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.
 5.1Exercise 13.1
 5.2Exercise 13.2
 5.3Miscellaneous Exercise
 6.1Class 11 Maths NCERT Solutions Chapter 13
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.
This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 13  Statistics, which you can download as PDFs.
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 Miscellaneous Exercise 
Exercises Under NCERT Solutions for Class 11 Maths Chapter 13 Statistics
Exercise 13.1: In this exercise, students will learn about measures of central tendency, including mean, median, and mode, and their properties. They will also practice finding the measures of central tendency for grouped data and ungrouped data.
Exercise 13.2: This exercise focuses on measures of dispersion, including range, quartile deviation, mean deviation, and standard deviation. Students will learn about the properties of these measures and practice finding them for grouped data and ungrouped data.
Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of statistics to solve various problems and answer questions. They will also practice finding measures of central tendency and dispersion for grouped data and ungrouped data, as well as organizing data using frequency distribution tables.
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 $20  30$
$\therefore $$20  30$ is the median class
Median = $l + \left( {\frac{{\left( {\left( {\frac{N}{2}} \right)  c} \right)}}{f}} \right) \times h$ …….(1)
Where $l = 20,\,\,c = 14,\,\,f = 14,\,\,h = 10,\,\,n = 50$
Substitute the above values in (1), we get
Median = $20 + \left( {\left( {\left( {\frac{{25  14}}{{14}}} \right)} \right)} \right) \times 10$
$\begin{gathered} = 20 + \left( {\left( {\frac{{11}}{{14}}} \right)} \right) \times 10 \hfill \\ \hfill \\ \end{gathered} $ $\begin{gathered} = 20 + \left( {\frac{{110}}{{14}}} \right) \hfill \\ \hfill \\ \end{gathered} $
\[ = 20 + 7.85\]
$ = 27.85$
So, \[\sum\limits_{i = 1}^6 {{f_i}\left {{x_i}  \bar x} \right} \, = \,517.1\]
Therefore, the mean deviation about the median is,
M.D = \[\begin{gathered} \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left {{x_i}  \bar x} \right} \hfill \\ \, \hfill \\ \end{gathered} \]
$ = \,\frac{1}{{50}} \times 517.1$
$ = \,10.34$
12. Calculate the mean deviation about the median age for the age distribution of ${\text{100}}$ persons.
Age  Number 
${\text{16  20}}$  ${\text{5}}$ 
${\text{21  25}}$  ${\text{6}}$ 
${\text{26  30}}$  ${\text{12}}$ 
${\text{31  35}}$  ${\text{14}}$ 
${\text{36  40}}$  ${\text{26}}$ 
${\text{41  45}}$  ${\text{12}}$ 
${\text{46  50}}$  ${\text{16}}$ 
${\text{51  55}}$  ${\text{9}}$ 
Ans: It can be clearly observed that the given data is not continuous and therefore it needs to be converted into a continuous frequency distribution, which can be done by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval. Now, observe the table as shown below, which is,
Age  Number$\left( {{f_i}} \right)$  Cumulative Frequency (c.f.)  Midpoint $\left( {{x_i}} \right)$  $\left {{x_i}  M} \right$  ${f_i}\left {{x_i}  M} \right$ 
$15.5  20.5$  $5$  $5$  $18$  $20$  $100$ 
$20.5  25.5$  $6$  $11$  $23$  $15$  $90$ 
$25.5  30.5$  $12$  $23$  $28$  $10$  $120$ 
$30.5  35.5$  $14$  $37$  $33$  $5$  $70$ 
$35.5  40.5$  $26$  $63$  $38$  $0$  $0$ 
$40.5  45.5$  $12$  $75$  $43$  $5$  $60$ 
$45.5  50.5$  $16$  $91$  $48$  $10$  $160$ 
$50.5  55.5$  $9$  $100$  $53$  $15$  $135$ 
$100$  $735$ 
Observe that the class interval containing the ${\left( {\dfrac{N}{2}} \right)^{th}}$ item or the ${50^{th}}$ item is $35.5  40.5$. Thus, $35.5  40.5$ is the median class.
Therefore, median of the data can be calculated as shown below,
\[\therefore M = l + \dfrac{{\dfrac{N}{2}  C}}{f} \times h\] ( where l = 35.5, C = 37, f = 26, h = 5 and N = 100)
\[ \Rightarrow M = 35.5 + \dfrac{{50  37}}{{26}} \times 5 = 35.5 + \dfrac{{13}}{{26}} \times 5 = 35.5 + 2.5 = 38\]
Therefore, the mean deviation about the median is,
\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left {{x_i}  M} \right} }}{N} = \dfrac{{735}}{{100}} = 7.35\]
Exercise 13.2
1. Find the mean and variance for the data \[{\text{6,}}\,{\text{7,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13, 4,}}\,{\text{8,}}\,{\text{12}}\].
Ans: Consider the given data which is, \[6,\,7,\,10,\,12,\,13,{\text{ }}4,\,8,\,12\].
The mean of the data can be calculated as shown below,
\[\therefore \overline x = \dfrac{1}{n}\sum\limits_{i = 1}^8 {{x_i}} = \dfrac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8} = \dfrac{{72}}{8} = 9\]
Now, observe the table as shown below, which is,
${x_i}$  $\left( {{x_i}  \overline x } \right)$  ${\left( {{x_i}  \overline x } \right)^2}$ 
$6$  $  3$  $9$ 
$7$  $  2$  $4$ 
$10$  $  1$  $1$ 
$12$  $3$  $9$ 
$13$  $4$  $16$ 
$4$  $  5$  $25$ 
$8$  $  1$  $1$ 
$12$  $3$  $9$ 
$74$ 
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i}  \overline x } \right)}^2}} }}{n} = \dfrac{{74}}{8} = 9.25\]
2. Find the mean and variance for the first ${\text{n}}$ natural numbers.
Ans: The mean of the first $n$ natural numbers can be calculated as shown below,
\[\therefore \overline x = \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n} = \dfrac{{n + 1}}{2}\]
Now, the variance can be calculated as,
\[\therefore Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i}  \overline x } \right)}^2}} }}{n}\]
\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left[ {{x_i}  {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \right]} \]
\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2}  \dfrac{1}{n}\sum\limits_{i = 1}^n {2\left( {\dfrac{{n + 1}}{n}} \right){x_i}} + \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \]
\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{\left( {n + 1} \right)(2n + 1)}}{6}  \dfrac{{{{\left( {n + 1} \right)}^2}}}{2} + \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}\]
\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \left( {n + 1} \right)\left[ {\dfrac{{4n + 2  3n  3}}{{12}}} \right]\]
\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{{n^2}  1}}{{12}}\]
3. Find the mean and variance for the first ${\text{10}}$ multiples of ${\text{3}}$.
Ans: Observe that the first $10$ multiples of $3$ are, \[3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27,\,30\].
The mean of the data can be calculated as shown below,
\[\therefore \overline x = \dfrac{1}{n}\sum\limits_{i = 1}^{10} {{x_i}} = \dfrac{{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}}{{10}} = \dfrac{{165}}{{10}} = 16.5\]
Now, observe the table as shown below, which is,
${x_i}$  $\left( {{x_i}  \overline x } \right)$  ${\left( {{x_i}  \overline x } \right)^2}$ 
$3$  $  13.5$  \[182.25\] 
$6$  $  10.5$  \[110.25\] 
$9$  $  7.5$  \[56.25\] 
$12$  $  4.5$  $20.25$ 
$15$  $  1.5$  $2.25$ 
$18$  $1.5$  $2.25$ 
$21$  $4.5$  $20.25$ 
$24$  $7.5$  \[56.25\] 
$27$  $10.5$  \[110.25\] 
$30$  $13.5$  \[182.25\] 
\[742.5\] 
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i}  \overline x } \right)}^2}} }}{n} = \dfrac{{742.5}}{{10}} = 74.25\]
4. Find the mean and variance for the data.
${{\text{x}}_{\text{i}}}$  ${\text{6}}$  ${\text{10}}$  ${\text{14}}$  ${\text{18}}$  ${\text{24}}$  ${\text{28}}$  ${\text{30}}$ 
${{\text{f}}_{\text{i}}}$  ${\text{2}}$  ${\text{4}}$  ${\text{7}}$  ${\text{12}}$  ${\text{8}}$  ${\text{4}}$  ${\text{3}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
${x_i}$  ${f_i}$  ${f_i}{x_i}$  $\left( {{x_i}  \overline x } \right)$  ${\left( {{x_i}  \overline x } \right)^2}$  ${f_i}{\left( {{x_i}  \overline x } \right)^2}$ 
$6$  $2$  $12$  $  13$  $169$  $338$ 
$10$  $4$  $40$  $  9$  $81$  $324$ 
$14$  $7$  $98$  $  5$  $25$  $175$ 
$18$  $12$  $216$  $  1$  $1$  $12$ 
$24$  $8$  $192$  $5$  $25$  $200$ 
$28$  $4$  $112$  $9$  $81$  $324$ 
$30$  $3$  $90$  $11$  $121$  $363$ 
$30$  $760$  $1736$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}} = \dfrac{1}{{40}} \times 760 = 19\]
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i}  \overline x } \right)}^2}} }}{N} = \dfrac{{1736}}{{40}} = 43.4\]
5. Find the mean and variance for the data.
${{\text{x}}_{\text{i}}}$  ${\text{92}}$  ${\text{93}}$  ${\text{97}}$  ${\text{98}}$  ${\text{102}}$  ${\text{104}}$  ${\text{109}}$ 
${{\text{f}}_{\text{i}}}$  ${\text{3}}$  ${\text{2}}$  ${\text{3}}$  ${\text{2}}$  ${\text{6}}$  ${\text{3}}$  ${\text{3}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
${x_i}$  ${f_i}$  ${f_i}{x_i}$  $\left( {{x_i}  \overline x } \right)$  ${\left( {{x_i}  \overline x } \right)^2}$  ${f_i}{\left( {{x_i}  \overline x } \right)^2}$ 
$92$  $3$  $276$  $  8$  $64$  $192$ 
$93$  $2$  $186$  $  7$  $49$  $98$ 
$97$  $3$  $291$  $  3$  $9$  $27$ 
$98$  $2$  $196$  $  2$  $4$  $8$ 
$102$  $6$  $612$  $2$  $4$  $24$ 
$104$  $3$  $312$  $4$  $16$  $48$ 
$109$  $3$  $327$  $9$  $81$  $243$ 
$22$  $2200$  $640$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}} = \dfrac{1}{{22}} \times 2200 = 100\]
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i}  \overline x } \right)}^2}} }}{N} = \dfrac{{640}}{{22}} = 29.09\]
6. Find the mean, variance and standard deviation using the shortcut method.
${{\text{x}}_{\text{i}}}$  ${\text{60}}$  ${\text{61}}$  ${\text{62}}$  ${\text{63}}$  ${\text{64}}$  ${\text{65}}$  ${\text{66}}$  ${\text{67}}$  ${\text{68}}$ 
${{\text{f}}_{\text{i}}}$  ${\text{2}}$  ${\text{1}}$  ${\text{12}}$  ${\text{29}}$  ${\text{25}}$  ${\text{12}}$  ${\text{10}}$  ${\text{4}}$  ${\text{5}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
${x_i}$  ${f_i}$  ${y_i} = \dfrac{{{x_i}  64}}{1}$  ${y_i}^2$  ${f_i}{y_i}$  ${f_i}{y_i}^2$ 
$60$  $2$  $  4$  $16$  $  8$  $32$ 
$61$  $1$  $  3$  $9$  $  3$  $9$ 
$62$  $12$  $  2$  $4$  $  24$  $48$ 
$63$  $29$  $  1$  $1$  $  29$  $29$ 
$64$  $25$  $0$  $0$  $0$  $0$ 
$65$  $12$  $1$  $1$  $12$  $12$ 
$66$  $10$  $2$  $4$  $20$  $40$ 
$67$  $5$  $3$  $9$  $12$  $36$ 
$68$  $4$  $4$  $16$  $20$  $80$ 
$100$  $0$  $286$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 64 + \dfrac{0}{{100}} \times 1 = 64\]
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2  \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{1}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 286  0} \right] = 2.86\]
Now the standard deviation can be calculated as shown below,
\[\therefore \sigma = \sqrt {2.86} = 1.69\]
7. Find the mean and variance for the following frequency distribution.
Classes  ${\text{0  30}}$  ${\text{30  60}}$  ${\text{60  90}}$  ${\text{90  120}}$  ${\text{120  150}}$  ${\text{150  180}}$  ${\text{180  210}}$ 
Frequencies  ${\text{2}}$  ${\text{3}}$  ${\text{5}}$  ${\text{10}}$  ${\text{3}}$  ${\text{5}}$  ${\text{2}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
Class  Frequency ${f_i}$  Midpoint $\left( {{x_i}} \right)$  ${y_i} = \dfrac{{{x_i}  105}}{{30}}$  ${y_i}^2$  ${f_i}{y_i}$  ${f_i}{y_i}^2$ 
$0  30$  $2$  $15$  $  3$  $9$  $  6$  $18$ 
$30  60$  $3$  $45$  $  2$  $4$  $  6$  $12$ 
$60  90$  $5$  $75$  $  1$  $1$  $  5$  $5$ 
$90  120$  $10$  $105$  $0$  $0$  $0$  $0$ 
$120  150$  $3$  $135$  $1$  $1$  $3$  $3$ 
$150  180$  $5$  $165$  $2$  $4$  $10$  $20$ 
$180  210$  $2$  $195$  $3$  $9$  $6$  $18$ 
$30$  $2$  $76$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = A + \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h = 105 + \dfrac{2}{{30}} \times 30 = 107\]
Therefore, the variance is,
\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2  {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {30} \right)}^2}}}{{{{\left( {30} \right)}^2}}}\left[ {30 \times 76  {{\left( 2 \right)}^2}} \right] = 2280  4 = 2276\]
8. Find the mean and variance for the following frequency distribution.
Classes  ${\text{0  10}}$  ${\text{10  20}}$  ${\text{20  30}}$  ${\text{30  40}}$  ${\text{40  50}}$ 
Frequencies  ${\text{5}}$  ${\text{8}}$  ${\text{15}}$  ${\text{16}}$  ${\text{6}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
Class  Frequency ${f_i}$  Midpoint $\left( {{x_i}} \right)$  ${y_i} = \dfrac{{{x_i}  25}}{{10}}$  ${y_i}^2$  ${f_i}{y_i}$  ${f_i}{y_i}^2$ 
$0  10$  $5$  $5$  $  2$  $4$  $  10$  $20$ 
$10  20$  $8$  $15$  $  1$  $1$  $  8$  $8$ 
$20  30$  $15$  $25$  $0$  $0$  $0$  $0$ 
$30  40$  $16$  $35$  $1$  $1$  $16$  $16$ 
$40  50$  $6$  $45$  $2$  $4$  $12$  $24$ 
$50$  $10$  $68$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 25 + \dfrac{{10}}{{50}} \times 10 = 27\]
Therefore, the variance is,
$ Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2  {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {50} \right)}^2}}}\left[ {50 \times 68  {{\left( {10} \right)}^2}} \right] \\$
$ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{{25}}\left[ {3400  100} \right] = 132 \\ $
9. Find the mean, variance and standard deviation using the shortcut method.
Height in Cms  Number of Children 
${\text{70  75}}$  ${\text{3}}$ 
${\text{75  80}}$  ${\text{4}}$ 
${\text{80  85}}$  ${\text{7}}$ 
${\text{85  90}}$  ${\text{7}}$ 
${\text{90  95}}$  ${\text{15}}$ 
${\text{95  100}}$  ${\text{9}}$ 
${\text{100  105}}$  ${\text{6}}$ 
${\text{105  110}}$  ${\text{6}}$ 
${\text{110  115}}$  ${\text{3}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
Class Interval  Frequency ${f_i}$  Midpoint $\left( {{x_i}} \right)$  ${y_i} = \dfrac{{{x_i}  92.5}}{5}$  ${y_i}^2$  ${f_i}{y_i}$  ${f_i}{y_i}^2$ 
$70  75$  $3$  $72.5$  $  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$  $7$ 
$90  95$  $15$  $92.5$  $0$  $0$  $0$  $0$ 
$95  100$  $9$  $97.5$  $1$  $1$  $9$  $9$ 
$100  105$  $6$  $102.5$  $2$  $4$  $12$  $24$ 
$105  110$  $6$  $107.5$  $3$  $9$  $18$  $54$ 
$110  115$  $3$  $112.5$  $4$  $16$  $12$  $48$ 
$60$  $6$  $254$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 92.5 + \dfrac{6}{{60}} \times 5 = 92.5 + 0.5 = 93\]
Therefore, the variance is,
$Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2  \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 5 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {60 \times 254  {{\left( 6 \right)}^2}} \right] \\$
$ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{25}}{{3600}}\left( {15204} \right) = 105.58 \\ $
Now the standard deviation can be calculated as shown below,
\[\therefore \sigma = \sqrt {105.58} = 10.27\]
10. The diameters of circles (in mm) drawn in a design are given below. Find the mean, variance and standard deviation using the shortcut method.
Diameters  No. of Circles 
${\text{33  36}}$  ${\text{15}}$ 
${\text{37  40}}$  ${\text{17}}$ 
${\text{41  44}}$  ${\text{21}}$ 
${\text{45  48}}$  ${\text{22}}$ 
${\text{49  52}}$  ${\text{25}}$ 
Ans: Consider the given data and observe the table as shown below, which is,
Class Interval  Frequency ${f_i}$  Midpoint $\left( {{x_i}} \right)$  ${y_i} = \dfrac{{{x_i}  92.5}}{5}$  ${y_i}^2$  ${f_i}{y_i}$  ${f_i}{y_i}^2$ 
$32.5  36.5$  $15$  $34.5$  $  2$  $4$  $  30$  $60$ 
$36.5  40.5$  $17$  $38.5$  $  1$  $1$  $  17$  $17$ 
$40.5  44.5$  $21$  $42.5$  $0$  $0$  $0$  $0$ 
$44.5  48.5$  $22$  $46.5$  $1$  $1$  $22$  $22$ 
$48.5  52.5$  $25$  $50.5$  $2$  $4$  $50$  $100$ 
$100$  $25$  $199$ 
The mean of the data can be calculated as shown below,
\[\therefore \overline x = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 42.5 + \dfrac{{25}}{{100}} \times 4 = 43.5\]
Therefore, the variance is,
$ Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^5 {{f_i}{y_i}^2  \left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 4 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 199  {{\left( {25} \right)}^2}} \right] \\$
$ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{16}}{{10000}}\left( {19900  625} \right) = \dfrac{{16}}{{10000}}\left( {19275} \right) = 30.84 \\\ $
Now the standard deviation can be calculated as shown below,
\[\therefore \sigma = \sqrt {30.84} = 5.55\]
Miscellaneous Exercise
1. The mean and variance of eight observations are ${\text{9}}$ and ${\text{9}}{\text{.25}}$ respectively. If six of the observations are \[{\text{6,}}\,{\text{7,10,}}\,{\text{12,}}\,{\text{12}}\] and \[{\text{13}}\], find the remaining two observations.
Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[6,\,7,10,\,12,\,12,\,13,\,x,\,y\].
Therefore, from the mean of the data it can be obtained that,
\[\overline x = \dfrac{{6 + 7 + \,10 + 12 + \,12 + 13 + x + y}}{8} = 9\]
\[ \Rightarrow 60 + x + y = 72\]
\[ \Rightarrow x + y = 12\] ......(i)
Again, from the variance of the data it can be obtained that,
\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i}  \overline x } \right)}^2}} }}{N} = 9.25\]
\[\therefore \dfrac{1}{8}\left[ {{{\left( {  3} \right)}^2} + {{\left( {  2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2}  (2)(9)\left( {x + y} \right) + \left( {2{{\left( 9 \right)}^2}} \right)} \right] = 9.25\]
\[ \Rightarrow \dfrac{1}{8}\left[ {9 + 4 + 9 + 16 + {x^2} + {y^2}  (18)\left( {12} \right) + 162} \right] = 9.25\] [By, using (i)]
\[ \Rightarrow \dfrac{1}{8}\left[ {{x^2} + {y^2}  6} \right] = 9.25\]
\[ \Rightarrow {x^2} + {y^2} = 80\] ......(ii)
Observe that form equation (i), it can be obtained that,
${x^2} + {y^2} + 2xy = 144$ ......(iii)
Also, from equations (ii) and (iii), it can be obtained that,
\[2xy = 64\] ......(iv)
Now, on subtracting equation (iv) from equation (ii), it can be obtained that,
\[{x^2} + {y^2}  2xy = 16\]
\[ \Rightarrow x  y = \pm 4\] ......(v)
It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 4$, when \[x  y = 4\] and $x = 4$ and $y = 8$, when \[x  y =  4\]. Henceforth, the remaining two observations are $4$ and $8$.
2. The mean and variance of ${\text{7}}$ observations are ${\text{8}}$ and ${\text{16}}$ respectively. If five of the observations are \[{\text{2,}}\,{\text{4,10,}}\,{\text{12}}\] and \[{\text{14}}\]. Find the remaining two observations.
Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[2,\,4,10,\,12,\,14,\,x,\,y\].
Therefore, from the mean of the data it can be obtained that,
\[\overline x = \dfrac{{2 + 4 + \,10 + 12 + \,14 + x + y}}{7} = 8\]
\[ \Rightarrow 42 + x + y = 56\]
\[ \Rightarrow x + y = 14\] ......(i)
Again, from the variance of the data it can be obtained that,
\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i}  \overline x } \right)}^2}} }}{N} = 16\]
\[\therefore \dfrac{1}{7}\left[ {{{\left( {  6} \right)}^2} + {{\left( {  4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2}  (2)(8)\left( {x + y} \right) + \left( {2{{\left( 8 \right)}^2}} \right)} \right] = 16\]
\[ \Rightarrow \dfrac{1}{7}\left[ {36 + 16 + 4 + 16 + {x^2} + {y^2}  (16)\left( {14} \right) + 128} \right] = 16\] [By, using (i)]
\[ \Rightarrow \dfrac{1}{7}\left[ {{x^2} + {y^2} + 12} \right] = 16\]
\[ \Rightarrow {x^2} + {y^2} = 100\] ......(ii)
Observe that form equation (i), it can be obtained that,
${x^2} + {y^2} + 2xy = 196$ ......(iii)
Also, from equations (ii) and (iii), it can be obtained that,
\[2xy = 96\] ......(iv)
Now, on subtracting equation (iv) from equation (ii), it can be obtained that,
\[{x^2} + {y^2}  2xy = 4\]
\[ \Rightarrow x  y = \pm 2\] ......(v)
It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 6$, when \[x  y = 2\]and $x = 6$ and $y = 8$, when \[x  y =  2\]. Henceforth, the remaining two observations are $6$ and $8$.
3. The mean and standard deviation of six observations are ${\text{8}}$ and ${\text{4}}$ respectively. If each observation is multiplied by ${\text{3}}$, find the new mean and new standard deviation of the resulting observations.
Ans: Assume the observations to be \[{x_1},\,{x_2},\,{x_3},\,{x_4},\,{x_5},\,{x_6}\].
Therefore, the mean of the data is,
\[\overline x = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\]
\[ \Rightarrow \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] ......(i)
Observe that when each of the observation is multiplied with $3$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,
$\therefore {y_1} = 3{x_1}$
$ \Rightarrow {x_1} = \dfrac{1}{3}{y_1},\,\forall i = 1,\,2,\,3,\,....,\,6\& i \in {\mathbb{Z}^ + }$
Now, the mean of the new data is,
\[\therefore \overline y = \dfrac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}\]
\[ \Rightarrow \overline y = \dfrac{{3({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6})}}{6}\]
\[ \Rightarrow \overline y = 3 \times 8\] [By using (i)]
\[ \Rightarrow \overline y = 24\]
Therefore, the standard deviation of the data is,
\[\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i}  \overline x } \right)}^2}} }}{n}} = 4\]
\[ \Rightarrow {\left( 4 \right)^2} = \dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i}  \overline x } \right)}^2}} }}{6}\]
\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{x_i}  \overline x } \right)}^2}} = 96\] ......(ii)
Again, it can be observed from equations (i) and (ii) that \[\overline y = 3\overline x \] and \[\overline x = \dfrac{{\overline y }}{3}\] and hence on substituting the values of ${x_i}$ and $\overline x $ in equation (ii) it can be clearly obtained as shown below,
\[\therefore \sum\limits_{i = 1}^6 {{{\left( {\dfrac{{{y_i}}}{3}  \dfrac{{\overline y }}{3}} \right)}^2}} = 96\]
\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_i}  \overline y } \right)}^2}} = 864\]
Henceforth, the standard deviation of the new data can be calculated as shown below,
\[\therefore \sigma = \sqrt {\dfrac{{864}}{6}} = \sqrt {144} = 12\]
4. Given that $\overline {\text{x}} $ is the mean and $\sigma^{2}$ is the variance of ${\text{n}}$ observations ${{\text{x}}_{\text{1}}}{\text{,}}\,{{\text{x}}_{\text{2}}}{\text{,}}\,......{\text{,}}\,{{\text{x}}_{\text{n}}}$.Prove that the mean and variance of observations ${\text{a}}{{\text{x}}_{\text{1}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{2}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{3}}}{\text{, }}.....{\text{,}}\,{\text{a}}{{\text{x}}_{\text{n}}}$ are ${\text{a}}\overline {\text{x}} $ and ${{\text{a}}^{\text{2}}}{{\sigma}}^{\text{2}}$, respectively $\left( {{\text{a}} \ne {\text{0}}} \right)$.
Ans: Observe that the given observations are ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.
\[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{y_i}{{\left( {{x_i}  \overline x } \right)}^2}} }}{n}\]......(i)
Observe that when each of the observation is multiplied with $a$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,
$\therefore {y_i} = a{x_i}$
$ \Rightarrow {x_i} = \dfrac{1}{a}{y_i},\,\forall i = 1,\,2,\,3,\,....,\,n\& i \in {\mathbb{Z}^ + }$
Now, the mean of the new data is,
$\therefore \overline y = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}} = \dfrac{1}{n}\sum\limits_{i = 1}^n {a{x_i}} = \dfrac{a}{n}\sum\limits_{i = 1}^n {{x_i}} = a\overline x \,\,\,\,\,\,\,\left[ {\because \overline x = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right]$
Again on substituting the values of ${x_i}$ and $\overline x $ in equation (i) it can be clearly obtained as shown below,
\[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {\dfrac{{{y_i}}}{a}  \dfrac{{\overline y }}{a}} \right)}^2}} }}{n}\]
\[ \Rightarrow {a^2}{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{y_i}  \overline y } \right)}^2}} }}{n}\]
Henceforth, it can be clearly proved that the mean and variance of the new data is $a\overline x $ and ${a^2}{\sigma ^2}$, respectively.
5. The mean and standard deviation of ${\text{20}}$ observations are found to be ${\text{10}}$ and ${\text{2}}$, respectively. On rechecking it was found that an observation ${\text{8}}$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
Ans: Observe that the incorrect number of observations, incorrect mean and the incorrect standard deviation are $20$, $10$ and $2$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.
\[\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n} = 10\]
\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 10\]
\[ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}} = 200\]
Now, the correct mean is,
\[\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{19} {{x_i}} }}{n}\]
\[ \Rightarrow \overline x = \dfrac{{192}}{{19}} = 10.1\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{19} {{x_i}} = 192} \right]\]
Again observe as shown below,
\[\therefore \sigma = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2  {{\left( {\overline x } \right)}^2}} } = 2\]
\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]
Therefore, the correct standard deviation of the data is,
\[\sum\limits_{i = 1}^{20} {{x_i}^2}  {\left( 8 \right)^2}\]
\[ \Rightarrow 2080  64\]
\[ \Rightarrow 2016\]
\[\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{19} {{x_i}^2} }}{n}  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {\dfrac{{2016}}{{19}}  {{\left( {10.1} \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {1061.1  102.1} = \sqrt {4.09} = 2.02\]
(ii) If it is replaced by ${\text{12}}$.
Ans: Observe that the incorrect sum of observations is $200$.
\[\therefore \sum\limits_{i = 1}^{20} {{x_i}} = 200  8 + 12 = 204\]
Now, the correct mean is,
\[\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n}\]
\[ \Rightarrow \overline x = \dfrac{{204}}{{20}} = 10.2\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{20} {{x_i}} = 204} \right]\]
Again observe as shown below,
\[\therefore \sigma = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2  {{\left( {\overline x } \right)}^2}} } = 2\]
\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]
Therefore, the correct standard deviation of the data is,
\[\sum\limits_{i = 1}^{20} {{x_i}^2}  {\left( 8 \right)^2} + {\left( {12} \right)^2}\]
\[ \Rightarrow 2080  64 + 144\]
\[ \Rightarrow 2160\]
\[\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{x_i}^2} }}{n}  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {\dfrac{{2160}}{{20}}  {{\left( {10.2} \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {108  104.04} = \sqrt {3.96} = 1.98\]
6. The mean and standard deviation of a group of ${\text{100}}$ observations were found to be ${\text{20}}$ and ${\text{3}}$, respectively. Later on it was found that ${\text{3}}$ observations are incorrect which were recorded as ${\text{21,}}\,{\text{21}}$ and ${\text{18}}$. Find the mean and standard deviation if the incorrect observations are omitted.
Ans: Observe that the number of observations, incorrect mean and the incorrect standard deviation are $100$, $20$ and $3$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.
\[\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{n} = 20\]
\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{{100}} = 20\]
\[ \Rightarrow \sum\limits_{i = 1}^{100} {{x_i}} = 2000\]
Now, the correct mean is,
\[\therefore \overline x = \dfrac{{\sum\limits_{i = 1}^{97} {{x_i}} }}{n}\]
\[ \Rightarrow \overline x = \dfrac{{2000  60}}{{97}} = 20\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{97} {{x_i}} = 1940} \right]\]
Again observe as shown below,
\[\therefore \sigma = \sqrt {\dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2  {{\left( {\overline x } \right)}^2}} } = 3\]
\[ \Rightarrow 9 = \dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow 40900 = \sum\limits_{i = 1}^{100} {{x_i}^2} \]
Therefore, the correct standard deviation of the data is,
\[\sum\limits_{i = 1}^{100} {{x_i}^2}  (2){\left( {21} \right)^2}  {(18)^2}\]
\[ \Rightarrow 40900  1206\]
\[ \Rightarrow 39694\]
\[\therefore \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{97} {{x_i}^2} }}{n}  {{\left( {\overline x } \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {\dfrac{{39694}}{{97}}  {{\left( {20} \right)}^2}} \]
\[ \Rightarrow \sigma = \sqrt {409.22  400} = \sqrt {9.22} = 3.04\].
Class 11 Maths NCERT Solutions Chapter 13  Summary
Statistics: It is defined as the process of collection and classification of data, interpretation, and presentation of the data, and analysis of data. Statistics also is defined as concluding the sample data that is collected using experiments. Statistics is applied in various fields such as sociology, psychology, geology, probability, and so on.
Measures of Dispersion in Statistics: The dispersion in the data is measured based on the observations data and measure of central tendency type.
There are different types to represent the measures of dispersion. They are Range, Mean deviation, Quartile deviation, and Standard deviation.
Range in Statistics: The range is the difference between the maximum value and the minimum value in the given data set.
The formula of Range = Maximum Value – Minimum Value
Mean Deviation in Statistics: The term “mean deviation” is defined as the difference between the observed value of a data point and the expected value.
Variance and Standard Deviation: These are the two important measurements in statistics. Variance is a measure of how data values vary from the mean, on the other hand the standard deviation is the measure of the distribution of statistical data. The variance and the standard deviation are measured in different units.
Analysis of Frequency Distributions: A frequency distribution represents the frequency of items of a data set in a graphical format or a tabular format. With the help of a frequency distribution, we will get a visual display of the frequency of data items which represents the number of times they repeated.
Measures of dispersion Range, Quartile deviation, mean deviation, variance, and standard deviation are measures of dispersion. Range $=$ Maximum Value  Minimum Value
Mean deviation for ungrouped data
$$ \begin{aligned} & M D .(\bar{x})=\frac{\sum f_i\left(x_i\bar{x}\right)}{N}, \\ & M D .(M)=\frac{\sum f_i\left(x_iM\right)}{N}, \end{aligned} $$
where $N=\sum f_i$
Variance and standard deviation for ungrouped data
$$ \begin{aligned} & \sigma^2=\frac{1}{n} \sum\left(x_l\bar{x}\right)^2, \\ & \sigma=\sqrt{\frac{1}{n} \sum\left(x_l\bar{x}\right)^2}, \end{aligned} $$
Variance and standard deviation of a discrete frequency distribution
$$ \begin{aligned} & \sigma^2=\frac{1}{N} \sum f_i\left(x_i\bar{x}\right)^2, \\ & \sigma=\sqrt{\frac{1}{N} \sum f_i\left(x_i\bar{x}\right)^2} \end{aligned} $$
Variance and standard deviation of a continuous frequency distribution
$$ \begin{aligned} & \sigma^2=\frac{1}{N} \sum f_i\left(x_i\bar{x}\right)^2, \\ & \sigma=\frac{1}{N} \sqrt{N \sum f_i x_i^2\left(\sum f_i x_i\right)^2} \end{aligned} $$
Shortcut method to find variance and standard deviation.
$$ \begin{aligned} & \sigma^2=\frac{h^2}{N^2}\left[N \sum f_i y_i^2\left(\sum f_i y_i\right)^2\right], \\ & \sigma=\frac{h}{N} \sqrt{N \sum f_i x_i^2\left(\sum f_i x_i\right)^2} \text { where } \frac{x_iA}{h} \end{aligned} $$
Coefficient of variation(C.V.) $=\frac{\sigma}{\bar{x}} \times 100, \bar{x} \neq 0$.
Class 11 Maths NCERT Solutions Chapter 13
Class 11 Maths Ch 15 NCERT Solutions shows the sums on a stepbystep basis. The concepts touched up in the chapter and the example sums are indicated below:
Mean Deviation About Mean
Ungrouped
Example: Finding mean deviation about the mean of 6, 7, 10, 12, 13, 12, 8, 4
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 
Continuous Frequency Distribution
Example: Finding the mean deviation about the mean of the below data:
Obtained Marks  Number of Students (fi)  Midpoint (xi)  fixi 
1020  2  (10+20)/2 = 15  2 x 15 = 30 
2030  3  15+10 = 25  3 x 25 = 75 
3040  8  35  8 x 35 = 280 
4550  14  45  14 x 45 = 630 
5060  8  55  8 x 55 = 440 
6070  3  65  3 x 65 = 195 
7080  2  75  2 x 75 = 150 
Mean Deviation About Median
Ungrouped
Example: Finding the mean deviation about the median of 3, 5, 9, 3, 10, 12, 4, 18, 7, 19, 21
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 
Continuous Frequency Distribution
Example: Calculation of the mean deviation about median of the below data:
Class  010  1020  2030  3040  4050  5060 
Frequency  6  7  15  16  4  2 
Standard Deviation and Variance
Ungrouped Data
Example: Finding the variance of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
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 
Continuous Frequency
Example: Calculation of the mean, variance and standard deviation of the following distribution:
Class  3040  4050  5060  6070  7080  8090  90100 
Frequency (fi)  3  7  12  15  8  3  2 
Coefficient of Variation
Example: Determine which group is more variable from the below data:
Marks  1020  2030  3040  4050  5060  6070  7080 
Group M  9  17  32  33  40  10  9 
Group N  10  20  30  25  43  15  7 
Indirect Questions
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?
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?
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 Distribution and 
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 realworld data for meaningful insights. This chapter is significant in exams, with an average of 34 questions being asked in previous years.
Other Study Material for CBSE Class 11 Maths Chapter 13
S. No  Important Links for Chapter 13 Statistics 
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ChapterSpecific NCERT Solutions for Class 11 Maths
Given below are the chapterwise NCERT Solutions for Class 11 Maths. Go through these chapterwise solutions to be thoroughly familiar with the concepts.
S. No  NCERT Solutions Class 11 Maths All Chapters 
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4  Chapter 4  Complex Numbers and Quadratic Equations Solutions 
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11  Chapter 11  Introduction to Three Dimensional Geometry Solutions 
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Important Related Links for CBSE Class 11 Maths
FAQs on NCERT Solutions Class 11 Maths Chapter 13 Statistics
1. What are the Different SubTopics of Chapter 13 Maths Class 11?
The subtopics 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 stepbystep 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.