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# NCERT Solutions Class 11 Maths Chapter 13 Statistics

Last updated date: 11th Sep 2024
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## NCERT Solutions Class 11 Maths Chapter 13 Statistics - FREE PDF Download

Class 11 Maths NCERT Solutions for Chapter 13 Statistics class 11 Maths, students will focus on the fundamental concepts of statistics, which are essential for analyzing and interpreting data. This chapter covers key topics such as measures of central tendency (mean, median, mode), measures of dispersion (range, variance, standard deviation), and the concept of mean deviation. By exploring these topics in statistics class 11 solutions, you will learn how to summarize large data sets, understand variability, and make informed decisions based on data analysis. Access the latest Class 11 Maths Syllabus here.

Table of Content
1. NCERT Solutions Class 11 Maths Chapter 13 Statistics - FREE PDF Download
2. Glance on Maths Chapter 13 Class 11 - Statistics
3. Access Exercise wise NCERT Solutions for Chapter 13 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Chapter 13 Statistics
5. Access NCERT Solutions for Class 11 Maths Chapter 13 – Statistics
5.1Exercise 13.1
5.2Exercise 13.2
5.3Miscellaneous Exercise
6. Class 11 Maths NCERT Solutions Chapter 13 - Summary
6.1Class 11 Maths NCERT Solutions Chapter 13
7. Overview of Deleted Syllabus for CBSE Class 11 Maths Statistics
8. Class 11 Maths Chapter 13: Exercises Breakdown
9. Other Study Material for CBSE Class 11 Maths Chapter 13
10. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

## Glance on Maths Chapter 13 Class 11 - Statistics

• Chapter 13 of the Class 11 NCERT Maths textbook delves into the fundamental concepts of Statistics, which is a critical branch of mathematics dealing with data collection, analysis, interpretation, and presentation.

• This chapter provides students with the tools necessary to understand and work with various statistical measures.

• Each section of the chapter includes detailed examples followed by exercises that help students practice and solidify their understanding of the concepts.

• There are two exercises (22 fully solved questions) in Class 11 Statistics Solutions.

## Access Exercise wise NCERT Solutions for Chapter 13 Maths Class 11

 Current Syllabus Exercises of Class 11 Maths Chapter 13 NCERT Solutions of Class 11 Maths Statistics Exercise 13.1 NCERT Solutions of Class 11 Maths Statistics Exercise 13.2 NCERT Solutions of Class 11 Maths Statistics Miscellaneous Exercise
Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 13 – Statistics

### Exercise 13.1

1. Find the mean deviation about the mean for the data ${\text{4,}}\,{\text{7,}}\,{\text{8,}}\,{\text{9,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13,}}\,{\text{17}}$.

Ans: Consider the given data, which is, $4,\,7,\,8,\,9,\,10,\,12,\,13,\,17$.

Therefore, the mean of the data is,

$\overline x = \dfrac{{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}}{8} = \dfrac{{80}}{8} = 10$

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x$ can be calculated as, $- 6,\, - 3, - 2,\, - 1,\,0,\,2,\,3,\,7$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $6,\,3,2,\,1,\,0,\,2,\,3,\,7$.

Now, the mean deviation about the mean is,

$\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {\left| {{x_i} - \overline x } \right|} }}{8} = \dfrac{{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}}{8} = \dfrac{{24}}{8} = 3$

2. Find the mean deviation about the mean for the data ${\text{38,}}\,{\text{70,}}\,{\text{48,40,}}\,{\text{42,}}\,{\text{55,}}\,{\text{63,}}\,{\text{46,}}\,{\text{54,}}\,{\text{44}}$.

Ans: Consider the given data, which is, $38,\,70,\,48,40,\,42,\,55,\,63,\,46,\,54,\,44$.

Therefore, the mean of the data is,

$\overline x = \dfrac{{38\, + 70 + \,48 + 40 + \,42 + \,55 + \,63 + \,46 + \,54 + \,44}}{{10}} = \dfrac{{500}}{{10}} = 50$

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x$ can be calculated as, $- 12,\,20, - 2,\, - 10,\, - 8,\,5,\,13,\, - 4,\,4,\, - 6$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $12,\,20,2,\,10,\,8,\,5,\,13,\,4,\,4,\,6$.

Now, the mean deviation about the mean is,

$\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - \overline x } \right|} }}{{10}} = \dfrac{{12 + \,20 + 2 + \,10 + \,8 + \,5 + \,13 + \,4 + \,4 + \,6}}{{10}} = \dfrac{{84}}{{10}} = 8.4$

3. Find the mean deviation about the median for the data ${\text{13,1}}\,{\text{7,}}\,{\text{16,14,}}\,{\text{11,}}\,{\text{13,}}\,{\text{10,}}\,{\text{16,}}\,{\text{11,}}\,{\text{18,}}\,{\text{12,}}\,{\text{17}}$.

Ans: Consider the given data, which is, $13,1\,7,\,16,14,\,11,\,13,\,10,\,16,\,11,\,18,\,12,\,17$.

Observe that the number of observations in this case is $12$, that is, even and on arranging the data in ascending order it can be obtained as, $10,1\,1,\,11,12,\,13,\,13,\,14,\,16,\,16,\,17,\,17,\,18$  Therefore, the median of the data is the average of the ${6^{th}}$ and the ${7^{th}}$ observations,

$\therefore M = \dfrac{{13 + 14}}{2} = \dfrac{{27}}{2} = 13.5$

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $- 3.5,\, - 2.5, - 2.5,\, - 1.5,\, - 0.5,\, - 0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $3.5,\,2.5,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$.

Now, the mean deviation about the median is,

$\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{12} {\left| {{x_i} - M} \right|} }}{{12}} = \dfrac{{3.5 + \,2.5 + 2.5 + \,1.5 + \,0.5 + \,0.5 + \,0.5 + \,2.5 + \,2.5 + \,3.5 + 3.5 + 4.5}}{{12}}$$\Rightarrow M.D.\left( M \right) = \dfrac{{28}}{{12}} = 2.33$

4. Find the mean deviation about the median for the data ${\text{36,72,}}\,{\text{46,42,}}\,{\text{60,}}\,{\text{45,}}\,{\text{53,}}\,{\text{46,}}\,{\text{51,}}\,{\text{49}}$.

Ans: Consider the given data, which is, $36,72,\,46,42,\,60,\,45,\,53,\,46,\,51,\,49$.

Observe that the number of observations in this case is $10$, that is, even and on arranging the data in ascending order it can be obtained as, $36,42,\,45,46,\,46,\,49,\,51,\,53,\,60,\,72$  Therefore, the median of the data is the average of the ${5^{th}}$ and the ${6^{th}}$ observations,

$\therefore M = \dfrac{{46 + 49}}{2} = \dfrac{{95}}{2} = 47.5$

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $- 11.5,\, - 5.5, - 2.5,\, - 1.5,\, - 1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $11.5,\,5.5,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Now, the mean deviation about the median is,

$\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - M} \right|} }}{{10}} = \dfrac{{11.5 + \,5.5 + 2.5 + \,1.5 + \,1.5 + \,1.5 + \,3.5 + \,5.5 + \,12.5 + \,24.5}}{{10}}$$\Rightarrow M.D.\left( M \right) = \dfrac{{70}}{{10}} = 7$

5. Find the mean deviation about the mean for the data.

 ${{\text{x}}_{\text{i}}}$ ${\text{5}}$ ${\text{10}}$ ${\text{15}}$ ${\text{20}}$ ${\text{25}}$ ${{\text{f}}_{\text{i}}}$ ${\text{7}}$ ${\text{4}}$ ${\text{6}}$ ${\text{3}}$ ${\text{5}}$

Ans: Consider the given data and observe the table as shown below, which is,

 ${x_i}$ ${f_i}$ ${f_i}{x_i}$ $\left| {{x_i} - \overline x } \right|$ ${f_i}\left| {{x_i} - \overline x } \right|$ $5$ $7$ $35$ $9$ $63$ $10$ $4$ $40$ $4$ $16$ $15$ $6$ $90$ $1$ $6$ $20$ $3$ $60$ $6$ $18$ $25$ $5$ $125$ $11$ $55$ $25$ $350$ $158$

The mean of the data can be calculated as shown below,

$\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} = \dfrac{1}{{25}} \times 350 = 14$

Therefore, the mean deviation about the mean is,

$M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{158}}{{25}} = 6.32$

6. Find the mean deviation about the mean for the data.

 ${{\text{x}}_{\text{i}}}$ ${\text{10}}$ ${\text{30}}$ ${\text{50}}$ ${\text{70}}$ ${\text{90}}$ ${{\text{f}}_{\text{i}}}$ ${\text{4}}$ ${\text{24}}$ ${\text{28}}$ ${\text{16}}$ ${\text{8}}$

Ans: Consider the given data and observe the table as shown below, which is,

 ${x_i}$ ${f_i}$ ${f_i}{x_i}$ $\left| {{x_i} - \overline x } \right|$ ${f_i}\left| {{x_i} - \overline x } \right|$ $10$ $4$ $40$ $40$ $160$ $30$ $24$ $720$ $20$ $480$ $50$ $28$ $1400$ $0$ $0$ $70$ $16$ $1120$ $20$ $320$ $90$ $8$ $720$ $40$ $320$ $80$ $4000$ $1280$

The mean of the data can be calculated as shown below,

$\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}} = \dfrac{1}{{80}} \times 4000 = 50$

Therefore, the mean deviation about the mean is,

$M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1280}}{{80}} = 16$

7. Find the mean deviation about the median for the data.

 ${{\text{x}}_{\text{i}}}$ ${\text{5}}$ ${\text{7}}$ ${\text{9}}$ ${\text{10}}$ ${\text{12}}$ ${\text{15}}$ ${{\text{f}}_{\text{i}}}$ ${\text{8}}$ ${\text{6}}$ ${\text{2}}$ ${\text{2}}$ ${\text{2}}$ ${\text{6}}$

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

 ${x_i}$ ${f_i}$ $c.f.$ $5$ $8$ $8$ $7$ $6$ $14$ $9$ $2$ $16$ $10$ $2$ $18$ $12$ $2$ $20$ $15$ $6$ $26$

Observe that the number of observations in this case is $26$, that is, even and therefore, the median is the mean of the ${13^{th}}$ and the ${14^{th}}$ observations. Observe that both of these observations lie in the cumulative frequency $14$, for which the corresponding observation is obtained as $7$,

$\therefore M = \dfrac{{7 + 7}}{2} = \dfrac{{14}}{2} = 7$

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,

 $\left| {{x_i} - M} \right|$ $2$ $0$ $2$ $3$ $5$ $8$ ${f_i}$ $8$ $6$ $2$ $2$ $2$ $6$ ${f_i}\left| {{x_i} - M} \right|$ $16$ $0$ $4$ $6$ $10$ $48$

Therefore, the mean deviation about the mean is,

$M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{16 + 4 + 6 + 10 + 48}}{{26}} = \dfrac{{84}}{{26}} = 3.23$

8. Find the mean deviation about the median for the data.

 ${{\text{x}}_{\text{i}}}$ ${\text{15}}$ ${\text{21}}$ ${\text{27}}$ ${\text{30}}$ ${\text{35}}$ ${{\text{f}}_{\text{i}}}$ ${\text{3}}$ ${\text{5}}$ ${\text{6}}$ ${\text{7}}$ ${\text{8}}$

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

 ${x_i}$ ${f_i}$ $c.f.$ $15$ $3$ $3$ $21$ $5$ $8$ $27$ $6$ $14$ $30$ $7$ $21$ $35$ $8$ $29$

Observe that the number of observations, in this case, is $29$, which is, odd and therefore, the median is the ${15^{th}}$ observation. Observe that this observation lie in the cumulative frequency $21$, for which the corresponding observation is obtained as $30$,

$\therefore M = 30$

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,

 $\left| {{x_i} - M} \right|$ $15$ $9$ $3$ $0$ $5$ ${f_i}$ $3$ $5$ $6$ $7$ $8$ ${f_i}\left| {{x_i} - M} \right|$ $45$ $45$ $18$ $0$ $40$

Therefore, the mean deviation about the median is,

$M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{45 + 45 + 18 + 0 + 40}}{{29}} = \dfrac{{148}}{{29}} = 5.1$

9. Find the mean deviation about the mean for the data.

 Income Per Day Number of Persons ${\text{0 - 100}}$ ${\text{4}}$ ${\text{100 - 200}}$ ${\text{8}}$ ${\text{200 - 300}}$ ${\text{9}}$ ${\text{300 - 400}}$ ${\text{10}}$ ${\text{400 - 500}}$ ${\text{7}}$ ${\text{500 - 600}}$ ${\text{5}}$ ${\text{600 - 700}}$ ${\text{4}}$ ${\text{700 - 800}}$ ${\text{3}}$

Ans: Consider the given data and observe the table as shown below, which is,

 Income per day Number of persons $\left( {{f_i}} \right)$ Midpoint $\left( {{x_i}} \right)$ ${f_i}{x_i}$ $\left| {{x_i} - \overline x } \right|$ ${f_i}\left| {{x_i} - \overline x } \right|$ $0 - 100$ $4$ $50$ $200$ $308$ $1232$ $100 - 200$ $8$ $150$ $1200$ $208$ $1664$ $200 - 300$ $9$ $250$ $2250$ $108$ $972$ $300 - 400$ $10$ $350$ $3500$ $8$ $80$ $400 - 500$ $7$ $450$ $3150$ $92$ $644$ $500 - 600$ $4$ $550$ $2750$ $192$ $960$ $600 - 700$ $5$ $650$ $2600$ $292$ $1168$ $700 - 800$ $3$ $750$ $2250$ $392$ $1176$ $50$ $17900$ $7896$

The mean of the data can be calculated as shown below,

$\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^8 {{f_i}{x_i}} = \dfrac{1}{{50}} \times 17900 = 358$

Therefore, the mean deviation about the mean is,

$M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{7896}}{{50}} = 157.92$

10. Find the mean deviation about the mean for the data.

 Height in cms Number of boys ${\text{95 - 105}}$ ${\text{9}}$ ${\text{105 - 115}}$ ${\text{13}}$ ${\text{115 - 125}}$ ${\text{26}}$ ${\text{125 - 135}}$ ${\text{30}}$ ${\text{135 - 145}}$ ${\text{12}}$ ${\text{145 - 155}}$ ${\text{10}}$

Ans: Consider the given data and observe the table as shown below, which is,

 Height in cms Number of boys $\left( {{f_i}} \right)$ Midpoint $\left( {{x_i}} \right)$ ${f_i}{x_i}$ $\left| {{x_i} - \overline x } \right|$ ${f_i}\left| {{x_i} - \overline x } \right|$ $95 - 105$ $9$ $100$ $900$ $25.3$ $227.7$ $105 - 115$ $13$ $110$ $1430$ $15.3$ $198.9$ $115 - 125$ $26$ $120$ $3120$ $5.3$ $137.8$ $125 - 135$ $30$ $130$ $3900$ $4.7$ $141$ $135 - 145$ $12$ $140$ $1680$ $14.7$ $176.4$ $145 - 155$ $10$ $150$ $1500$ $24.7$ $247$ $100$ $12530$ $1128.8$

The mean of the data can be calculated as shown below,

$\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}} = \dfrac{1}{{100}} \times 12530 = 125.3$

Therefore, the mean deviation about the mean is,

$M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1128.8}}{{100}} = 11.28$

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

Ans: Consider the given data and observe the table as shown below, which is,

 Marks Number of girls$\left( {{f_i}} \right)$ Cumulative Frequency(c.f) Midpoint $\left( {{x_i}} \right){\text{\;}}$ $\left| {{x_i} - \bar x} \right|$ ${f_i}\left| {{x_i} - \bar x} \right|$ $0 - 10$ 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 $\frac{{{N^{th}}}}{2}$ or ${25^{th}}$ item is $20 - 30$

### Class 11 Maths NCERT Solutions Chapter 13

Class 11 Maths Ch 15 NCERT Solutions shows the sums on a step-by-step basis. The concepts touched up in the chapter and the example sums are indicated below:

1. Ungrouped

Example: Finding mean deviation about the mean of 6, 7, 10, 12, 13, 12, 8, 4

1. Discrete Frequency

Example: Finding mean deviation about the mean of the below data:

 xi 2 5 6 8 10 12 fi 2 8 10 7 8 5
1. Continuous Frequency Distribution

Example: Finding the mean deviation about the mean of the below data:

 Obtained Marks Number of Students (fi) Mid-point(xi) fixi 10-20 2 (10+20)/2 = 15 2 x 15 = 30 20-30 3 15+10 = 25 3 x 25 = 75 30-40 8 35 8 x 35 = 280 45-50 14 45 14 x 45 = 630 50-60 8 55 8 x 55 = 440 60-70 3 65 3 x 65 = 195 70-80 2 75 2 x 75 = 150

1. Ungrouped

Example: Finding the mean deviation about the median of 3, 5, 9, 3, 10, 12, 4, 18, 7, 19, 21

1. Discrete Frequency

Example: Finding mean deviation about the median of the below data:

 xi 3 6 9 12 13 15 21 22 fi 3 4 5 2 4 5 4 3
1. Continuous Frequency Distribution

Example: Calculation of the mean deviation about median of the below data:

 Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 6 7 15 16 4 2

• Standard Deviation and Variance

1. Ungrouped Data

Example: Finding the variance of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

1. Discrete Frequency

Example: Finding the variance and standard deviation of the below data:

 xi 4 8 11 17 20 24 32 fi 3 5 9 5 4 4 1
1. Continuous Frequency

Example: Calculation of the mean, variance and standard deviation of the following distribution:

 Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency (fi) 3 7 12 15 8 3 2
• Coefficient of Variation

Example: Determine which group is more variable from the below data:

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group M 9 17 32 33 40 10 9 Group N 10 20 30 25 43 15 7
• Indirect Questions

1. Multiplication of Observation

Example: Given that the mean and standard deviation of six observations are 8 and 4 respectively, if each observation is multiplied by 3, what would be the new mean and new standard deviation of the resultant observations?

1. Finding Remaining Observations

Example: The variance and mean of 7 observations are 16 and 8 respectively. If 5 of the observations amount to 2, 4, 10, 12, 14, what are the remaining 2 observations?

1. Incorrect Observation

Example: The standard deviation and mean of 100 observations are 5.1 and 40 respectively. However, a mistake was detected in case of one observation where instead of 40, 50 was taken. What would be the correct standard deviation and mean?

## Overview of Deleted Syllabus for CBSE Class 11 Maths Statistics

 Chapter Dropped Topics Statistics 13.6 - Analysis of Frequency Distributionand Question No. 6 Miscellaneous Exercise The last point in the Summary

## Class 11 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 12 Questions & Solutions Exercise 13.2 10 Questions & Solutions Miscellaneous Exercise 6 Questions & Solutions

## Conclusion

Class 11 Maths Statistics Solutions, we have thoroughly examined the principles and methods of statistics, including measures of dispersion, mean deviation, variance, and standard deviation. Understanding these concepts is crucial for interpreting and analyzing data effectively. By mastering these tools, Students can calculate and interpret measures of central tendency and dispersion, applying them to real-world data for meaningful insights. This chapter is significant in exams, with an average of 3-4 questions being asked in previous years.

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Maths Chapter 13 Statistics

1. What are the Different Sub-Topics of Chapter 13 Maths Class 11?

• The sub-topics of Class 11 Statistics include:

• Measure of Dispersion

• Range

• Mean Deviation

• Variance and Standard Deviation

• Frequency Distribution Analysis.

2. What is Understood by Standard Deviation?

Standard deviation in solution of Statistics Class 11 deals with the measurement of the amount of dispersion or variation of a given set of values. Depending on the level of standard deviation, the range is determined.

3. Why are Class 11 Maths Chapter 13 NCERT Solutions Useful to Students?

Class 11 Statistics NCERT solutions include accurate explanations to the problems and also, it is indicated in a step-by-step method. It becomes easier for students to understand how to understand similar questions.

4. What are the topics discussed in Chapter 13 of NCERT Solutions for Class 11 Maths?

The Class 11 Statistics Maths Solutions is the branch that deals with the collection, organization, and interpretation of data. This chapter talks about the concepts of central tendency, mean, and median. You can find the NCERT Solutions Class 11 Maths Chapter 13 on Vedantu. It will help you clear your doubts and concepts.

5. What are the important formulae of Statistics Class 11?

Formulae of mean, median, variance and standard deviation are important in Class 11. Once you finish them, you will be able to develop an explicit understanding of the topics covered in the CBSE Class 11 Maths syllabus. To do this, you need to solve and practice each sum from the examples and exercises given in the NCERT Class 11 Maths CBSE book. You can find the solutions for these exercises in Vedantu’s NCERT Solutions for Class 11 Maths Statistics. The solutions are free.

6. Name the important concepts discussed in Chapter 13 of NCERT Solutions for Class 11 Maths.

Vedantu offers study material designed in a simple, straightforward language, which is easy to memorize. Class 11 Statistics Maths Solutions, provides you with an idea of graphs, histograms, pie charts, and bar graphs. It works on your analytical skills. Relevant diagrams, graphs, and illustrations are provided along with the answers wherever required.

7. How can I make a Study Plan for Class 11?

You may find Class 11 a little difficult as you are introduced to new streams and subjects. It may be messy to dwell in the new study pattern because it requires hard and smart work both to achieve good marks in exams. Setting a routine and working according to it, and dividing the time for study and relaxing will help you. You need to cut off the distractions and focus on studies to excel in exams. Use the Vedantu Mobile app for better comfort and portability.