# NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.3) Exercise 15.3

Throughout a student’s academic career, Mathematics serves as one of the most important subjects. Mathematics is one of the subjects for Class 11 students studying in the Science and Commerce stream under the Central Board of Secondary Education (CBSE). Also, it is frequently regarded as the most difficult subject by Class 11 students. Not only do students need to fully grasp the theories in Mathematics, but they also need to practise answering the questions. Many times, despite studying and comprehending every topic in the curriculum, students do not complete enough exercises. They are therefore at a disadvantage when taking the annual exam. They must address this issue in order to secure better marks in the exams. To help solve this problem, Extramarks provides NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3.

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## NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.3) Exercise 15.3

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### NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

The Central Board of Secondary Education (CBSE) is one of India’s major school education boards. It is the board with the largest enrollment of students. Moreover, thousands of schools are associated with it. The CBSE is responsible for formulating the syllabus for Class 11. The board prescribes the NCERT textbooks for students to follow in Class 11. The NCERT textbook contains numerous questions that are very important for the exams. These questions help students become thorough with the topics that not only help them in the examinations but also in their higher studies. To succeed in their higher education, students need to have a solid grasp of the concepts of Mathematics. However, Class 11 students find it challenging to study for their exams. The Class 11 Mathematics curriculum includes a lot of fresh concepts. Students need to have access to all the required study materials in order to perform well on the annual exams. One of the most important study tools for preparing for the exams is the NCERT textbook. The exercises include different types of questions. Students are required to correctly answer these questions prior to taking the annual exams. As they try to solve these questions and find answers, they come across a number of challenges. Students can get help in this area from the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3, which are available on Extramarks.

Class 11 Maths Exercise 15.3 deals with the topic Analysis of Frequency Distributions. This exercise deals with the important concept of coefficient of variation(C.V.). The exercise contains 5 questions. These types of questions often appear in examinations. The proper examples and visuals are used in the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3. It boosts students’ confidence as a result, which helps them perform well in the final exam. Offering sufficient examples for practice helps students develop a conceptual understanding of each topic. The CBSE curriculum is completely followed by NCERT textbooks. The NCERT textbook is the source of most of the exam questions. To perform well on their exams, students must practise the exercises diligently. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3 are also accessible on Extramarks. The in-house subject specialists have put up the solutions and study materials on the platform. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3 given to the students, have been verified as accurate and factual. The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3 will help them to solidify their fundamentals and play a significant role during their whole study period, whether it be for the Class 11 examination or any competitive examination.

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### NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

There are many ways in which students can benefit from using the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3 on Extramarks. Using the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3, students can evaluate their own learning and determine their strong and weak areas. Additionally, the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.3 give evaluated and analysed learning while assisting students in the development of a wide range of skills, including logical and reasoning skills. The NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3 also assist students in successfully preparing for the annual exams.It is regarded as a vital stage in test preparation. Conceptual understanding and application are essential when practising mathematical problems. The detailed, step-by-step solutions provided in the NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3 allow students to solve difficult questions.

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Q.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

Ans.
For group A:

$\begin{array}{l}\begin{array}{|ccccccc|}\hline \mathrm{Marks}\left(\mathrm{A}\right)& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{10}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 10-20& 9& 15& -3& 9& -27& 81\\ 20-30& 17& 25& -2& 4& -34& 68\\ 30-40& 32& 35& -1& 1& -32& 32\\ 40-50& 33& 45=\mathrm{A}& 0& 0& 0& 0\\ 50-60& 40& 55& 1& 1& 40& 40\\ 60-70& 10& 65& 2& 4& 20& 40\\ 70-80& 9& 75& 3& 9& 27& 81\\ & 150& & & & -6& 342\\ \hline\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}\\ \text{\hspace{0.17em}}=45+\frac{-6}{150}×10\\ \text{\hspace{0.17em}}=45-0.4\text{\hspace{0.17em}}=44.6\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\end{array}$

$\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\left(10\right)}^{2}}{{\left(150\right)}^{2}}\left[150×342-{\left(-6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}\left(51300-36\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}×51264\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=227.84\\ \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{227.84}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15.09\end{array}$ $\begin{array}{|ccccccc|}\hline \mathrm{Marks}\left(\mathrm{A}\right)& {\mathrm{f}}_{\mathrm{i}}& {\mathrm{x}}_{\mathrm{i}}& {\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-\mathrm{A}}{10}& {{\mathrm{u}}_{\mathrm{i}}}^{2}& {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}& {\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\\ 10-20& 10& 15& -3& 9& -30& 90\\ 20-30& 20& 25& -2& 4& -\text{\hspace{0.17em}}40& 80\\ 30-40& 30& 35& -1& 1& -30& 30\\ 40-50& 25& 45=\mathrm{A}& 0& 0& 0& 0\\ 50-60& 43& 55& 1& 1& 43& 43\\ 60-70& 15& 65& 2& 4& 30& 60\\ 70-80& 7& 75& 3& 9& 21& 63\\ & 150& & & & -6& 366\\ \hline\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\mathrm{A}+\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}}{\mathrm{n}}×\mathrm{h}\\ \text{\hspace{0.17em}}=45+\frac{-6}{150}×10\\ \text{\hspace{0.17em}}=45-0.4\text{\hspace{0.17em}}=44.6\end{array}$

$\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\mathrm{h}}^{2}}{{\mathrm{n}}^{2}}\left[\mathrm{n}\sum _{\mathrm{i}}^{\mathrm{n}}\left({\mathrm{f}}_{\mathrm{i}}{{\mathrm{u}}_{\mathrm{i}}}^{2}\right)-{\left(\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\right)}^{2}\right]\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{{\left(10\right)}^{2}}{{\left(150\right)}^{2}}\left[150×366-{\left(-6\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}\left(54900-36\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{225}×54864\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=243.84\\ \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{243.84}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15.61\end{array}$

Since, both groups have equal mean. Group B has greater standard deviation. So, group B is more variable than group A.

Q.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

Ans.
For prices of share X:

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{x}}\right)=\frac{35+54+52+53+56+58+52+50+51+49}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{510}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=51\end{array}$

$\begin{array}{l}\begin{array}{|ccc|}\hline {\mathrm{x}}_{\mathrm{i}}& \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)& {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}\\ 35& 35-51=-16& 256\\ 54& 54-51=3& 9\\ 52& 52-51=1& 1\\ 53& 53-51=2& 4\\ 56& 56-51=5& 25\\ 58& 58-51=7& 49\\ 52& 52-51=1& 1\\ 50& 50-51=-1& 1\\ 51& 51-51=0& 0\\ 49& 49-51=-2& 4\\ & & 350\\ \hline\end{array}\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{N}}\\ =\frac{350}{10}\\ =35\\ \therefore \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ =\sqrt{35}\\ =5.91\\ \mathrm{C}.\mathrm{V}.\left(\mathrm{share}\text{X}\right)=\frac{\mathrm{\sigma }}{\overline{\mathrm{x}}}×100\\ =\frac{5.91}{51}×100\\ =11.59\left(\mathrm{Appox}.\right)\end{array}$

For prices of share Y:

$\begin{array}{l}\mathrm{Mean}\left(\overline{\mathrm{y}}\right)=\frac{108+107+105+105+106+107+104+103+104+101}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1050}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=105\\ \begin{array}{|ccc|}\hline {\mathrm{y}}_{\mathrm{i}}& \left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)& {\left({\mathrm{y}}_{\mathrm{i}}-\overline{\mathrm{y}}\right)}^{2}\\ 108& 108-105=3& 9\\ 107& 107-105=2& 4\\ 105& 105-105=0& 0\\ 105& 105-105=0& 0\\ 106& 106-105=1& 1\\ 107& 107-105=2& 4\\ 104& 104-105=-1& 1\\ 103& 103-105=-2& 4\\ 104& 104-105=-1& 1\\ 101& 101-105=-4& 16\\ & & 40\\ \hline\end{array}\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^{2}}{\mathrm{N}}\\ =\frac{40}{10}\\ =4\\ \therefore \text{Standard deviation}\left(\mathrm{\sigma }\right)\\ =\sqrt{4}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\end{array}$ $\begin{array}{l}=2\\ \mathrm{}\end{array}$

$\begin{array}{l}\mathrm{C}.\mathrm{V}.\left(\mathrm{share}\text{Y}\right)=\frac{\mathrm{\sigma }}{\overline{\mathrm{y}}}×100\\ =\frac{2}{105}×100\\ =1.9<11.59\\ \mathrm{C}.\mathrm{V}.\text{of prices of share Y is lesser than that of share X.}\\ \text{Thus, the prices of share Y are more stable than that}\\ \text{of share X.}\end{array}$

Q.3 An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A Firm B
No. of wage earners 586 648
Mean of monthly wages Rs. 5253 Rs. 5253
Variance of distribution of wages 100 121

1. Which firm A or B pays larger amount as monthly wages?
2. Which firm, A or B, shows greater variability in individual wages?

Ans.

$\begin{array}{l}\mathrm{i}.\text{ Monthly wage of firm A = Rs. 5253}\\ \text{ Number of workers in firm A = 586}\\ \text{ Total paid amount in wages = Rs. 5253 × 586}\\ \text{ = Rs. 3078258}\\ \text{ Monthly wage of firm B = Rs. 5253}\\ \text{ Number of workers in firm A = 648}\\ \text{ Total paid amount in wages = Rs. 5253 x 648}\\ \text{ = Rs.3403944}\\ \text{Thus, firm B pays the larger amount as monthly wages.}\\ \text{ii. Variance of distribution of wages in firm A = 100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Standard deviation of wages in firm A = }\sqrt{\text{100}}\\ \text{ = 10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}Variance of distribution of wages in firm B = 121}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Standard deviation of wages in firm B = }\sqrt{\text{121}}\\ \text{ = 11}\\ \text{Since the average monthly wages in both the firms is same,}\\ \text{i.e., Rs. 5253, therefore, the firm with greater standard deviation}\\ \text{will have more variability.}\\ \text{Thus, the firm B has greater variability in the individual wages.}\end{array}$

Q.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 matched 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Ans.

$\begin{array}{|ccc|}\hline \begin{array}{l}No.\text{of goals}\\ \text{}\left({\text{x}}_{\text{i}}\right)\end{array}& \begin{array}{l}No.\text{of matches}\\ \text{}{\left(\text{f}\right)}_{\text{i}}\end{array}& {}_{}\\ \hline\end{array}$

$\begin{array}{|ccc|}\hline {x}_{i}-\overline{x}& {\left({x}_{i}-\overline{x}\right)}^{2}& {f}_{i}{\left({x}_{i}-\overline{x}\right)}^{2}\\ 0& 1& 0-2=-2& 4& 4\\ 1& 9& 1-2=-1& 1& 9\\ 2& 7& 2-2=0& 0& 0\\ 3& 5& 3-2=1& 1& 5\\ 4& 3& 4-2=2& 4& 12\\ & 25& & & 30\\ \hline\end{array}$

$\begin{array}{l}\text{Mean = }\frac{\sum _{\text{i=1}}^{\text{n}}{\text{f}}_{\text{i}}{\text{x}}_{\text{i}}}{\text{n}}\\ \text{ = }\frac{\text{50}}{\text{25}}\\ \text{ = 2}\\ \text{Variation}\left({\text{σ}}^{\text{2}}\right)\\ \text{= }\frac{\sum _{\text{i=1}}^{\text{n}}{\text{f}}_{\text{i}}{\left({\text{x}}_{\text{i}}\text{–}\overline{\text{x}}\right)}^{\text{2}}}{\text{n}}\\ \text{= }\frac{\text{30}}{\text{25}}\\ \text{= 1.2}\\ \text{Standard deviation}\\ \text{= }\sqrt{\text{1.2}}\\ \text{= 1.09}\\ \text{Standard deviation of team B}\\ \text{= 1.25 goals}\\ \text{Mean of team B}\\ \text{= 2}\\ \text{The team with lower standard deviation is more consistent.}\\ \text{Therefore, team A is more consistent.}\end{array}$

Q.5 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

$\begin{array}{l}\sum _{i=1}^{50}{x}_{i}=212,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{x}_{i}}^{2}=902.8,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{y}_{i}=261,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{y}_{i}}^{2}=1457.6\\ \mathbf{Which}\mathbf{}\mathbf{is}\mathbf{}\mathbf{more}\mathbf{}\mathbf{varying}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{length}\mathbf{}\mathbf{or}\mathbf{}\mathbf{weight}\mathbf{?}\end{array}$

Ans.

$\begin{array}{l}\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}=212,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{\mathrm{i}=1}^{50}{{\mathrm{x}}_{\mathrm{i}}}^{2}=902.8,\\ \mathrm{Mean}=\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{212}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4.24\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{{\mathrm{x}}_{\mathrm{i}}}^{2}}{50}-{\left(\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{x}}_{\mathrm{i}}}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{902.8}{50}-{\left(\frac{212}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=18.056-17.978\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.078\\ \text{Standandard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{0.078}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.28\\ \mathrm{C}.\mathrm{V}.\text{of length}=\frac{\mathrm{\sigma }}{\overline{\mathrm{x}}}×100\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6.6\\ \text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}=261,\text{\hspace{0.17em}\hspace{0.17em}}\sum _{\mathrm{i}=1}^{50}{{\mathrm{y}}_{\mathrm{i}}}^{2}=1457.6\\ \mathrm{Mean}\left(\overline{\mathrm{y}}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}}{50}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{261}{50}\text{\hspace{0.17em}\hspace{0.17em}}=5.22\\ \mathrm{Variance}\left({\mathrm{\sigma }}^{2}\right)=\frac{\sum _{\mathrm{i}=1}^{50}{{\mathrm{y}}_{\mathrm{i}}}^{2}}{50}-{\left(\frac{\sum _{\mathrm{i}=1}^{50}{\mathrm{y}}_{\mathrm{i}}}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1457.6}{50}-{\left(\frac{261}{50}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=29.15-27.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.89\\ \text{Standandard deviation}\left(\mathrm{\sigma }\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{1.89}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.37\\ \mathrm{C}.\mathrm{V}.\text{of weight}=\frac{\mathrm{\sigma }}{\overline{\mathrm{y}}}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1.37}{5.22}×100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=26.24\end{array}$

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}C.V. of weights is greater than the C.V. of lengths.}\\ \text{Therefore, weights are more varing than lengths.}\end{array}$