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NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics

NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics

Edited By Sumit Saini | Updated on Jul 18, 2022 12:03 PM IST

NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 mainly deals with the mean deviation topics and other exercises include important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range, etc. In Class 9 and 10 Maths NCERT syllabus, students have already read the basic Statistics like mean, mode, and median. In this NCERT book Class 11 chapter applications of Statistics with advanced concepts are discussed.

This Story also Contains
  1. Statistics Class 11 Chapter 15-Exercise: 15.1
  2. More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exemplar solutions

Exercise 15.1 Class 11 Maths have easy questions. Solving NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 is a must to solve upcoming exercises. Direct questions are asked from this exercise as well in Boards. Also students can refer to the following exercise of NCERT for further information about upcoming exercises.

Statistics Class 11 Chapter 15-Exercise: 15.1

Question:1 . Find the mean deviation about the mean for the data.

\small 4,7,8,9,10,12,13,17

Answer:

Mean ( \overline{x} ) of the given data:

\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10

The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

6, 3, 2, 1, 0, 2, 3, 7

\therefore \sum_{i=1}^{8}|x_i - 10| = 24

\therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

= \frac{24}{8} = 3

Hence, the mean deviation about the mean is 3.

Question:2. Find the mean deviation about the mean for the data.

\small 38,70,48,40,42,55,63,46,54,44

Answer:

Mean ( \overline{x} ) of the given data:

\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50

The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

\therefore \sum_{i=1}^{8}|x_i - 50| = 84

\therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

= \frac{84}{10} = 8.4

Hence, the mean deviation about the mean is 8.4.

Question:3. Find the mean deviation about the median.

\small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median (M)

\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5

The respective absolute values of the deviations from median, |x_i - M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

\therefore \sum_{i=1}^{8}|x_i - 13.5| = 28

\therefore M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|

= \frac{28}{12} = 2.33

Hence, the mean deviation about the median is 2.33.

Question:4. Find the mean deviation about the median.

\small 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Now, Median (M)

\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5

The respective absolute values of the deviations from median, |x_i - M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

\therefore \sum_{i=1}^{8}|x_i - 47.5| = 70

\therefore M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|

= \frac{70}{10} = 7

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

\small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25\\\small f_i\hspace{1cm}7\hspace{1cm}4\hspace{1.15cm}6\hspace{1.22cm}3\hspace{1.3cm}5

Answer:

x_i
f_i
f_ix_i
|x_i - \overline{x}|
f_i|x_i - \overline{x}|
5
7
35
9
63
10
4
40
4
16
15
6
90
1
6
20
3
60
6
18
25
5
125
11
55

\sum{f_i}
= 25
\sum f_ix_i
= 350

\sum f_i|x_i - \overline{x}|
=158

N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

\sum f_i|x_i - \overline{x}| = 158

\therefore M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|

= \frac{158}{25} = 6.32

Hence, the mean deviation about the mean is 6.32

Question:6. Find the mean deviation about the mean.

\small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90\\\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8

Answer:

x_i
f_i
f_ix_i
|x_i - \overline{x}|
f_i|x_i - \overline{x}|
10
4
40
40
160
30
24
720
20
480
50
28
1400
0
0
70
16
1120
20
320
90
8
720
40
320

\sum{f_i}
= 80
\sum f_ix_i
= 4000

\sum f_i|x_i - \overline{x}|
=1280

N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

\sum f_i|x_i - \overline{x}| = 1280

\therefore M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|

= \frac{1280}{80} = 16

Hence, the mean deviation about the mean is 16

Question:7. Find the mean deviation about the median.

\small x_i \small 5 \small 7 \small 9 \small 10 \small 12 \small 15

\small f_i \small 8 \small 6 \small 2 \small 2 \small 2 \small 6

Answer:

x_i
f_i
c.f.
|x_i - M|
f_i|x_i - M|
5
8
8
2
16
7
6
14
0
0
9
2
16
2
4
10
2
18
3
6
12
2
20
5
10
15
6
26
8
48

Now, N = 26 which is even.

Median is the mean of 13^{th} and 14^{th} observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7

Now, we calculate the absolute values of the deviations from median, |x_i - M| and

\sum f_i|x_i - M| = 84

\therefore M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|

= \frac{84}{26} = 3.23

Hence, the mean deviation about the median is 3.23

Question:8 Find the mean deviation about the median.

\small x_i \small 15 \small 21 \small 27 \small 30 \small 35

\small f_i \small 3 \small 5 \small 6 \small 7 \small 8

Answer:

x_i
f_i
c.f.
|x_i - M|
f_i|x_i - M|
15
3
3
13.5
40.5
21
5
8
7.5
37.5
27
6
14
1.5
9
30
7
21
1.5
10.5
35
8
29
6.5
52

Now, N = 30, which is even.

Median is the mean of 15^{th} and 16^{th} observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30

Now, we calculate the absolute values of the deviations from median, |x_i - M| and

\sum f_i|x_i - M| = 149.5

\therefore M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|

= \frac{149.5}{29} = 5.1

Hence, the mean deviation about the median is 5.1

Question:9. Find the mean deviation about the mean.

Income per day in Rs
\small 0-100
\small 100-200
\small 200-300
\small 300-400
\small 400-500
\small 500-600
\small 600-700
\small 700-800
Number of persons
\small 4
\small 8
\small 9
\small 10
\small 7
\small 5
\small 4
\small 3





Answer:

Income
per day
Number of
Persons f_i
Mid
Points x_i
f_ix_i
|x_i - \overline{x}|
f_i|x_i - \overline{x}|
0 -100
4
50
200
308
1232
100 -200
8
150
1200
208
1664
200-300
9
250
2250
108
972
300-400
10
350
3500
8
80
400-500
7
450
3150
92
644
500-600
5
550
2750
192
960
600-700
4
650
2600
292
1168
700-800
3
750
2250
392
1176

\sum{f_i}
=50

\sum f_ix_i
=17900

\sum f_i|x_i - \overline{x}|
=7896


N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900

\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

\sum f_i|x_i - \overline{x}| = 7896

\therefore M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|

= \frac{7896}{50} = 157.92

Hence, the mean deviation about the mean is 157.92

Question:10. Find the mean deviation about the mean.

Height in cms
\small 95-105
\small 105-115
\small 115-125
\small 125-135
\small 135-145
\small 145-155
Number of person
\small 9
\small 13
\small 26
\small 30
\small 12
\small 10

Answer:

Height
in cms
Number of
Persons f_i
Mid
Points x_i
f_ix_i
|x_i - \overline{x}|
f_i|x_i - \overline{x}|
95 -105
9
100
900
25.3
227.7
105 -115
13
110
1430
15.3
198.9
115-125
26
120
3120
5.3
137.8
125-135
30
130
3900
4.7
141
135-145
12
140
1680
14.7
176.4
145-155
10
150
1500
24.7
247

\sum{f_i}
=100

\sum f_ix_i
=12530

\sum f_i|x_i - \overline{x}|
=1128.8


N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530

\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3

Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

\sum f_i|x_i - \overline{x}| = 1128.8

\therefore M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|

= \frac{1128.8}{100} = 11.29

Hence, the mean deviation about the mean is 11.29

Question:11. Find the mean deviation about median for the following data :

Marks
\small 0-10
\small 10-20
\small 20-30
\small 30-40
\small 40-50
\small 50-60
Number of girls
\small 6
\small 8
\small 14
\small 16
\small 4
\small 2

Answer:

Marks
Number of
Girls f_i
Cumulative
Frequency c.f.
Mid
Points x_i
|x_i - M|
f_i|x_i - M|
0-10
6
6
5
22.85
137.1
10-20
8
14
15
12.85
102.8
20-30
14
28
25
2.85
39.9
30-40
16
44
35
7.15
114.4
40-50
4
48
45
17.15
68.6
50-60
2
50
55
27.15
54.3





\sum f_i|x_i - M|
=517.1

Now, N = 50, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 25^{th} item is 20-30. Therefore, 20-30 is the median class.

We know,

Median = l + \frac{\frac{N}{2}- C}{f}\times h

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85

Now, we calculate the absolute values of the deviations from median, |x_i - M| and

\sum f_i|x_i - M| = 517.1

\therefore M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|

= \frac{517.1}{50} = 10.34

Hence, the mean deviation about the median is 10.34

Question:12 Calculate the mean deviation about median age for the age distribution of \small 100 persons given below:

Age (in years)
\small 16-20
\small 21-25
\small 26-30
\small 31-35
\small 36-40
\small 41-45
\small 46-50
\small 51-55
Number
\small 5
\small 6
\small 12
\small 14
\small 26
\small 12
\small 16
\small 9

[ Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Answer:

Age
(in years)
Number
f_i
Cumulative
Frequency c.f.
Mid
Points x_i
|x_i - M|
f_i|x_i - M|
15.5-20.5
5
5
18
20
100
20.5-25.5
6
11
23
15
90
25.5-30.5
12
23
28
10
120
30.5-35.5
14
37
33
5
70
35.5-40.5
26
63
38
0
0
40.5-45.5
12
75
43
5
60
45.5-50.5
16
91
48
10
160
50.5-55.5
9
100
53
15
135





\sum f_i|x_i - M|
=735

Now, N = 100, which is even.

The class interval containing \left (\frac{N}{2} \right)^{th} or 50^{th} item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median = l + \frac{\frac{N}{2}- C}{f}\times h

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38

Now, we calculate the absolute values of the deviations from median, |x_i - M| and

\sum f_i|x_i - M| = 735

\therefore M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|

= \frac{735}{100} = 7.35

Hence, the mean deviation about the median is 7.35



More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1

The NCERT Class 11 Maths chapter Statistics is very scoring chapters and questions are asked on the expected concepts only. It does not have much application in other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.1 Class 11 Maths Mainly discusses the mean deviation about mean and median. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 can be seen as a scoring exercise from the exam perspective.

Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1

  • The Class 11 Maths chapter 15 exercise is prepared by expert faculties with rich experience.

  • Exercise 15.1 Class 11 Maths can help one enhance their marks in the exam as one question of 5 marks is mostly asked from this.

  • Class 11 Maths chapter 15 exercise 15.1 solutions provided here are comprehensive in manner.

Also see-

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exemplar solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. Which topics are covered in Exercise 15.1 Class 11 Maths?

Exercise 15.1 Class 11 Maths includes mean deviation about mean and median.  

2. Is this chapter 15 related to other chapters ?

No, mostly concepts are new and can be understood if other chapters are not read.

3. What is the difficulty level of the questions asked in this chapter ?

Questions are easier to moderate level of difficulty if concepts are memorized. 

4. Is it necessary to remember the formulas ?

Yes, some basic formulas must be remembered to solve the questions.  

5. How many questions are there in the Exercise 15.1 Class 11 Maths ?

Total 12 questions are discussed in this exercise.

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