in cm) of 10 rods in a shop are given below:

40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2

(i) Find the mean deviation from median

(ii) Find mean deviation from the mean also.

$\displaystyle \text{(i) Formula used for mean deviation: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = x_i - M \text{ where } M = \text{ Median }$

$\displaystyle \text{Arranging the data in ascending order } 15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79$

$\displaystyle \text{Here, Median } (M) = \frac{40+ 52.3}{2} = 46.15 \text{ and } n = 10$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 40 6.15 52.3 6.15 55.2 9.05 72.9 26.75 52.8 6.65 79 32.85 32.5 13.65 15.2 30.95 29.9 18.25 30.2 15.95 Total 166.4

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 166.4 =16.64$

$\displaystyle \text{(ii) Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{40.0+ 52.3+ 55.2+ 72.9+ 52.8+ 79.0+ 32.5+ 15.2+ 27.9+ 30.2}{10} = 45.98$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 3020 |$$\displaystyle |d_i| = | x_i - 3020 |$ 40 5.98 52.3 6.32 55.2 9.22 72.9 26.92 52.8 6.82 79 33.02 32.5 13.48 15.2 30.78 27.9 18.08 32 13.98 Total 164.6

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 164.6 =16.46$

$\\$

Question 5: In question 1 (iii), (iv), (v) find the number of observations lying between $\overline{X} - M.D.$ and $\overline{X} + M.D.$, where M.D. is the mean deviation from the mean.

$\displaystyle \text{(i) Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{34+66+30+38+44+50+40+60+42+51}{10} = 45.5$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 45.5 |$$\displaystyle |d_i| = | x_i - 45.5 |$ 34 11.5 66 20.5 30 15.5 38 7.5 44 1.5 50 4.5 40 5.5 60 14.5 42 3.5 51 5.5 Total 90

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 90 =9$

$\displaystyle \overline{x} - M.D. = 45.5 - 9 = 36.5$

Also $\displaystyle \overline{x} + M.D. = 45.5 + 9 = 54.5$

Hence there are 6 observations between 36.5 and 54.5.

$\\$

$\displaystyle \text{(ii) Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{22+24+30+27+29+31+25+28+41+42}{10} =29.9$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 29.9 |$$\displaystyle |d_i| = | x_i - 29.9 |$ 22 7.9 24 5.9 30 0.1 27 2.9 29 0.9 31 1.1 25 4.9 28 1.9 41 11.9 42 12.1 Total 48.8

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 48.8 =4.88$

$\displaystyle \overline{x} - M.D. = 29.9 - 4.88 = 25.02$

Also $\displaystyle \overline{x} + M.D. = 29.9 + 4.88 = 34.78$

Hence there are 5 observations between 25.02 and 34.78.

$\\$

$\displaystyle \text{(iii) Formula used for mean deviation about mean: } MD = \frac{1}{n} \sum_{i=0}^{n} | d_i |$

$\displaystyle \text{Here, } d_i = |x_i - \overline{x} | \text{ where } \overline{x} = \text{ Mean of the given data }$

$\displaystyle \overline{x} = \frac{ 38+ 70+ 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47 }{10} =49.4$

 $\displaystyle x_i$$\displaystyle x_i$ $\displaystyle |d_i| = | x_i - 49.4 |$$\displaystyle |d_i| = | x_i - 49.4 |$ 38 11.4 70 20.6 48 1.4 34 15.4 63 13.6 42 7.4 55 5.6 44 5.4 53 3.6 47 2.4 Total 86.8

$\displaystyle MD = \frac{1}{n} \sum_{i=0}^{n} | d_i | \hspace{1.0cm} \Rightarrow MD = \frac{1}{10} \times 86.6 =8.68$

$\displaystyle \overline{x} - M.D. = 49.4 - 8.68 = 40.72$

Also $\displaystyle \overline{x} + M.D. =49.4 + 8.68 = 58.08$

Hence there are 6 observations between 40.72 and 58.08.