NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 - Statistics

NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 - Statistics

Komal MiglaniUpdated on 07 May 2025, 02:16 PM IST

Understanding and interpretation of data in a meaningful way can be done with the help of statistics. Statistics are all around us, whether it is counting the steps moved daily on a fitness app or analysing the average scores of cricket players. In statistics, topics such Introduction to Statistics, measures of central tendency, such as mean, median, and mode, and measures of dispersion, such as range, mean deviation, variance, and standard deviation, are discussed in detail.

This Story also Contains

  1. NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1
  2. Topics covered in Chapter 13 Statistics Exercise 13.1
  3. NCERT Solutions of Class 11 Subject Wise
  4. Subject-Wise NCERT Exemplar Solutions

In this exercise 13.1 of Class 11 Maths Chapter 13 of the NCERT, you will learn how to find the mean deviation about mean and median for various data sets provided in the questions. The NCERT solutions discuss step-by-step methodology to apply these techniques effectively. If you are looking for NCERT Solutions, you can click on the given link to get NCERT solutions for Classes 6 to 12.

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1

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.

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.

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 :

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:

[ 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

Also read,

Topics covered in Chapter 13 Statistics Exercise 13.1

  1. Introduction to Statistics: This section deals witht collecting, organizing, analyzing, and interpreting numerical data to make informed decisions.
  2. Measures of Dispersion: In this part, we study the statistical tools used to describe the spread or variability in a data set, showing how much the values differ from the average.
  3. Range: The range is the simplest measure of dispersion and is calculated as the difference between the highest and lowest values in a data set.
  4. Mean Deviation: Mean deviation is the average of the absolute differences between each data point and the mean of the data set, showing how spread out the values are around the mean.

Also Read

NCERT Solutions of Class 11 Subject Wise

Students can also access the NCERT solutions for other subjects and make their learning feasible.

Subject-Wise NCERT Exemplar Solutions

Use the links provided in the table below to get your hands on the NCERT exemplar solutions available for all the subjects.

Frequently Asked Questions (FAQs)

Q: How many questions are there in the Exercise 13.1 Class 11 Maths ?
A:

Total 12 questions are discussed in this exercise.

Q: Which topics are covered in Exercise 13.1 Class 11 Maths?
A:

Exercise 13.1 Class 11 Maths includes mean deviation about mean and median.  

Q: Is this chapter 13 related to other chapters ?
A:

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

Q: What is the difficulty level of the questions asked in this chapter ?
A:

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

Q: Is it necessary to remember the formulas ?
A:

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

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