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NCERT Solutions for Class 11 Maths Chapter 15 Statistics are provided here. You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous Class NCERT syllabus. In this NCERT book chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In class 11 maths chapter 15 question answer, you will get questions related to all the above topics. These NCERT solutions are prepared by expert team at Careers360 keeping in my the latest syllabus of CBSE 2023-24, easy and simple language, step by step comprehensive coverage of the concepts.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
Suggested: JEE Main: high scoring chapters | Past 10 year's papers
This statistics class 11 solutions 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 in 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. Class 11 maths chapter 15 NCERT solutions 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. Interested students can find all NCERT solutions for class 11 at one place which will help to understand the concepts in an easy way.
Also Read :
Measure of Dispersion:
Dispersion measures the degree of variation in the values of a variable. It quantifies how scattered observations are around the central value in a distribution.
Range:
Range is the simplest measure of dispersion. It is defined as the difference between the largest and smallest observations in a distribution.
Range of distribution = Largest observation – Smallest observation.
Mean Deviation:
Mean Deviation for Ungrouped Data:
For n observations x1, x2, x3, ..., xn, the mean deviation about their mean x¯ is calculated as:
MD(\bar x) = (Σ |xᵢ - x¯|) / n
The mean deviation about its median M is calculated as:
MD(M) = (Σ |xᵢ - M|) / n
Mean Deviation for Discrete Frequency Distribution:
For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the mean deviation is calculated as:
MD(\bar x) = (Σ fᵢ |xᵢ - x¯|) / (Σ fᵢ) = (Σ fᵢ |xᵢ - x¯|) / N
Variance:
Variance is the average of the squared deviations from the mean x¯. If x_{₁}, x_{₂}, …, xₙ are n observations with mean x¯, the variance denoted by σ² is calculated as:
σ^{²} = (Σ(xᵢ - x¯)^{²}) / n
Standard Deviation:
Standard deviation, denoted as σ, is the square root of the variance σ^{²}. If σ^{²} is the variance, then the standard deviation is given by:
σ = √(σ^{²})
For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the standard deviation is calculated as:
σ = √((Σ fᵢ(xᵢ - x¯)^{²}) / N)
Coefficient of Variation:
The coefficient of variation (CV) is used to compare two or more frequency distributions. It is defined as:
Coefficient of Variation = (Standard Deviation / Mean) × 100
CV = (σ / x¯) × 100
Free download NCERT Solutions for Class 11 Maths Chapter 15 Statistics for CBSE Exam.
Class 11 maths chapter 15 question answer - Exercise: 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 |
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 |
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 |
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 |
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 | ||
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 |
=517.1 |
Now, N = 50, which is even.
The class interval containing or item is 20-30. Therefore, 20-30 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:
Age (in years) | ||||||||
[ Hint Convert the given data into continuous frequency distribution by subtracting from the lower limit and adding to the upper limit of each class interval]
Answer:
Age (in years) | Number | Cumulative Frequency c.f. | Mid Points | ||
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 |
=735 |
Now, N = 100, which is even.
The class interval containing or item is 35.5-40.5. Therefore, 35.5-40.5 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
Class 11 maths chapter 15 ncert solutions - Exercise: 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.
First n natural numbers.
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
First 10 multiples of 3
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 short-cut 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.
Classes | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Answer:
Classes | Frequency | Mid point | ||||
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 |
= 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.
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer:
Classes | Frequency | Mid-point | ||||
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 |
= N = 50 | = 1350 | =6600 |
We know, Variance,
Hence, Mean = 27 and Variance = 132
Question:9. Find the mean, variance and standard deviation using short-cut method.
Height in cms | 70-75 | 75-80 | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
No. of students | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Answer:
Let the assumed mean, A = 92.5 and h = 5
Height in cms | Frequency | Midpoint | ||||
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 |
=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:
Diameters | 33-36 | 37-40 | 41-44 | 45-48 | 49-52 |
No. of circles | 15 | 17 | 21 | 22 | 25 |
Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as and then proceed.]
Answer:
Let the assumed mean, A = 92.5 and h = 5
Diameters | No. of circles | Midpoint | ||||
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 |
=N = 100 | = 25 | =199 |
Mean,
We know, Variance,
We know, Standard Deviation =
Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553
Statistics class 11 NCERT solutions - Exercise: 15.3
Question:1. 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:
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 | ||||
10-20 | 9 | 15 | -3 | 9 | -27 | 81 |
20-30 | 17 | 25 | -2 | 4 | -34 | 68 |
30-40 | 32 | 35 | -1 | 1 | -32 | 32 |
40-50 | 33 | 45 | 0 | 0 | 0 | 0 |
50-60 | 40 | 55 | 1 | 1 | 40 | 40 |
60-70 | 10 | 65 | 2 | 4 | 20 | 40 |
70-80 | 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 | ||||
10-20 | 10 | 15 | -3 | 9 | -30 | 90 |
20-30 | 20 | 25 | -2 | 4 | -40 | 80 |
30-40 | 30 | 35 | -1 | 1 | -30 | 30 |
40-50 | 25 | 45 | 0 | 0 | 0 | 0 |
50-60 | 43 | 55 | 1 | 1 | 43 | 43 |
60-70 | 15 | 65 | 2 | 4 | 30 | 60 |
70-80 | 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:
X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |
Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |
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:
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 goals. Find which team may be considered more consistent?
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.
Statistics maths chapter 15 class 11 - Miscellaneous 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.
If wrong item is omitted.
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
If it is replaced by
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 - 21-21-18 = 1940
New number of observation, n' = 100-3 =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
X represents values or scores,
N represents the number of values or scores.
interested students can practice class 11 maths ch 15 question answer using the following exercises.
chapter-1 | |
chapter-2 | |
chapter-3 | |
chapter-4 | |
chapter-5 | |
chapter-6 | |
chapter-7 | |
chapter-8 | |
chapter-9 | |
chapter-10 | |
chapter-11 | |
chapter-12 | |
chapter-13 | |
chapter-14 | |
chapter-15 | Statistics |
chapter-16 |
Easy to Understand: The ch 15 maths class 11 solutions are written in simple language and are easy to understand, which makes them accessible for students who may not have a strong background in statistics.
Comprehensive Coverage: The statistics NCERT solutions cover all the important concepts and topics in Statistics Class 11 NCERT curriculum. This ensures that students are well-prepared and have a thorough understanding of the subject.
Structured Format: The ch 15 maths class 11 solutions are presented in a structured format that helps students to organize their thoughts and approach problems in a systematic manner.
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 NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.
The concept of standard deviation involves measuring the amount of variation or deviation present in a given set of values, and the range is determined by considering the level of standard deviation. For a clearer understanding of this term, students are encouraged to download the PDF solutions provided by Careers360, which are available in both chapter-wise and exercise-wise formats to cater to their specific needs.
The topics discussed in class 11 maths statistics are
1. Introduction
2. Methods of Dispersion
3. Range
4. Mean Deviation
5. Variance and Standard Deviation
6. Analysis of Frequency Distributions
The NCERT Solutions for class 11 chapter 15 maths offer precise and straightforward explanations that aid students in achieving good grades in their exams. The solutions' methodical approach to problem-solving provides students with a clear understanding of the marks allocation as per the latest CBSE Syllabus 2023. By using these class 11th statistics solutions , students can identify their weak areas and work towards improving them, resulting in better academic performance.
Students can find a detailed NCERT solutions for class 11 maths by clicking on the link. they can practice these solutions to get confidence in the concepts and in-depth understanding that ultimately lead to high score in the exam.
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