Class 11: Statistics – Exercise 32.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

Answer:

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - \overline{x} \|$
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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

Answer:

1st Data Set

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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.

Answer:

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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

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

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 3020 \|$
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.

Answer:

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 45.5 \|$
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.

$\\$

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 29.9 \|$
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.

$\\$

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

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

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

$\displaystyle x_i$	$\displaystyle \|d_i\| = \| x_i - 49.4 \|$
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.

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Statistics – Exercise 32.1

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