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# NCERT Solutions for Class 10 Maths Chapter 13 - Statistics Exercise 13.3

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Maths Class 10 Chapter 13 Statistics Exercise 13.3 - FREE PDF Download

NCERT Solutions for Class 10 Maths Chapter 13 - Statistics, Exercise 13.3, by Vedantu! This exercise focuses on essential statistical measures such as mean, median, and mode. Mastering these concepts is crucial as they form the basis for understanding data analysis and interpretation. In this chapter, students will learn how to calculate these measures from grouped and ungrouped data. Pay special attention to the methods and formulas used for each calculation, as they are vital for solving various problems efficiently. With Vedantu's Class 10 Maths NCERT Solutions step-by-step.

Table of Content
1. NCERT Solutions for Maths Class 10 Chapter 13 Statistics Exercise 13.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 13 Exercise 13.3 Class 10 | Vedantu
3. Formulas Used in Class 10 Chapter 13 Exercise 13.3
4. Access NCERT Solutions for Maths Class 10 Chapter 13 - Statistics
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

## Glance on NCERT Solutions Maths Chapter 13 Exercise 13.3 Class 10 | Vedantu

• NCERT Solutions Maths Chapter 13 Exercise 13.3 Class 10 is about the Median of Grouped Data.

• The median splits the data set into two equal parts, with half of the values above it and half below.

• In grouped data, the median falls within a specific class interval called the median class.

• The median class is found by locating the class interval where the cumulative frequency surpasses half of the total observations.

• To identify the median class, calculate cumulative frequencies to see how data accumulates across intervals.

• The median is a useful measure of central tendency, especially in fields like economics and sociology, for analysing distributions such as income or age.

• There are seven questions in Maths Chapter 13 ex 13.3 class 10 solutions which are fully solved by experts at Vedantu.

## Formulas Used in Class 10 Chapter 13 Exercise 13.3

• Mean of Grouped Data: $Mean\left ( \bar{x} \right ) = \frac{\sum fixi}{\sum fi}$

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

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

Exercise: 13.3

1. The following frequency distribution gives the monthly consumption of electricity of $\mathbf{68}$  consumers of a locality. Find the median, mean and mode of the data and compare them.

 Monthly Consumption (in units) Number of Consumers $65-85$ $4$ $85-105$ $5$ $105-125$ $13$ $125-145$ $20$ $145-165$ $14$ $165-185$ $8$ $185-205$ $4$

Ans: 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:

To find the class mark for each interval, the following relation is used.

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

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

$h=85-65$

$h=20$

[${{d}_{i}},$${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ can be calculated according to step deviation method as follows:

 Monthlyconsumption(in units) Number of Consumers${{f}_{i}}$ Class Mark ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-135$ ${{u}_{i}}=\frac{{{d}_{i}}}{20}$ ${{f}_{i}}{{u}_{i}}$ $65-85$ $4$ $75$ $-60$ $-3$ $-12$ $85-105$ $5$ $95$ $-40$ $-2$ $-10$ $105-125$ $13$ $115$ $-20$ $-1~$ $-13$ $125-145$ $20$ $135$ $0~$ $0~$ $0~$ $145-165$ $14$ $155$ $20$ $1$ $14$ $165-185$ $8$ $175$ $40$ $2$ $16$ $185-205$ $4$ $195$ $60$ $3$ $12$ Total $68$ $7$

It can be observed from the above table

$\sum{{{f}_{i}}=68}$

$\sum{{{f}_{i}}{{u}_{i}}}=7$

Class size $\left( h \right)\text{ }=\text{ }20$

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

The required mean:

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

$\Rightarrow\overline{X}=135+\left(\frac{7}{68} \right)\times (20)$

$\Rightarrow\overline{X}=135+\frac{140}{68}$

$\Rightarrow\overline{X}=137.058$

Hence, the mean of given data is $137.058$.

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

From the table, it can be noticed that the maximum class frequency is $20$ ,

Belongs to class interval $125-145$.

Modal class = $125-145$

$l=125$

Class size $\text{h}=20$

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

${{f}_{0}}=13$

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

Substituting these values in the formula of mode we get:

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

$\Rightarrow M=125+\left( \frac{20-13}{2(20)-13-14} \right)\times 20$

$\Rightarrow M=125+\frac{7}{13}\times 20$

$\Rightarrow M=135.76$

Hence, the value of mode is $135.76$

For Median

We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

To find the median of the given data, cumulative frequency is calculated as follows.

 Monthly consumption (in units) Number of Consumers Cumulative Frequency $65-85$ $4$ $4$ $85-105$ $5$ $4+5=9$ $105-125$ $13$ $9+13=22$ $125-145$ $20$ $22+20=42$ $145-165$ $14$ $42+14=56$ $165-185$ $8$ $56+8=64$ $185-205$ $4$ $64+4=68$

It can be observed from the given table

$n\text{ }=\text{ }68$

$\frac{\text{n}}{2}=34$

Cumulative frequency just greater than $\frac{\text{n}}{2}$ is  $42$ , belonging to

interval $125-145$.

Therefore, median class = $125-145$.

$l=\text{ }125$

$\text{h }=\text{ }20$

$f=20$

$cf=22$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$\Rightarrow m=125+\left( \frac{34-22}{20} \right)\times 20$

$\Rightarrow m=125+12$

$\Rightarrow m=137$

Hence, median, mode, mean of the given data is $137,\text{ }135.76,$ and $137.05$ respectively.

Mean, mode and median are almost equal in this case.

2. If the median of the distribution is given below is$\mathbf{28}.\mathbf{5}$, find the values of $\mathbf{x}$ and $\mathbf{y}$.

 Class interval Frequency $0-10$ $5$ $10-20$ $X$ $20-30$ $20$ $30-40$ $15$ $40-50$ $Y$ $50-60$ $5$ Total $60$

Ans: We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

The cumulative frequency for the given data is calculated as follows.

 Class interval Frequency Cumulative frequency $0-10$ $5$ $5$ $10-20$ $X$ $5+x$ $20-30$ $20$ $25+x$ $30-40$ $15$ $40+x$ $40-50$ $Y$ $40+x+y$ $50-60$ $5$ $45+x+y$ Total $(n)$ $60$

It is given that the value of $\text{n}$ is $60$

From the table, it can be noticed that the cumulative frequency of last entry is $45+x+y$

Equating $45+x+y$ and $\text{n}$, we get:

$45\text{ }+\text{ }x\text{ }+\text{ }y\text{ }=\text{ }60$

$x\text{ }+\text{ }y\text{ }=\text{ }15\text{ }$……(1)

It is given that.

Median of the data is given $28.5$ which lies in interval $20-30$.

Therefore, median class = $20-30$

$l=\text{ 20}$

$cf=5+x$

$f=\text{ }20$

$h=10$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$\Rightarrow 28.5=20+\left( \frac{\frac{60}{2}-(5+x)}{20} \right)\times 10$

$\Rightarrow 8.5=\left( \frac{25-x}{2} \right)$

$\Rightarrow 17=25-x$

$\Rightarrow x=8$

Substituting $x=8$ in equation (1), we get:

$8\text{ }+\text{ }y\text{ }=\text{ }15$

$y\text{ }=\text{ }7$

Hence, the values of $x$ and $y$ are $8$  and $7$  respectively.

3. A life insurance agent found the following data for distribution of ages of $\mathbf{100}$  policy holders. Calculate the median age, if policies are given only to persons having age $\mathbf{18}$  years onwards but less than $\mathbf{60}$  year.

 Age (in years) Number of Policy Holders Below $20$ $2$ Below $25$ $6$ Below $30$ $24$ Below $35$ $45$ Below $40$ $78$ Below $45$ $89$ Below $50$ $92$ Below $55$ $98$ Below $60$ $100$

Ans: We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

In this case, class width is not the constant. We are not required to adjust the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age $18$  years onwards but less than $60$  years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

 Age (in years) Number of Policy Holders $({{f}_{i}})$ Cumulative Frequency$(cf)$ $18-20$ $2$ $2$ $20-25$ $6-2=4$ $6$ $25-30$ $24-6=18$ $24$ $30-35$ $45-24=21$ $45$ $35-40$ $78-45=33$ $78$ $40-45$ $89-78=11$ $89$ $45-50$ $92-89=3$ $92$ $50-55$ $98-92=6$ $98$ $55-60$ $100-98=2$ $100$ Total $(n)$

From the table, it can be observed that $n\text{ }=\text{ }100$ .

Thus,

$\frac{\text{n}}{2}=50$

Cumulative frequency ($cf$) just greater than $\frac{n}{2}$ is$78$ ,

belongs interval $35\text{ }-\text{ }40$ .

Therefore, median class = $35\text{ }-\text{ }40$

$l=35$

$\text{h}=5$

$f=33$

$cf=45$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$\Rightarrow m=35+\left( \frac{50-45}{33} \right)\times 5$

$\Rightarrow m=35+\left( \frac{25}{33} \right)$

$\Rightarrow m=35.76$

Hence, median age of people who get the policies is $35.76$ years.

4. The lengths of $\mathbf{40}$  leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

 Length (in mm) Number of leaves ${{f}_{i}}$ $118-126$ $3$ $127-135$ $5$ $136-144$ $9$ $145-153$ $12$ $154-162$ $5$ $163-171$ $4$ $172-180$ $2$

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to$\mathbf{117}.\mathbf{5}\text{ }-\text{ }\mathbf{126}.\mathbf{5},\text{ }\mathbf{126}.\mathbf{5}\text{ }-\text{ }\mathbf{135}.\mathbf{5}...\text{ }\mathbf{171}.\mathbf{5}\text{ }-\text{ }\mathbf{180}.\mathbf{5}$)

Ans: We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

The given data does not have continuous class intervals. It can be noticed that the difference between two class intervals is$1$ . Therefore, we will add $0.5$ in the upper class and subtract $0.5$ in the lower class.

Continuous class intervals with respective cumulative frequencies can be represented as follows.

 Length (in mm) Number of leaves ${{f}_{i}}$ Cumulative frequency $117.5-126.5$ $3$ $3$ $126.5-135.5$ $5$ $3+5=8$ $135.5-144.5$ $9$ $8+9=17$ $144.5-153.5$ $12$ $17+12=29$ $153.5-162.5$ $5$ $29+5=34$ $162.5-171.5$ $4$ $34+4=38$ $171.5-180.5$ $2$ $38+2=40$

It can be observed from the given table

$n=\text{40}$

$\frac{\text{n}}{2}=20$

From the table, it can be noticed that the cumulative frequency just greater than

$\frac{n}{2}$ is $29$ , Belongs to interval $144.5-153.5$ .

median class = $144.5-153.5$

$l=144.5$

$\text{h}=9$

$f=12$

$cf=17$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$m=144.5+\left( \frac{20-17}{12} \right)\times 9$

$m=144.5+\left( \frac{9}{4} \right)$

$m=146.75$

Hence, median length of leaves is $146.75$ mm.

5. The following table gives the distribution of the life time of $\mathbf{400}$  neon lamps:

 Lifetime (inhours) Number ofLamps $1500-2000$ $14$ $2000-2500$ $56$ $2500-3000$ $60$ $3000-3500$ $86$ $3500-4000$ $74$ $4000-4500$ $62$ $4500-5000$ $48$

Find the median life time of a lamp.

Ans: We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows.

 Lifetime (inhours) Number ofLamps Cumulative Frequency $1500-2000$ $14$ $14$ $2000-2500$ $56$ $14+56=70$ $2500-3000$ $60$ $70+60=130$ $3000-3500$ $86$ $130+86=216$ $3500-4000$ $74$ $216+74=290$ $4000-4500$ $62$ $290+62=352$ $4500-5000$ $48$ $352+48=400$ Total$(n)$ $400$

It can be observed from the given table

$n\text{ }=400$

$\frac{\text{n}}{2}=200$

It can be observed that the cumulative frequency just greater than

$\frac{n}{2}$ is $290$ ,Belongs to interval $3000-3500$ .

Median class = $3000-3500$

$l=3000$

$f=86$

$cf=130$

$h=500$

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$m=3000+\left( \frac{200-130}{86} \right)\times 500$

$m=3000+\left( \frac{70\times 500}{86} \right)$

$m=3406.976$

Hence, median life time of lamps is $3406.98$ hours.

6.$\mathbf{100}$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

 Numberof letters $1-4$ $4-7$ $7-10$ $10-13$ $13-16$ $16-19$ Number ofSurnames $6$ $30$ $40$ $16$ $4$ $4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Ans: For median

We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows.

 Numberof Letters Number ofSurnames Cumulative Frequency $1-4$ $6$ $6$ $4-7$ $30$ $30+6=36$ $7-10$ $40$ $36+40=76$ $10-13$ $16$ $76+16=92$ $13-16$ $4$ $92+4=96$ $16-19$ $4$ $96+4=100$ Total$(n)$ $100$

It can be observed from the given table

$n\text{ }=100$

$\frac{\text{n}}{2}=50$

It can be noticed that the cumulative frequency just greater than

$\frac{n}{2}$ is $76$ , Belongs to interval $7-10$ .

Median class = $7-10$

$l=7$

$cf=36$

$f=40$

$\text{h}=3$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$m=7+\left( \frac{50-36}{40} \right)\times 3$

$m=7+\left( \frac{14\times 3}{40} \right)$

$m=8.05$

Hence, the median number of letters in the surnames is $8.05$.

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 $11.5$.

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

Class mark $\left( {{x}_{i}} \right)=\frac{\text{Upper class limit+Lower class limit}}{2}$

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

$h=4-1$

$h=3$

${{d}_{i}},$${{u}_{i}},$and ${{f}_{i}}{{u}_{i}}$ can be calculated according to step deviation method as follows:

 Number of letters Number of surnames${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-11.5$ ${{u}_{i}}=\frac{{{d}_{i}}}{20}$ ${{f}_{i}}{{u}_{i}}$ $1-4$ $6$ $2.5$ $-9$ $-3$ $-18$ $4-7$ $30$ $5.5$ $-6$ $-2$ $-60$ $7-10$ $40$ $8.5$ $-3$ $-1~$ $-40$ $10-13$ $16$ $11.5$ $0~$ $0~$ $0~$ $13-16$ $4$ $14.5$ $3$ $1$ $4$ $16-19$ $4$ $17.5$ $6$ $2$ $8$ Total $100$ $-106$

It can be observed from the above table

$\sum{{{f}_{i}}{{u}_{i}}}=-106$

$\sum{{{f}_{i}}=100}$

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

The required mean:

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

$\overline{X}=11.5+\left( \frac{-106}{100} \right)\times 3$

$\Rightarrow \overline{X}=11.5-3.18$

$\Rightarrow \overline{X}=8.32$

Hence the mean of number of letters in the surnames is $8.32$.

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

The data in the given table can be written as

 Numberof Letters Frequency $({{f}_{i}})$ $1-4$ $6$ $4-7$ $30$ $7-10$ $40$ $10-13$ $16$ $13-16$ $4$ $16-19$ $4$ Total$(n)$ $100$

From the table, it can be observed that the maximum class frequency is $40$

Belongs to $7-10$ class intervals.

Therefore, modal class = $7-10$

$l=7$

$h=3$

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

${{f}_{0}}=30$

${{f}_{2}}=16$

Substituting these values in the formula of mode we get:

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

$m=7+\left( \frac{40-30}{2(40)-30-16} \right)\times 3$

$m=7+\left( \frac{10}{34} \right)\times 3$

$m=7+\frac{30}{34}$

$m=7.88$

Hence, modal size of surname is $7.88$.

7.The distribution below gives the weights of $\mathbf{30}$  students of a class. Find the median weight of the students.

 Weight(in kg) $40-45$ $45-50$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$ Numberofstudents $2$ $3$ $8$ $6$ $6$ $3$ $2$

Ans: We can calculate the median as given below:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

Where

$l=$ Lower limit of median class

$h=$ Class size

$f=$ Frequency of median class

$cf=$ cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows :

 Weight (in kg) Numberof Students Cumulative Frequency $40-45$ $2$ $2$ $45-50$ $3$ $2+3=5$ $50-55$ $8$ $5+8=13$ $55-60$ $6$ $13+6=19$ $60-65$ $6$ $19+6=25$ $65-70$ $3$ $25+3=28$ $70-75$ $2$ $28+2=30$ Total$(n)$ $30$

It can be observed from the given table

$n\text{ }=30$

$\frac{\text{n}}{2}=15$

Cumulative frequency just greater than $\frac{n}{2}$ is $19$, Belongs to class interval $55-60$.

Median class = $55-60$

$l=55$

$f=6$

$cf=13$

$h=5$

Substituting these values in the formula of median we get:

$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$

$m=55+\left( \frac{15-13}{6} \right)\times 5$

$m=55+\left( \frac{10}{6} \right)$

$m=56.67$

Hence, median weight is $56.67$ kg.

## Conclusion

Class 10 maths chapter 13 exercise 13.3, students will deal with problems focused on determining the median of grouped data, an important measure of central tendency that identifies the middle value in a dataset. The provided grouped data will help students find the median for both odd and even numbers of observations, reinforcing their grasp of statistical methods. These solutions follow the guidelines and formulas illustrated in the example questions, offering clear and structured problem-solving approaches. For a thorough understanding and additional study resources, students can access notes, books, and question papers. NCERT solutions for class 10 ex 13.3 are invaluable for those looking for detailed and accurate answers across all subjects, organized by class and chapter.

## Class 10 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 9 Questions & Solutions (9 Long Answers) Exercise 13.2 6 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 13.3

1. What do you Mean by Median, Mean, and Mode?

Mean:-  It refers to average. Mean is the average of a given set of data. We can calculate by dividing the sum of all values by several values.

Mean = (n1 + n2 +........n10)/10.

Median:- Simply, the middle term of the set is known as the set’s median. If it has an even number of terms in the set, the average of two middle terms is the median.

Mode:- The highest repetitive term of the set is called the mode of the set.

2. Explain the Relationship Between Mean, Median, and Mode?

As these three are statistical measures, all are interrelated but can't say dependent all the time. The relation can be understood easily by plotting graphs. Apart from that, after several observations, it is clear that the difference between mean and mode is equal to the thrice of the difference between mean and median. It can be represented as,

Mean – Mode = 3 (Mean – Median).

We will also get a symmetrical curve when we plot the graph while showing the relation between them. It also considers frequency distribution on the other side.

3. Can you please brief me about the Class 10 Maths Exercise 13.3?

Exercise 13.3 of Class 10 Maths basically covers the concepts of mean, median and mode. There are a total of seven questions in Exercise 13.3. All these questions ask for the application of mean, median, and mode. Practising these questions will help you to understand the concept and how to apply them practically. Also, these types of questions are most likely to appear in Board exams as well.

4. How can I score excellent grades in Maths in Class 10 Boards?

Scoring a good set of marks is something every 10th grader dreams off. To achieve the same, it is important to devise a study plan. First, list down the important chapters which hold the maximum weightage of marks. Choose the study materials wisely. Practice the question given in NCERT Solutions. Along with them also revise the concept by practising previous years papers and mock tests. You can refer to the official website of Vedantu for the study materials. They are also available on Vedantu App, and all the resources are available free of cost.

5. How many questions are there in Class 10 Maths Chapter 13 Exercise 13.3?

Exercise 13.3 of Class 10 Maths Chapter 13 'Statistics' consists of seven questions. These questions revolve around the concept of mean, median and mode. The questions require the application of the formulas of mean, median and mode. If you have any doubts or queries regarding the exercise, you can visit Vedantu’s website to get the solutions to the NCERT Questions. The NCERT Solutions Class 10 Chapter 13 can help you solve the exercises easily.

6. What is Statistics Class 10 Maths?

Statistics Class 10 Maths is a branch of mathematics concerned with the collection, classification, and representation of any type of data to facilitate analysis and comprehension. There are various data representations in statistics which include bar graphs, pie charts, histograms, and frequency polygons. Class 10 Statistics comprises calculating the central tendency of grouped data with the help of three measures, i.e. mean, median and mode.

7. Which topics should I focus on most for the Class 10 Chapter 13 CBSE board exam?

CBSE Class 10 Maths Chapter 13 'Statistics' is an important chapter that can assure you with good grades. This chapter is easy to understand as well as fun to study. The main topics that are covered in this chapter are data analysis and calculation of central tendency of grouped data through the measurement of mean, median and mode. You can expect questions that ask the application of these concepts.

8. What topics are covered in class 10 maths chapter 13 exercise 13.3?

Class 10 Maths Ch 13 Ex 13.3 focuses on calculating the median of grouped data, which is a key measure of central tendency. It involves identifying the median class and understanding how to use cumulative frequency. The exercise teaches students how to apply the median formula to find the middle value in grouped data. This topic is essential for analysing statistical data. Understanding the median helps in interpreting the distribution of data sets in exercise 13.3 class 10.

9. How do I find the median of grouped data in class 10 maths ex 13.3?

To find the median of grouped data in class 10 maths ex 13.3, you need to identify the median class, which is the class interval where the cumulative frequency exceeds half of the total observations. Once the median class is determined, use the formula provided in the NCERT solutions to calculate the median. This formula involves the lower boundary of the median class, the cumulative frequency, and the class width. The NCERT solutions of class 10 maths ch 13 ex 13.3 offer detailed steps and examples to guide you through the process. Practising these steps will enhance your problem-solving skills.

10. Are there any prerequisite concepts needed for solving class 10 maths 13.3?

Yes, understanding cumulative frequency and determining class intervals are crucial for solving class 10 maths 13.3. You need to know how to calculate cumulative frequency to identify the median class. Familiarity with statistical terms and measures of central tendency is also important. These concepts are fundamental for accurately finding the median in grouped data. A solid grasp of these prerequisites will help you solve the problems effectively in class 10 maths chapter 13.3.