NCERT Solutions Class 11 Mathematics Chapter 15 Statistics have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 11 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 11 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 11 Mathematics are an important part of exams for Class 11 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 11 Mathematics and also download more latest study material for all subjects. Chapter 15 Statistics is an important topic in Class 11, please refer to answers provided below to help you score better in exams
Chapter 15 Statistics Class 11 Mathematics NCERT Solutions
Class 11 Mathematics students should refer to the following NCERT questions with answers for Chapter 15 Statistics in Class 11. These NCERT Solutions with answers for Class 11 Mathematics will come in exams and help you to score good marks
Chapter 15 Statistics NCERT Solutions Class 11 Mathematics
Exercise 15.1
Question. Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17
Answer :
The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data,
Question. Find the mean deviation about the mean for the data
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer :
The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data,
Question. Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer :
The given data is
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
The deviations of the respective observations 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
The absolute values of the deviations, |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
The required mean deviation about the median is
Question. Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer :
The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, the number of observations is 10, which is even.
Arranging the data in ascending order, we obtain
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Question. Find the mean deviation about the mean for the data.
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
Answer :
xi | fi | fixi | ||
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 |
Question. Find the mean deviation about the mean for the data
xi | 10 | 30 | 50 | 70 | 90 |
fi | 4 | 24 | 28 | 16 | 8 |
Answer :
xi | fi | fi xi | ||
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 |
Question. Find the mean deviation about the median for the data.
Answer :
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
xi | fi | c.f. |
5 | 8 | 8 |
7 | 6 | 14 |
9 | 2 | 16 |
10 | 2 | 18 |
12 | 2 | 20 |
15 | 6 | 26 |
Here, N = 26, which is even.
Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
The absolute values of the deviations from median, i.e, |xi – M|, are
Question. Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |
Answer :
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
xi | fi | c.f. |
15 | 3 | 3 |
21 | 5 | 8 |
27 | 6 | 14 |
30 | 7 | 21 |
35 | 8 | 29 |
Here, N = 29, which is odd.
∴ Median = [(29 + 1)/2 ]th observation = 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
Median = 30
The absolute values of the deviations from median, i.e. |xi – M|, are
|xi – M| | 15 | 9 | 3 | 0 | 5 |
fi | 3 | 5 | 6 | 7 | 8 |
fi |xi – M| | 45 | 45 | 18 | 0 | 40 |
Question. Find the mean deviation about the mean for the data.
Income per day | Number of persons |
0-100 | 4 |
100-200 | 8 |
200-300 | 9 |
300-400 | 10 |
400-500 | 7 |
500-600 | 5 |
600-700 | 4 |
700-800 | 3 |
Answer :
The following table is formed.
Income per day | Number of person fi | Mid – point xi | fi xi | ||
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 |
50 | 17900 | 7896 |
Question. Find the mean deviation about the mean for the data
Height in cms | Number of boys |
95-105 | 9 |
105-115 | 13 |
115-125 | 26 |
125-135 | 30 |
135-145 | 12 |
145-155 | 10 |
Answer :
The following table is formed.
Height in cms | Number of boys fi | Mid – point xi | fi xi |
|
|
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 |
Question. Find the mean deviation about median for the following data:
Marks | Number of girls |
0-10 | 6 |
10-20 | 8 |
20-30 | 14 |
30-40 | 16 |
40-50 | 4 |
50-60 | 2 |
Answer :
The following table is formed.
Marks | Number of boys fi | Cumulative frequency (c. f.) | Mid – point 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 |
Question. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age | Number |
16-20 | 5 |
21-25 | 6 |
26-30 | 12 |
31-35 | 14 |
36-40 | 26 |
41-45 | 12 |
46-50 | 16 |
51-55 | 9 |
Answer :
The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.
Age | Number fi | Cumulative frequency (c.f.) | Mid – point 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 | 37 | 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 the (N/2)th or 50th item is 35.5 - 40.5 .
Therefore, 35.5 - 40.5 is the median class.
It is known that,
Exercise 15.2
Question. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.
Answer :
6, 7, 10, 12, 13, 4, 8, 12
The following table is obtained.
xi | (xi – x̅) | (xi – x̅)2 |
6 | –3 | 9 |
7 | –2 | 4 |
10 | –1 | 1 |
12 | 3 | 9 |
13 | 4 | 16 |
4 | –5 | 25 |
8 | –1 | 1 |
12 | 3 | 9 |
|
| 74 |
Question. Find the mean and variance for the first n natural numbers
Answer :
The mean of first n natural numbers is calculated as follows.
Mean = (Sum of all observations)/(Number of observations)
Question. Find the mean and variance for the first 10 multiples of 3.
Answer :
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
The following table is obtained.
xi | (xi – x̅) | (xi – x̅)2 |
3 | –13.5 | 182.25 |
6 | –10.5 | 110.25 |
9 | –7.5 | 56.25 |
12 | –4.5 | 20.25 |
15 | –1.5 | 2.25 |
18 | 1.5 | 2.25 |
21 | 4.5 | 20.25 |
24 | 7.5 | 56.25 |
27 | 10.5 | 110.25 |
30 | 13.5 | 182.25 |
|
| 742.5 |
Question. Find the mean and variance for the data
xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
f i | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Answer :
The data is obtained in tabular form as follows.
xi | f i | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
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 | 11 | 121 | 363 |
40 | 760 | 1736 |
Question. Find the mean and variance for the data
xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
f i | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer :
The data is obtained in tabular form as follows.
xi | f i | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
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 |
Question. Find the mean and standard deviation using short-cut method.
xi | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
fi | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |
Answer :
The data is obtained in tabular form as follows.
xi | fi | Yi = (xi – A)/h | yi2 | fiyi | fiyi2 |
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 | 220 |
| 0 | 286 |
Question. Find the mean and variance for the following frequency distribution.
Classes | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Answer :
Class | Frequency fi | Mid-point xi | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
0-30 | 2 | 15 | 30 | -92 | 8464 | 16928 |
30-60 | 3 | 45 | 135 | -62 | 3844 | 11532 |
60-90 | 5 | 75 | 375 | -32 | 1024 | 5120 |
90-120 | 10 | 105 | 1050 | -2 | 4 | 40 |
120-150 | 3 | 135 | 405 | 28 | 784 | 2352 |
150-180 | 5 | 165 | 825 | 58 | 3364 | 16820 |
180-210 | 2 | 195 | 390 | 88 | 7744 | 15488 |
30 | 3210 | 2 | 68280 |
Question. Find the mean and variance for the following frequency distribution.
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer :
Class | Frequency fi | Mid-point xi | fixi | (xi – x̅) | (xi – x̅)2 | fi(xi – x̅)2 |
0-10 | 5 | 5 | 25 | -22 | 484 | 2420 |
10-20 | 8 | 15 | 120 | -12 | 144 | 1152 |
20-30 | 15 | 25 | 375 | -2 | 4 | 60 |
30-40 | 16 | 35 | 560 | 8 | 64 | 1024 |
40-50 | 6 | 45 | 270 | 18 | 324 | 1944 |
50 | 1350 | 6600 |
Question. Find the mean, variance and standard deviation using short-cut method
Height in cms | No. of children |
70-75 | 3 |
75-80 | 4 |
80-85 | 7 |
85-90 | 7 |
90-95 | 15 |
95-100 | 9 |
100-105 | 6 |
105-110 | 6 |
110-115 | 3 |
Answer :
Class Interval | Frequency fi | Mid-point xi | Yi = (xi – A)/h | yi2 | fiyi | fiyi2 |
70-75 | 3 | 72.5 | –4 | 16 | –12 | 48 |
75-80 | 4 | 77.5 | –3 | 9 | –12 | 36 |
80-85 | 7 | 82.5 | –2 | 4 | –14 | 28 |
85-90 | 7 | 87.5 | –1 | 1 | –7 | 7 |
90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
95-100 | 9 | 97.5 | 1 | 1 | 9 | 9 |
100-105 | 6 | 102.5 | 2 | 4 | 12 | 24 |
105-110 | 6 | 107.5 | 3 | 9 | 18 | 54 |
110-115 | 3 | 112.5 | 4 | 16 | 12 | 48 |
60 | 6 | 254 |
Question. The diameters of circles (in mm) drawn in a design are given below:
Diameters | No. of children |
33-36 | 15 |
37-40 | 17 |
41-44 | 21 |
45-48 | 22 |
49-52 | 25 |
Answer :
Class Interval | Frequency fi | Mid-point xi | Yi = (xi – A)/h | fi2 | fiyi | fiyi2 |
32.5-36.5 | 15 | 34.5 | –2 | 4 | –30 | 60 |
36.5-40.5 | 17 | 38.5 | –1 | 1 | –17 | 17 |
40.5-44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
44.5-48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
48.5-52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
| 100 |
|
|
| 25 | 199 |
Exercise 15.3
Question. From the data given below state which group is more variable, A or B?
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
Answer :
Firstly, the standard deviation of group A is calculated as follows.
Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.
Question. From the prices of shares X and Y below, find out which is more stable in value:
X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |
Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |
Answer :
The prices of the shares X are
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10
Question. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results::
Firm A | Firm B | |
No. of wage earners | 586 | 648 |
Mean of monthly wages | Rs 5253 | Rs 5253 |
Variance of the distribution of wages | 100 | 121 |
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer :
(i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
(ii) Variance of the distribution of wages in firm A (σ12) = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ1) = √100 = 10
Variance of the distribution of wages in firm B (σ22) = 121
∴ Standard deviation of the distribution of wages in firm B (σ22) = √121 = 11
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater
standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.
Question. The following is the record of goals scored by team A in a football session:
No. of goals scored | 0 | 1 | 2 | 3 | 4 |
No. of matches | 1 | 9 | 7 | 5 | 3 |
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Answer :
The Mean and the standard deviation of goals scored by team A are calculated as follows.
No. of goals scored | No. of matches | fi xi | xi2 | fixi2 |
0 | 1 | 0 | 0 | 0 |
1 | 9 | 9 | 1 | 9 |
2 | 7 | 14 | 4 | 28 |
3 | 5 | 15 | 9 | 45 |
4 | 3 | 12 | 16 | 48 |
| 25 | 50 |
| 130 |
The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.
Question. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer :
Thus, C.V. of weights is greater than the c.v. of lengths. Therefore, weights vary more than the lengths.
Miscellaneous Solutions
Question. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Answer :
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
From (1), we obtain
x2 + y2 + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x2 + y2 – 2xy = 80 – 64 = 16
⇒ x – y = ± 4 …(5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.
Question. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Answer :
Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.
Question. Given that x is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the mean and variance of the observations ax1, ax2, ax3 …axn are ax and a2 σ2, respectively (a ≠ 0).
Answer :
Question. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Answer :
(i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
That is, incorrect sum of observations = 200
Correct sum of observations = 200 - 8 = 192
Correct mean = (Correct sum )/19 = 192/19 = 10.1
(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
Correct sum of observations = 200 - 8 + 12 = 204
∴ Correct mean = (Correct sum)/20 = 204/20 = 10.2
Question. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject | Mathematics | Physics | Chemistry |
Mean | 42 | 32 | 40.9 |
Standard deviation | 12 | 15 | 20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Answer :
Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) is given by (Standard deviation )/Mean × 100
The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.
Question. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Answer :
Number of observations (n) = 100
Incorrect mean x = 20
Incorrect standard deviation (σ) = 3
NCERT Solutions Class 11 Mathematics Chapter 1 Sets |
NCERT Solutions Class 11 Mathematics Chapter 2 Relations and Functions |
NCERT Solutions Class 11 Mathematics Chapter 3 Trigonometric Functions |
NCERT Solutions Class 11 Mathematics Chapter 4 Principle of Mathematical Induction |
NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations |
NCERT Solutions Class 11 Mathematics Chapter 6 Linear Inequalities |
NCERT Solutions Class 11 Mathematics Chapter 7 Permutations and Combinations |
NCERT Solutions Class 11 Mathematics Chapter 8 Binomial Theorem |
NCERT Solutions Class 11 Mathematics Chapter 9 Sequences and Series |
NCERT Solutions Class 11 Mathematics Chapter 10 Straight Lines |
NCERT Solutions Class 11 Mathematics Chapter 11 Conic Sections |
NCERT Solutions Class 11 Mathematics Chapter 12 Introduction to Three Dimensional Geometry |
NCERT Solutions Class 11 Mathematics Chapter 13 Limits and Derivatives |
NCERT Solutions Class 11 Mathematics Chapter 14 Mathematical Reasoning |
NCERT Solutions Class 11 Mathematics Chapter 15 Statistics |
NCERT Solutions Class 11 Mathematics Chapter 16 Probability |
NCERT Solutions Class 11 Mathematics Chapter 15 Statistics
NCERT Solutions Class 11 Mathematics Chapter 15 Statistics is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 11 Mathematics textbook online or you can easily download them in pdf.
Chapter 15 Statistics Class 11 Mathematics NCERT Solutions
The Class 11 Mathematics NCERT Solutions Chapter 15 Statistics are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 15 Statistics of Mathematics Class 11 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 15 Statistics Class 11 chapter of Mathematics so that it can be easier for students to understand all answers.
NCERT Solutions Chapter 15 Statistics Class 11 Mathematics
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Chapter 15 Statistics Class 11 NCERT Solution Mathematics
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Class 11 NCERT Solution Mathematics Chapter 15 Statistics
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