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# NCERT Solutions for Class 10 Maths Chapter 13 - Statistics Exercise 13.2

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Maths Class 10 Chapter 13 Statistics Exercise 13.2 - FREE PDF Download

The NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 - Statistics by Vedantu provides complete solutions that help students understand statistical concepts. This exercise focuses on determining the mean, median, and mode of grouped data, which are important tools in data analysis. The step-by-step instructions help students follow and understand the steps.

Table of Content
1. NCERT Solutions for Maths Class 10 Chapter 13 Statistics Exercise 13.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 13 Exercise 13.2 Class 10 | Vedantu
3. Formulas Used in Class 10 Chapter 13 Exercise 13.2
4. Access NCERT Solutions for Maths Class 10 Chapter 13 - Statistics
4.1Exercise 13.2
5. Class 10 Maths Chapter 13: Exercises Breakdown
6. CBSE Class 10 Maths Chapter 13 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

Focusing on NCERT Solutions for Class 10 Maths answers is important because they give a strong statistical foundation, allowing students to handle a variety of data challenges. Students can improve their analytical skills by practising regularly, which is necessary for both examinations and practical uses. Practising these solutions could help students improve their understanding and solve problems in the CBSE Class 10 Maths Syllabus.

## Glance on NCERT Solutions Maths Chapter 13 Exercise 13.2 Class 10 | Vedantu

• Class 10 Chapter 13 Statistics Exercise 13.2 explains the mode of grouped data and how to determine it.

• The mode of grouped data is the value that appears most frequently within a given data set.

• Problems involve working with frequency distribution tables to find the mode, it requires understanding class intervals and frequencies to accurately identify the mode.

• The exercise helps compare the mode with other measures of central tendency like mean and median.

• It includes various practice problems to strengthen problem-solving skills related to grouped data.

• There are 6 fully solved questions in NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Statistics.

## Formulas Used in Class 10 Chapter 13 Exercise 13.2

• Mode of Grouped Data: $L+\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$

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## Access NCERT Solutions for Maths Class 10 Chapter 13 - Statistics

### Exercise 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

 Age(in years) $5-15$ $15-25$ $25-35$ $35-45$ $45-55$ $55-65$ Number of patients $6$ $11$ $21$ $23$ $14$ $5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Ans: The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)$

Suppose the assured mean $\left( a \right)$ of the data is $30$.

Class mark $\left( {{x}_{i}} \right)$ for each interval is calculated as follows:

${{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}$

${{d}_{i}},$${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ can be evaluated as follows:

 Age (in years) Number ofPatients${{f}_{i}}$ Class Mark${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-30$ ${{f}_{i}}{{d}_{i}}$ $5-15$ $6$ $10$ $-20$ $-120$ $15-25$ $11$ $20$ $-10$ $-110$ $25-35$ $21$ $30$ $0~$ $0~$ $35-45$ $23$ $40$ $10~$ $230$ $45-55$ $14$ $50$ $20$ $280$ $55-65$ $5$ $60$ $30$ $150$ Total $80$ $430$

It can be observed that from the above table

$\sum{{{f}_{i}}=80}$

$\sum{{{f}_{i}}{{d}_{i}}}=430$

Substituting the value of ${{u}_{i}},$ and ${{f}_{i}}{{u}_{i}}$ in the formula of mean we get:

The required mean:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)$

$\overline{X}=30+\left( \frac{430}{80} \right)$

$\overline{X}=30+5.375$

$\overline{X}=35.38$

Hence, the mean of this data is $35.38$. It demonstrates that the average age of a patient admitted to hospital was $35.38$years.

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

It can be noticed that the maximum class frequency is $23$  belonging to class interval$35\text{ }-\text{ }45$ .

Modal class $=35\text{ }-\text{ }45$

The values of unknowns is given as below as per given data:

$l=\text{ }35$

${{f}_{1}}=\text{ }23$

$h=15-5=10$

${{f}_{0}}=\text{ }21$

${{f}_{2}}=\text{ }14$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=35+\left( \frac{23-21}{2(23)-21-14} \right)\times 10$

$M=35+\left[ \frac{2}{46-35} \right]\times 10$

$M=35+\frac{20}{11}$

$\text{M=35+1}\text{.81}$

$\text{M=36}\text{.8}$

Hence, the Mode of the data is $36.8$. It demonstrates that the age of maximum number of patients admitted in hospital was $36.8$years.

2. The following data gives the information on the observed lifetimes (in hours) of $\mathbf{225}$ electrical components:

 LifeTimes(in hours) $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ Frequency $10$ $35$ $52$ $61$ $38$ $29$

Determine the modal lifetimes of the components.

Ans: Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

From the data given above, it can be noticed that the maximum class

frequency is$61$ belongs to class interval $60\text{ }-\text{ }80$.

Therefore, Modal class $=60-80$

The values of unknowns are given as below as per given data:

$l=\text{60}$

${{f}_{1}}=61$

$h=20$

${{f}_{0}}=52$

${{f}_{2}}=38$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=60+\left( \frac{61-52}{2(61)-52-38} \right)\times 20$

$M=60+\left[ \frac{9}{122-90} \right]\times 20$

$M=360+\left( \frac{9\times 20}{32} \right)$

$M=60+\frac{90}{16}=60+5.625$

$M=65.625$

Hence, the modal lifetime of electrical components is $65.625$ hours.

3. The following data gives the distribution of total monthly household expenditure of  $\mathbf{200}$  families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

 Expenditure (in Rs) Number of Families $1000-1500$ $24$ $1500-2000$ $40$ $2000-2500$ $33$ $2500-3000$ $28$ $3000-3500$ $30$ $3500-4000$ $22$ $4000-4500$ $16$ $4500-5000$ $7$

Ans: For mode

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

It can be observed from the given data that the maximum class frequency is$40$ ,

Belongs to $1500\text{ }-\text{ }2000$ intervals.

Therefore, modal class $=\text{ }1500\text{ }-\text{ }2000$

The values of unknowns are given as below as per given data:

$l=1500$

${{f}_{1}}=40$

${{f}_{0}}=24$

${{f}_{2}}=33$

$h=500$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=1500+\left( \frac{40-24}{2(40)-24-33} \right)\times 500$

$M=1500+\left[ \frac{16}{80-57} \right]\times 500$

$M=1500+\frac{8000}{23}$

$M=1500+347.826$

$M=1847.826$

$M\approx 1847.83$

Therefore, modal monthly expenditure was Rs $1847.83$ .

For mean,

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is $2750$.

Class mark $\left( {{x}_{i}} \right)$ for each interval is calculated as follows:

$\text{Class mark}=\frac{\text{Upper class limit+Lower class limit}}{2}$

Class size ($h$) of this data is:

$h=1500-1000$

$h=500$

${{d}_{i}},$${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ can be evaluated as follows:

 Expenditure (in Rs) Number of families ${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-2750$ ${{u}_{i}}=\frac{{{d}_{i}}}{500}$ ${{f}_{i}}{{u}_{i}}$ $1000-1500$ $24$ $1250$ $-1500$ $-3$ $-72$ $1500-2000$ $40$ $1750$ $-1000$ $-2$ $-80$ $2000-2500$ $33$ $2250$ $-500$ $-1~$ $-33$ $2500-3000$ $28$ $2750$ $0~$ $0~$ $0~$ $3000-3500$ $30$ $3250$ $500$ $1$ $30$ $3500-4000$ $22$ $3750$ $1000$ $2$ $44$ $4000-4500$ $16$ $4250$ $1500$ $3$ $48$ $4500-5000$ $7$ $4750$ $2000$ $4$ $28$ Total $200$ $-35$

It can be observed from the above table

$\sum{{{f}_{i}}=200}$

$\sum{{{f}_{i}}{{u}_{i}}}=-35$

Substituting ${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ in the formula of mean

The required mean:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h$

$\overline{X}=2750+\left( \frac{-35}{200} \right)\times (500)$

$\overline{X}=2750-87.5$

$\overline{X}=2662.5$

Hence, the mean monthly expenditure is Rs$2662.5$ .

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two Measures.

 Number of Students Per Teacher Number of States/U.T $15-20$ $3$ $20-25$ $8$ $25-30$ $9$ $30-35$ $10$ $35-40$ $3$ $40-45$ $0$ $45-50$ $0$ $50-55$ $2$

Ans: For mode

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

It can be observed from the given data that the maximum class frequency is $10$ which belongs to class interval$30\text{ }-\text{ }35$ .

Therefore, modal class $=\text{ }30\text{ }-\text{ }35$

$h=5$

$l=30$

${{f}_{1}}=10$

${{f}_{0}}=9$

${{f}_{2}}=3$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=30+\left( \frac{10-9}{2(10)-9-3} \right)\times 5$

$M=30+\left( \frac{1}{20-12} \right)\times 5$

$M=30.6$

Hence, the mode of the given data is $30.6$. It demonstrates that most of the states/U.T have a teacher-student ratio of $30.6$ .

For mean,

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is $32.5$.

Class mark $\left( {{x}_{i}} \right)$ for each interval is calculated as follows:

$\text{Class mark}=\frac{\text{Upper class limit+Lower class limit}}{2}$

Class size ($h$) of this data is:

$h=20-15$

$h=5$

${{d}_{i}},\text{ }{{u}_{i}},\text{ }and\text{ }{{f}_{i}}{{u}_{i}}$ can be calculated as follows:

 Number ofstudents perteacher Number ofstates /U.T${{f}_{i}}$ ${{x}_{i}}$ ${{D}_{i}}={{x}_{i}}-32.5$ ${{u}_{i}}=\frac{{{d}_{i}}}{5}$ ${{f}_{i}}{{u}_{i}}$ $15-20$ $3$ $17.5$ $-15$ $-3$ $-9$ $20-25$ $8$ $22.5$ $-10$ $-2$ $-16$ $25-30$ $9$ $27.5$ $-5$ $-1~$ $-9$ $30-35$ $10$ $32.5$ $0~$ $0~$ $0~$ $35-40$ $3$ $37.5$ $5$ $1$ $3$ $40-45$ $0~$ $42.5$ $10$ $2$ $0~$ $45-50$ $0~$ $47.5$ $15$ $3$ $0~$ $50-55$ $2$ $52.5$ $20$ $4$ $8$ Total $35$ $-23$

It can be observed from the above table

$\sum{{{f}_{i}}=35}$

$\sum{{{f}_{i}}{{u}_{i}}}=-23$

Substituting ${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ in the formula of mean

The required mean

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h$

$\overline{X}=32.5+\left( \frac{-23}{35} \right)\times 5$

$\overline{X}=32.5-\frac{23}{7}$

$\overline{X}=29.22$

Hence, the mean of the data is $29.2$ .

It demonstrates that the average teacher−student ratio was $29.2$.

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

 Runs Scored Number of Batsmen $3000-4000$ $4$ $4000-5000$ $18$ $5000-6000$ $9$ $6000-7000$ $7$ $7000-8000$ $6$ $8000-9000$ $3$ $9000-10000$ $1$ $10000-11000$ $1$

Find the mode of the data.

Ans: Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

From the given data, it can be observed that the maximum class frequency is $18$

Belongs to class interval $4000-5000$.

Therefore, modal class = $4000-5000$

$l=4000$

${{f}_{1}}=18$

${{f}_{0}}=4$

${{f}_{2}}=9$

$h=1000$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=4000+\left( \frac{18-4}{2(18)-4-9} \right)\times 1000$

$M=4000+\left( \frac{14000}{23} \right)$

$M=4608.695$

Hence, the mode of the given data is $4608.7$ runs.

6. A student noted the number of cars passing through a spot on a road for $\mathbf{100}$  periods each of $\mathbf{3}$  minutes and summarised it in the table given below. Find the mode of the data:

 Numberof cars $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $7$ $14$ $13$ $12$ $20$ $11$ $15$ $8$

Ans: Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

$l=$ Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

${{f}_{0}}=$ Frequency of class preceding the modal class

${{f}_{2}}=$ Frequency of class succeeding the modal class

$h=$Class size

From the given data, it can be observed that the maximum class frequency is $20$,

Belonging to $40-50$ class intervals.

Therefore, modal class = $40-50$

$l=40$

${{f}_{1}}=20$

${{f}_{0}}=12$

${{f}_{2}}=11$

$h=10$

Substituting these values in the formula of mode we get:

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

$M=40+\left( \frac{20-12}{2(20)-12-11} \right)\times 10$

$M=40+\left( \frac{80}{40-23} \right)$

$M=44.7$

Hence, the mode of this data is $44.7$ cars.

## Conclusion

Vedantu's NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 - Statistics explains how to calculate the mode for grouped data. It is important to focus on knowing the formula and correctly understanding frequency distribution tables. This practice is necessary for improving your data analysis qualities and knowing statistical metrics. By carefully practising these solutions, you will improve your understanding of statistics, which will help you with your exams and further studies.

## Class 10 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 9 Questions & Solutions (9 Long Answers) Exercise 13.3 7 Questions & Solutions (6 Long Answers)

## CBSE Class 10 Maths Chapter 13 Other Study Materials

 S.No Important Links for Chapter 13 Statistics 1 Class 10 Statistics Important Questions 2 Class 10 Statistics Revision Notes 3 Class 10 Statistics Important Formulas 4 Class 10 Statistics RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 13 - Statistics Exercise 13.2

1. What is statistics according to NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2?

Statistics is considered as a subject which is concerned with collection, organisation, analysis, interpretation and presentation of data. Statistics is used in different fields for data analysis purposes. People who are using statistics for data analysis and interpretation must have some technical and cognitive skills to achieve maximum success in his/her work. The government uses statistics in the census and for calculating or measuring different indicators that affect the economy of the country — indicators such as unemployment, stock market, etc. Statistics is not an easy task. One needs a lot of experience to complete it in an efficient and productive way according to NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2.

2. How much time do students need to complete NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2?

Providing at least one hour completely to the Maths Class 10 Chapter 13 Exercise 13.2 would be enough for the student to complete and understand the concepts. But, it is also to be remembered that continuous practice would help the student attain the desired results. Therefore it is advisable that the student come back to the exercise time and again and practice the numericals to get a better hold of the concepts. To make the studying process easier, the students can refer to the NCERT Solutions which they can easily download from the website of Vedantu.

3. What is ungrouped data according to Maths Class 10 Chapter 13 Exercise 13.2?

Data that is usually represented in its original or raw form and where the observations are not classified into groups is called ungrouped data. In Maths Class 10 Chapter 13 Exercise 13.2, for instance if the teacher of a kindergarten writes down the ages of her students in the following manner, then it gets called an ungrouped data

5, 3, 4, 3, 5, 4, 4, 3, 4, 4, 3, 3, 3, 4, 3, 27 .

It is easier to analyse and work upon ungrouped data if the set is small and limited. But it becomes a hassle if the set is in a larger quantity.

4. What is the direct method of finding the mean in Maths Class 10 Chapter 13 Exercise 13.2 ?

To find the mean through the direct method, the following steps are to be considered

• The first step includes the classification of the data into intervals in addition to finding the corresponding frequency of each class.

• The next step includes the finding of the classmark by considering and taking into account the midpoint of the upper-class limits and the lower class limits.

• The third step includes the grouping and tabulating of the product of the class mark and its corresponding frequency for each class. Their sum is also to be calculated.

• The last step involves dividing the acquired sum by the sum of the frequencies in order to attain the mean.

5. How to find the mean using the assumed mean method from Maths Class 10 Chapter 13 Exercise 13.2 ?

To find the mean using the assumed mean method, the following steps are to be taken into account:

• The first step involves the classification of the data into intervals and finding out the corresponding frequency of each class.

• The next step includes the finding of the classmark by considering the midpoints of the upper-class limits and the lower class limits.

• The third step involves considering one of the ‘s as the assumed mean.

• The next one needs to find out the derivation of “a” from each of the xi.

di=xi-a′

• The final step involves finding out the mean of the deviations and calculate the mean

6. Are the NCERT Solutions available for Class 10 Maths Chapter 13 Statistics Exercise 13.2?

Yes, the NCERT Solutions are easily feasible for Class 10 Maths Chapter 13 Statistics Exercise 13.2 and can be downloaded from the Vedantu website or from the Vedantu app at free of cost. These solutions are devised by experts keeping in mind the needs and requirements of the students. Every exercise that these solutions contain have detailed and explained answers that make the learning process smoother and simpler for the students. With the regular and adequate practice of these exercises, the student will be able to solve any question that gets asked from this chapter.

7. What is the mode of grouped data and how is it calculated in Class 10 Maths Chapter 13 Statistics Exercise 13.2?

From Class 10 Maths Chapter 13 Statistics Exercise 13.2, the mode of grouped data is the value that appears most frequently within a data set. In Exercise 13.2, it is calculated using the formula Mode = $L+\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$, where L is the lower limit of the modal class, $f_{1}$  is the frequency of the modal class, $f_{0}$ is the frequency of the class before the modal class, $f_{2}$ is the frequency of the class after the modal class, and h is the class width.

8. Why is finding the mode of grouped data important in Class 10 Maths Chapter 13 Exercise 13.2 Solutions?

Finding the mode of grouped data is important because it helps identify the most frequent occurrence in a dataset, which can be useful in various real-life applications such as market research, quality control, and social sciences from Class 10 Maths Chapter 13 Exercise 13.2 Solutions. It provides insights into the central tendency of the data.

9. What are the key steps to solve problems in Class 10 Maths Chapter 13 Exercise 13.2 Solutions?

The key steps to solve problems in Class 10 Maths Chapter 13 Exercise 13.2 Solutions include identifying the modal class from the frequency distribution table, applying the formula for mode, and performing the necessary calculations to find the mode. It is essential to accurately determine the class intervals and frequencies for correct results.

10. How do Vedantu's solutions help in understanding NCERT Class 10 Maths Chapter 13 Exercise 13.2?

Vedantu's solutions provide step-by-step explanations for each problem in NCERT Class 10 Maths Chapter 13 Exercise 13.2, making it easier for students to understand the process of finding the mode of grouped data. The clear and detailed approach helps students grasp the concepts and apply them effectively in their exams.

11. What common mistakes should be avoided while calculating the mode in NCERT Class 10 Maths Chapter 13 Exercise 13.2?

Common mistakes to avoid in NCERT Class 10 Maths Chapter 13 Exercise 13.2 which include incorrectly identifying the modal class, miscalculating the class width, and making errors in applying the mode formula. Paying close attention to the frequency distribution table and carefully following the steps in the formula can help prevent these errors.