NCERT Solutions for Class 11 Maths Chapter 15 Statistics
NCERT Solutions for Class 11 Maths Chapter 15 Statistics: You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous class. I n this chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In NCERT solutions for class 11 maths chapter 15 statistics, you will get questions related to all the above topics. You will also learn some methods for finding a representative value for the given data. This value is known as the measure of central tendency. This chapter is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc . As statistics deal with the collected data for specific purposes and i n the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data and make a conclusive decision based on the data. Solutions of NCERT for class 11 maths chapter 15 statistics will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. All these questions are explained in CBSE NCERT solutions for class 11 maths chapter 15 statistics in a detailed manner. Check all NCERT solutions which will help you to understand the concepts in an easy way. There are 3 exercises and a miscellaneous exercise explained below.
Topics of NCERT Grade 11 Maths Chapter15 Statistics
15.1 Introduction
15.2 Measures of Dispersion
15.3 Range
15.4 Mean Deviation
15.5 Variance and Standard Deviation
15.6 Analysis of Frequency Distributions
The complete Solutions of NCERT Class 11 Mathematics Chapter 15 is provided below:
NCERT solutions for class 11 maths chapter 15 statisticsExercise: 15.1
Question:1 . Find the mean deviation about the mean for the data.
Answer:
Mean ( ) of the given data:
The respective absolute values of the deviations from mean, are
6, 3, 2, 1, 0, 2, 3, 7
Hence, the mean deviation about the mean is 3.
Question:2. Find the mean deviation about the mean for the data.
Answer:
Mean ( ) of the given data:
The respective absolute values of the deviations from mean, are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Hence, the mean deviation about the mean is 8.4.
Question:3. Find the mean deviation about the median.
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)
The respective absolute values of the deviations from median, 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
Hence, the mean deviation about the median is 2.33.
Question:4. Find the mean deviation about the median.
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)
The respective absolute values of the deviations from median, are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Hence, the mean deviation about the median is 7.
Question:5 Find the mean deviation about the mean.
Answer:





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 
Now, we calculate the absolute values of the deviations from mean, and
= 158
Hence, the mean deviation about the mean is 6.32
Question:6. Find the mean deviation about the mean.
Answer:





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 
Now, we calculate the absolute values of the deviations from mean, and
= 1280
Hence, the mean deviation about the mean is 16
Question:7. Find the mean deviation about the median.
Answer:





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 and observations.
Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Therefore, Median, M
Now, we calculate the absolute values of the deviations from median, and
= 84
Hence, the mean deviation about the median is 3.23
Question:8 Find the mean deviation about the median.
Answer:





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 and observations.
Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.
Therefore, Median, M
Now, we calculate the absolute values of the deviations from median, and
= 149.5
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 
Mid Points 



0 100 
4 
50 
200 
308 
1232 
100 200 
8 
150 
1200 
208 
1664 
200300 
9 
250 
2250 
108 
972 
300400 
10 
350 
3500 
8 
80 
400500 
7 
450 
3150 
92 
644 
500600 
5 
550 
2750 
192 
960 
600700 
4 
650 
2600 
292 
1168 
700800 
3 
750 
2250 
392 
1176 

=50 

=17900 

=7896 
Now, we calculate the absolute values of the deviations from mean, and
= 7896
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 
Mid Points 



95 105 
9 
100 
900 
25.3 
227.7 
105 115 
13 
110 
1430 
15.3 
198.9 
115125 
26 
120 
3120 
5.3 
137.8 
125135 
30 
130 
3900 
4.7 
141 
135145 
12 
140 
1680 
14.7 
176.4 
145155 
10 
150 
1500 
24.7 
247 

=100 

=12530 

=1128.8 
Now, we calculate the absolute values of the deviations from mean, and
= 1128.8
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 
Cumulative Frequency c.f. 
Mid Points 


010 
6 
6 
5 
22.85 
137.1 
1020 
8 
14 
15 
12.85 
102.8 
2030 
14 
28 
25 
2.85 
39.9 
3040 
16 
44 
35 
7.15 
114.4 
4050 
4 
48 
45 
17.15 
68.6 
5060 
2 
50 
55 
27.15 
54.3 





=517.1 
Now, N = 50, which is even.
The class interval containing or item is 2030. Therefore, 2030 is the median class.
We know,
Median
Here, l = 20, C = 14, f = 14, h = 10 and N = 50
Therefore, Median
Now, we calculate the absolute values of the deviations from median, and
= 517.1
Hence, the mean deviation about the median is 10.34
Question:12 Calculate the mean deviation about median age for the age distribution of persons given below:
Answer:
Age (in years) 
Number

Cumulative Frequency c.f. 
Mid Points 


15.520.5 
5 
5 
18 
20 
100 
20.525.5 
6 
11 
23 
15 
90 
25.530.5 
12 
23 
28 
10 
120 
30.535.5 
14 
37 
33 
5 
70 
35.540.5 
26 
63 
38 
0 
0 
40.545.5 
12 
75 
43 
5 
60 
45.550.5 
16 
91 
48 
10 
160 
50.555.5 
9 
100 
53 
15 
135 





=735 
Now, N = 100, which is even.
The class interval containing or item is 35.540.5. Therefore, 35.540.5 is the median class.
We know,
Median
Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100
Therefore, Median
Now, we calculate the absolute values of the deviations from median, and
= 735
Hence, the mean deviation about the median is 7.35
NCERT solutions for class 11 maths chapter 15 statisticsExercise: 15.2
Question:1. Find the mean and variance for each of the data.
Answer:
Mean ( ) of the given data:
The respective values of the deviations from mean, are
3, 2, 1 3 4 5 1 3
Hence, Mean = 9 and Variance = 9.25
Question:2. Find the mean and variance for each of the data.
Answer:
Mean ( ) of first n natural numbers:
We know, Variance
We know that
Hence, Mean = and Variance =
Question:3. Find the mean and variance for each of the data
Answer:
First 10 multiples of 3 are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Mean ( ) of the above values:
The respective values of the deviations from mean, are
13.5, 10.5, 7.5, 4.5, 1.5, 1.5, 4.5, 7.5, 10.5, 13.5
Hence, Mean = 16.5 and Variance = 74.25
Question:4. Find the mean and variance for each of the data.
Answer:






6 
2 
12 
13 
169 
338 
10 
4 
40 
9 
81 
324 
14 
7 
98 
5 
25 
175 
18 
12 
216 
1 
1 
12 
24 
8 
192 
5 
25 
200 
28 
4 
112 
9 
81 
324 
30 
3 
90 
13 
169 
363 

= 40 
= 760 


=1736 
We know, Variance,
Hence, Mean = 19 and Variance = 43.4
Question:5. Find the mean and variance for each of the data.
Answer:






92 
3 
276 
8 
64 
192 
93 
2 
186 
7 
49 
98 
97 
3 
291 
3 
9 
27 
98 
2 
196 
2 
4 
8 
102 
6 
612 
2 
4 
24 
104 
3 
312 
4 
16 
48 
109 
3 
327 
9 
81 
243 

= 22 
= 2200 


=640 
We know, Variance,
Hence, Mean = 100 and Variance = 29.09
Question:6 Find the mean and standard deviation using shortcut method.
Answer:
Let the assumed mean, A = 64 and h = 1






60 
2 
4 
16 
8 
32 
61 
1 
3 
9 
3 
9 
62 
12 
2 
4 
24 
48 
63 
29 
1 
1 
29 
29 
64 
25 
0 
0 
0 
0 
65 
12 
1 
1 
12 
12 
66 
10 
2 
4 
20 
40 
67 
4 
3 
9 
12 
36 
68 
5 
4 
16 
20 
80 

=100 


= 0 
=286 
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 64 and Standard Deviation = 1.691
Question:âââââââ7. Find the mean and variance for the following frequency distributions.
Answer:
Classes 
Frequency

Mid point





030 
2 
15 
30 
92 
8464 
16928 
3060 
3 
45 
135 
62 
3844 
11532 
6090 
5 
75 
375 
32 
1024 
5120 
90120 
10 
105 
1050 
2 
4 
40 
120150 
3 
135 
405 
28 
784 
2352 
150180 
5 
165 
825 
58 
3364 
16820 
180210 
2 
195 
390 
88 
7744 
15488 

= N = 30 

= 3210 


=68280 
We know, Variance,
Hence, Mean = 107 and Variance = 2276
Question:âââââââ8. Find the mean and variance for the following frequency distributions.
Answer:
Classes 
Frequency

Midpoint





010 
5 
5 
25 
22 
484 
2420 
1020 
8 
15 
120 
12 
144 
1152 
2030 
15 
25 
375 
2 
4 
60 
3040 
16 
35 
560 
8 
64 
1024 
4050 
6 
45 
270 
18 
324 
1944 

= N = 50 

= 1350 


=6600 
We know, Variance,
Hence, Mean = 27 and Variance = 132
Question:9. Find the mean, variance and standard deviation using shortcut method.
Answer:
Let the assumed mean, A = 92.5 and h = 5
Height in cms 
Frequency

Midpoint





7075 
3 
72.5 
4 
16 
12 
48 
7580 
4 
77.5 
3 
9 
12 
36 
8085 
7 
82.5 
2 
4 
14 
28 
8590 
7 
87.5 
1 
1 
7 
7 
9095 
15 
92.5 
0 
0 
0 
0 
95100 
9 
97.5 
1 
1 
9 
9 
100105 
6 
102.5 
2 
4 
12 
24 
105110 
6 
107.5 
3 
9 
18 
54 
110115 
3 
112.5 
4 
16 
12 
48 

=N = 60 



= 6 
=254 
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275
Question:10. The diameters of circles (in mm) drawn in a design are given below:
Answer:
Let the assumed mean, A = 92.5 and h = 5
Diameters 
No. of circles 
Midpoint





32.536.5 
15 
34.5 
2 
4 
30 
60 
36.540.5 
17 
38.5 
1 
1 
17 
17 
40.544.5 
21 
42.5 
0 
0 
0 
0 
44.548.5 
22 
46.5 
1 
1 
22 
22 
48.552.5 
25 
50.5 
2 
4 
50 
100 

=N = 100 



= 25 
=199 
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553
CBSE NCERT solutions for class 11 maths chapter 15 statisticsExercise: 15.3
Question:âââââââ1. From the data given below state which group is more variable, A or B?
Answer:
The group having a higher coefficient of variation will be more variable.
Let the assumed mean, A = 45 and h = 10
For Group A
Marks 
Group A

Midpoint





1020 
9 
15 
3 
9 
27 
81 
2030 
17 
25 
2 
4 
34 
68 
3040 
32 
35 
1 
1 
32 
32 
4050 
33 
45 
0 
0 
0 
0 
5060 
40 
55 
1 
1 
40 
40 
6070 
10 
65 
2 
4 
20 
40 
7080 
9 
75 
3 
9 
27 
81 

=N = 150 



= 6 
=342 
Mean,
We know, Variance,
We know, Standard Deviation =
Coefficient of variation =
C.V.(A) =
Similarly,
For Group B
Marks 
Group A

Midpoint





1020 
10 
15 
3 
9 
30 
90 
2030 
20 
25 
2 
4 
40 
80 
3040 
30 
35 
1 
1 
30 
30 
4050 
25 
45 
0 
0 
0 
0 
5060 
43 
55 
1 
1 
43 
43 
6070 
15 
65 
2 
4 
30 
60 
7080 
7 
75 
3 
9 
21 
72 

=N = 150 



= 6 
=375 
Mean,
We know, Variance,
We know, Standard Deviation =
Coefficient of variation =
C.V.(B) =
Since C.V.(B) > C.V.(A)
Therefore, Group B is more variable.
Question:2 From the prices of shares X and Y below, find out which is more stable in value:
Answer:
X( ) 
Y( ) 


35 
108 
1225 
11664 
54 
107 
2916 
11449 
52 
105 
2704 
11025 
53 
105 
2809 
11025 
56 
106 
8136 
11236 
58 
107 
3364 
11449 
52 
104 
2704 
10816 
50 
103 
2500 
10609 
51 
104 
2601 
10816 
49 
101 
2401 
10201 
=510 
= 1050 
=26360 
=110290 
For X,
Mean ,
Variance,
We know, Standard Deviation =
C.V.(X) =
Similarly, For Y,
Mean ,
Variance,
We know, Standard Deviation =
C.V.(Y) =
Since C.V.(Y) < C.V.(X)
Hence Y is more stable.
Which firm A or B pays larger amount as monthly wages?
Answer:
Given, Mean of monthly wages of firm A = 5253
Number of wage earners = 586
Total amount paid = 586 x 5253 = 30,78,258
Again, Mean of monthly wages of firm B = 5253
Number of wage earners = 648
Total amount paid = 648 x 5253 = 34,03,944
Hence firm B pays larger amount as monthly wages.
Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Which firm, A or B, shows greater variability in individual wages?
Answer:
Given, Variance of firm A = 100
Standard Deviation =
Again, Variance of firm B = 121
Standard Deviation =
Since , firm B has greater variability in individual wages.
Question:4 The following is the record of goals scored by team A in a football session:
Answer:
No. of goals scored 
Frequency




0 
1 
0 
0 
0 
1 
9 
1 
9 
9 
2 
7 
4 
14 
28 
3 
5 
9 
15 
45 
4 
3 
16 
12 
48 

=N = 25 

= 50 
=130 
For Team A,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(A) =
For Team B,
Mean = 2
Standard deviation, = 1.25
C.V.(B) =
Since C.V. of firm B is more than C.V. of A.
Therefore, Team A is more consistent.
Which is more varying, the length or weight?
Answer:
For lenght x,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(x) =
For weight y,
Mean,
Mean,
We know, Variance,
We know, Standard Deviation =
C.V.(y) =
Since C.V.(y) > C.V.(x)
Therefore, weight is more varying.
NCERT solutions for class 11 maths chapter 15 statisticsMiscellaneous Exercise
Answer:
Given,
The mean and variance of 8 observations are 9 and 9.25, respectively
Let the remaining two observations be x and y,
Observations: 6, 7, 10, 12, 12, 13, x, y.
∴ Mean,
60 + x + y = 72
x + y = 12 (i)
Now, Variance
(Using (i))
(ii)
Squaring (i), we get
(iii)
(iii)  (ii) :
2xy = 64 (iv)
Now, (ii)  (iv):
(v)
Hence, From (i) and (v):
x – y = 4 x = 8 and y = 4
x – y = 4 x = 4 and y = 8
Therefore, The remaining observations are 4 and 8. (in no order)
Answer:
Given,
The mean and variance of 7 observations are 8 and 16, respectively
Let the remaining two observations be x and y,
Observations: 2, 4, 10, 12, 14, x, y
∴ Mean,
42 + x + y = 56
x + y = 14 (i)
Now, Variance
(Using (i))
(ii)
Squaring (i), we get
(iii)
(iii)  (ii) :
2xy = 96 (iv)
Now, (ii)  (iv):
(v)
Hence, From (i) and (v):
x – y = 2 x = 8 and y = 6
x – y = 2 x = 6 and y = 8
Therefore, The remaining observations are 6 and 8. (in no order)
Answer:
Given,
Mean = 8 and Standard deviation = 4
Let the observations be
Mean,
Now, Let _{ } be the the resulting observations if each observation is multiplied by 3:
New mean,
= 24
We know that,
Standard Deviation =
(i)
Now, Substituting the values of in (i):
Hence, the variance of the new observations =
Therefore, Standard Deviation = = = 12
Answer:
Given, Mean = and variance =
Now, Let _{ } be the the resulting observations if each observation is multiplied by a:
Hence the mean of the new observations is
We know,
Now, Substituting the values of :
Hence the variance of the new observations is
Hence proved.
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 = 192
Therefore, Correct Mean = (Correct Sum)/19
= 10.1
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum  (8x8)
= 2080 – 64
= 2016
Hence, Correct Standard Deviation =
=2.02
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 + 12 = 204
Therefore, Correct Mean = (Correct Sum)/20
= 10.2
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum  (8x8) + (12x12)
= 2080 – 64 + 144
= 2160
Hence, Correct Standard Deviation =
= 1.98
Which of the three subjects shows highest variability in marks and which shows the lowest?
Answer:
Given,
Standard deviation of physics = 15
Standard deviation of chemistry = 20
Standard deviation of mathematics = 12
We know ,
C.V.
The subject with greater C.V will be more variable than others.
Hence, Mathematics has lowest variability and Chemistry has highest variability.
Answer:
Given,
Initial Number of observations, n = 100
Thus, incorrect sum = 2000
Hence, New sum of observations = 2000  212118 = 1940
New number of observation, n' = 1003 =97
Therefore, New Mean = New Sum)/100
= 20
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old (Incorrect) sum  (21x21)  (21x21)  (18x18)
= 40900  441  441  324
= 39694
Hence, Correct Standard Deviation =
= 3.036
NCERT solutions for class 11 mathematics
NCERT solutions for class 11 Subject wise
Some important terms and formulas to be remembered form NCERT solutions for class 11 maths chapter 15 statistics 
 Measures of dispersion Range, mean deviation, Quartile deviation, standard deviation, variance are measures of dispersion.
 Range The difference between the maximum and the minimum values of each series is called the ‘Range’ of the data. Range = Maximum Value – Minimum Value.
 Mean The arithmetic average of a range of values or quantities, found by adding all values and dividing by the number of values. For example, the mean of 4, 10, and 1 is (4+10+1)/3=15/3=5.
X represents values or scores,
N represents the number of values or scores.
 Median If the total number of values(n) is an odd number, then the formula is
 Median If the total number of values(n) is an even number, then the formula is
 Mode It is the most frequently occurring value or number.
 Mean deviation for ungrouped data
 Mean deviation for grouped data
This chapter seems very easy but it very useful in the field of science, research, reliable predictions or forecasts for future use, decision making, etc. There are 7 questions are given in the miscellaneous exercise. Try to solve them also to get command in this chapter. In CBSE NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.
Happy Reading !!!