NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all exercise 15.1 questions. These solutions help students to clear their concepts and solving difficult problems at their own pace. Students can download the NCERT Solutions of Class 11 maths and practice offline to improve their skills.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

ncert sol class 11 maths ch 15 ex 1 01
ncert sol class 11 maths ch 15 ex 1 02
ncert sol class 11 maths ch 15 ex 1 03
ncert sol class 11 maths ch 15 ex 1 04
ncert sol class 11 maths ch 15 ex 1 05
ncert sol class 11 maths ch 15 ex 1 06
ncert sol class 11 maths ch 15 ex 1 07
ncert sol class 11 maths ch 15 ex 1 08
ncert sol class 11 maths ch 15 ex 1 09
ncert sol class 11 maths ch 15 ex 1 10
ncert sol class 11 maths ch 15 ex 1 11

 

Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.2 Solutions : 10 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15 Exercise 15.1

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 1

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 2

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:-

First we have to find (x̅) of the given data

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 3

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

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

Now absolute values of the deviations,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 4

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 12

Then,

Median = ((12/2)th observation + ((12/2)+ 1)th observation)/2

(12/2)th observation = 6th = 13

(12/2)+ 1)th observation = 6 + 1

= 7th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |xi – 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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 5

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 6

= (1/12) × 28

= 2.33

So, the mean deviation about the median for the given data is 2.33.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

First we have to arrange the given observations into ascending order,

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

The number of observations is 10

Then,

Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2

(10/2)th observation = 5th = 46

(10/2)+ 1)th observation = 5 + 1

= 6th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 7

Mean Deviation,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 8

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

xi 5 10 15 20 25
fi 7 4 6 3 5

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi fixi |xi – x̅| fi |xi – x̅|
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
25 350 158

The sum of calculated data,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 9

The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 10

6.

xi 10 30 50 70 90
fi 4 24 28 16 8

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi fixi |xi – x̅| fi |xi – x̅|
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
80 4000 1280

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 11

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 12

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi c.f. |xi – M| fi |xi – 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 the 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Then,

Median = (13th observation + 14th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 13

8.

xi 15 21 27 30 35
fi 3 5 6 7 8

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi c.f. |xi – M| fi |xi – 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 the 15th and 16th observations. Both of these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Then,

Median = (15th observation + 16th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 14

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Income per day in ₹ 0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800
Number of persons 4 8 9 10 7 5 4 3

Solution:-

Let us make the table of the given data and append other columns after calculations.

Income per day in ₹ Number of persons fi Mid – points

xi

fixi |xi – x̅| fi|xi – x̅|
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 1160
700 – 800 3 750 2250 392 1176
50 17900 7896

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 15

10.

Height in cms 95 – 105 105 – 115 115 – 125 125 – 135 135 – 145 145 – 155
Number of boys 9 13 26 30 12 10

Solution:-

Let us make the table of the given data and append other columns after calculations.

Height in cms Number of boys fi Mid – points

xi

fixi |xi – x̅| fi|xi – x̅|
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
100 12530 1128.8

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 16

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

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

Solution:-

Let us make the table of the given data and append other columns after calculations.

Marks Number of Girls fi Cumulative frequency (c.f.) Mid – points

xi

|xi – Med| fi|xi – Med|
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
50 517.1

The class interval containing Nth/2 or 25th item is 20-30

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 17

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

Age

(in years)

16 – 20 21 – 25 26 – 30 31 – 35 36 – 40 41 – 45 46 – 50 51 – 55
Number 5 6 12 14 26 12 16 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]

Solution:-

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

Age Number fi Cumulative frequency (c.f.) Mid – points

xi

|xi – Med| fi|xi – Med|
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
100 735

The class interval containing Nth/2 or 50th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 18

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class