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# NCERT Solutions Class 10 Maths Chapter 13 Statistics

Last updated date: 18th Sep 2024
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## NCERT Solutions for Class 10 Maths Chapter 13 Statistics - Free PDF Download

NCERT Solutions of Class 10 Maths Chapter 13 Statistics, is crucial for understanding how to collect, analyze, and interpret data. This chapter covers important concepts such as mean, median, mode, and the representation of data using various graphical methods. It also includes cumulative frequency, which helps in understanding data distribution.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 13 Statistics - Free PDF Download
2. Glance of NCERT Solutions for Class 10 Maths Chapter 13 Statistics | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 13 Maths Class 10
4. Exercises Under NCERT Solutions for Class 10 Maths Chapter 13 – Statistics
5. Access NCERT Solutions for Class 10 Maths  Chapter 13 - Statistics
5.1Exercise 13.1
5.2Exercise 13.2
5.3Exercise 13.3
6. Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 13 Statistics
7. Class 10 Maths Chapter 13 : Exercises Breakdown
8. Other Study Material for CBSE Class 10 Maths Chapter 13
9. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

Focusing on these concepts is essential as they form the foundation for statistical analysis, a critical skill in various fields. Vedantu’s NCERT Solutions for Class 10 Maths Chapter 13 provides detailed explanations and step-by-step solutions to all the important questions, helping students grasp the material effectively. These solutions are designed to boost confidence and improve exam performance by ensuring a solid understanding of statistical methods.

## Glance of NCERT Solutions for Class 10 Maths Chapter 13 Statistics | Vedantu

• In this article, the concepts of data and its types (grouped data and ungrouped data) will be discussed.

• This chapter covers what is mean, mode and median. And how to calculate mean, mode and median for both grouped and ungrouped data.

• The basic formulas to calculate mean, mode and median are as follows:

• $\text{Mean} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n}$

• $\text{Mode} = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$

• $\text{Median} = L + \left( \frac{\frac{n}{2} - F}{f} \right) \times h$

• There are three exercises (22 fully solved questions) in class 10th maths chapter 13 Statistics.

## Access Exercise Wise NCERT Solutions for Chapter 13 Maths Class 10

 S.No. Current Syllabus Exercises of Class 10 Maths Chapter 13 1 NCERT Solutions of Class 10 Maths Statistics Exercise 13.1 2 NCERT Solutions of Class 10 Maths Statistics Exercise 13.2 3 NCERT Solutions of Class 10 Maths Statistics Exercise 13.3

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

### 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

 Number of Plants $0-2$ $2-4$ $4-6$ $6-8$ $8-10$ $10-12$ $12-14$ Number of Houses $1$ $2$ $1$ $5$ $6$ $2$ $3$

Which method did you use for finding the mean, and why?

Ans: The number of houses denoted by ${{x}_{i}}$.

The mean can be found as given below:

$\overline{X}=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}$

Class mark $\left( {{x}_{i}} \right)$ for each interval is calculated as follows:$xi=\text{ }\frac{\text{Upper class limit+Lower class limit}}{2}$

${{x}_{i}}$ and ${{f}_{i}}{{x}_{i}}$ can be calculated as follows:

 Number of Plants Number of Houses ${{f}_{i}}$ ${{x}_{i}}$ ${{f}_{i}}{{x}_{i}}$ $0-2$ $1$ $1$ $1\times 1=1$ $2-4$ $2$ $3$ $2\times 3=6$ $4-6$ $1$ $5$ $1\times 5=5$ $6-8$ $5$ $7$ $5\times 7=35$ $8-10$ $6$ $9$ $6\times 9=54$ $10-12$ $2$ $11$ $2\times 11=22$ $12-14$ $3$ $13$ $3\times 13=39$ Total $20$ $162$

From the table, it can be observed that

$\sum{{{f}_{i}}=20}$

$\sum{{{f}_{i}}{{x}_{i}}}=162$

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

Mean number of plants per house $\left( \overline{X} \right)$:

$\overline{X}=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}$

$\overline{X}=\frac{162}{20}=8.1$

Therefore, the mean number of plants per house is $8.1$.

In this case, we will use the direct method because the value of ${{x}_{i}}$ and ${{f}_{i}}$ .

2. Consider the following distribution of daily wages of $\mathbf{50}$ worker of a factory.

 Daily wages (in Rs) $100\text{ }-120~$ $120\text{ - }140$ $140\text{ - }160~$ $160\text{ - }180~$ $180\text{ - }200~$ Number of  workers $12$ $14$ $8$ $6$ $10$

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

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

Take the assured mean $(a)$ of the given data

$a=150$

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

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

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

$h=120-100$

$h=20$

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

 Daily wages (In Rs) Number of  Workers (${{f}_{i}}$) ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-150$ ${{u}_{i}}=\frac{{{d}_{i}}}{20}$ ${{f}_{i}}{{u}_{i}}$ $100\text{ }-120~$ $12$ $110$ $-40$ $-2$ $-24$ $120\text{ - }140$ $14$ $130$ $-20$ $-1$ $-14$ $140\text{ - }160~$ $8$ $150$ $0$ $0$ $0$ $160\text{ - }180~$ $6$ $170$ $20$ $1$ $6$ $180\text{ - }200~$ $10$ $190$ $40$ $2$ $20$ Total $50$ $-12$

From the table, it can be observed that

$\sum{{{f}_{i}}=50}$

and

$\sum{{{f}_{i}}{{u}_{i}}}=-12$

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

The required mean:

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

$\overline{X}=150+\left( \frac{-12}{50} \right)20$

$\overline{X}=150-\frac{24}{5}$

$\overline{X}=150-4.8$

$\overline{X}=145.2$

Hence, the mean daily wage of the workers of the factory is Rs $145.20$.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.$18$. Find the missing frequency $f$.

 Daily  pocket  allowance (in Rs) $11\text{ - }13~$ $13\text{ - }15$ $15\text{ - }17$ $17\text{ - }19~$ $19\text{ - }21~$ $21\text{ - }23~$ $23\text{ - }25$ Number  of  workers $7$ $6$ $9$ $13$ $f~$ $5~$ $4$

Ans: Let the class size of the data be $h$.

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 $18$

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

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

It is given that, mean pocket allowance, $\overline{X}=\text{ }Rs\text{ }18$

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

$h=13-11$

$h=2$

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

 Daily  Pocket  Allowance (in Rs) Number  of  Workers (${{f}_{i}}$) Class Mark${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-18$ ${{f}_{i}}{{d}_{i}}$ $11\text{ }-13~$ $7$ $12$ $-6$ $-42$ $13\text{ - }15$ $6$ $14$ $-4$ $-24$ $15\text{ - }17$ $9$ $16$ $-2$ $-18$ $17\text{ - }19~$ $13$ $18$ $0$ $0$ $19\text{ - }21~$ $f~$ $20$ $2$ $2f$ $21\text{ - }23~$ $5~$ $22$ $4$ $20$ $23\text{ - }25$ $4$ $24$ $6$ $24$ Total $\sum{{{f}_{i}}}=44+f$ $2f-40$

From the table, it can be observed that

$\sum{{{f}_{i}}=44+f}$

$\sum{{{f}_{i}}{{u}_{i}}}=2f-40$

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

The required mean:

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

$18=18+\left( \frac{2f-40}{44+f} \right)2$

$0=\left( \frac{2f-40}{44+f} \right)$

$2f-40=0$

$f=20$

Hence, the value of frequency ${{f}_{i}}$ is $20$.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

 Number of Heart Beats Per Minute $65\text{ - }68~$ $68\text{ - }71~$ $71\text{ - }74~$ $74\text{ - }77~$ $77\text{ - }80$ $80\text{ - }83~$ $83\text{ - }86$ Number of Women $2$ $4~$ $3~$ $8$ $7~$ $4~$ $2$

Ans: Let the class size of the data be $h$.

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 $75.5$

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

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

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

$h=68-65$

$h=3$

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

 Number of  heart beats  per minute Number of  women ${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-75.5$ ${{u}_{i}}=\frac{{{d}_{i}}}{3}$ ${{f}_{i}}{{u}_{i}}$ $65\text{ - }68~$ $2$ $66.5~$ $-9~$ $-3$ $-6$ $68\text{ - }71~$ $4$ $69.5$ $-6~$ $-2$ $-8$ $71\text{ - }74~$ $3$ $72.5~$ $-3~$ $-1$ $-3$ $74\text{ - }77~$ $8$ $75.5~$ $0~$ $0~$ $0~$ $77\text{ -}80$ $7$ $78.5~$ $3$ $1$ $7$ $80\text{ - }83~$ $4$ $81.5~$ $6~$ $2$ $8$ $83\text{ - }86$ $2$ $84.5$ $3~$ $3~$ $6$ Total $30$ $4$

From the table, it can be observed that

$\sum{{{f}_{i}}=30}$

$\sum{{{f}_{i}}{{u}_{i}}}=4$

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

The required mean:

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

$\overline{X}=75.5+\left( \frac{4}{30} \right)3$

$\overline{X}=75.5+0.4$

$\overline{X}=75.9$

Therefore, mean heart beats per minute for these women are $75.9$ beats per minute.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Number of mangoes $\text{50-52}$ $53-55$ $56-58$ $59-61$ $62-64$ Number of  Boxes $15$ $110$ $135$ $115$ $25$

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ans:

 Number of Mangoes Number of Boxes${{f}_{i}}$ $\text{50-52}$ $15$ $53-55$ $110$ $56-58$ $135$ $59-61$ $115$ $62-64$ $25$

It can be noticed that class intervals are not continuous in the given data. There is a gap of $1$  between two class intervals. Therefore, we have to subtract $\frac{1}{2}$ to lower class and have to add $\frac{1}{2}$ to upper to make the class intervals continuous.

Let the class size of the data be $h$.

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 $57$.

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

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

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

$h=52.5-49.5$

$h=3$

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

 Class interval ${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-57$ ${{u}_{i}}=\frac{{{d}_{i}}}{3}$ ${{f}_{i}}{{u}_{i}}$ $49.5-52.5$ $15$ $51$ $-6$ $-2$ $-30$ $52.5-55.5$ $110$ $54$ $-3$ $-1$ $-110$ $55.5-58.5$ $135$ $57$ $0~$ $0~$ $0~$ $58.5-61.5$ $115$ $60$ $3$ $1$ $115$ $61.5-64.5$ $25$ $63$ $6$ $2$ $50$ Total $400$ $25$

It can be observed that from the above table

$\sum{{{f}_{i}}=400}$

$\sum{{{f}_{i}}{{u}_{i}}}=25$

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

The required mean:

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

$\overline{X}=57+\left( \frac{25}{400} \right)3$

$\overline{X}=57+\frac{3}{16}$

$\overline{X}=57.1875$

Hence, the mean number of mangoes kept in a packing box is $57.1875$.

In the above case, we used step deviation method as the values of ${{f}_{i}},\text{ }{{d}_{i}}$ are large and the class interval is not continuous.

6. The table below shows the daily expenditure on food of $\mathbf{25}$ households in a locality.

 DailyExpenditure(In Rs) $100-150$ $150-200$ $200-250$ $250-300$ $300-350$ Number of Households $4$ $5$ $12$ $2$ $2$

Find the mean daily expenditure on food by a suitable method.

Ans: Let the class size of the data be $h$.

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 $225$.

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

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

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

$h=150-100$

$h=50$

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

 Daily expenditure (in Rs) ${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-225$ ${{u}_{i}}=\frac{{{d}_{i}}}{50}$ ${{f}_{i}}{{u}_{i}}$ $100-150$ $4$ $125$ $-100$ $-2$ $-8$ $150-200$ $5$ $175$ $-6~$ $-1$ $-5$ $200-250$ $12$ $225$ $0~$ $0~$ $0~$ $250-300$ $2$ $275$ $50$ $1$ $2$ $300-350$ $2$ $325$ $100$ $2$ $4$ Total $25$ $-7$

It can be observed that from the above table

$\sum{{{f}_{i}}=25}$

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

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

The required mean:

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

$\overline{X}=225+\left( \frac{-7}{25} \right)\times 50$

$\overline{X}=221$

Hence, mean daily expenditure on food is Rs$211$.

7. To find out the concentration of $\mathbf{S}{{\mathbf{O}}_{\mathbf{2}}}$ in the air (in parts per million, i.e., ppm), the data was collected for $\mathbf{30}$ localities in a certain city and is presented below:

 Concentration of $\mathbf{S}{{\mathbf{O}}_{\mathbf{2}}}$ (in ppm) Frequency $0.00-0.04$ $4$ $0.04-0.08$ $9$ $0.08-0.12$ $9$ $0.12-0.16$ $2$ $0.16-0.20$ $4$ $0.20-0.24$ $2$

Find the mean concentration of $\mathbf{S}{{\mathbf{O}}_{\mathbf{2}}}$ in the air.

Ans: Let the class size of the data be $h$.

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 $0.14$.

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

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

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

$h=0.04-0.00$

$h=0.04$

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

 Concentrationof $\text{S}{{\text{O}}_{\text{2}}}$ (in ppm) Frequency${{f}_{i}}$ Class mark  ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-0.14$ ${{u}_{i}}=\frac{{{d}_{i}}}{0.04}$ ${{f}_{i}}{{u}_{i}}$ $0.00-0.04$ $4$ $0.02$ $-0.12$ $-3$ $-12$ $0.04-0.08$ $9$ $0.06$ $-0.08$ $-2$ $-5$ $0.08-0.12$ $9$ $0.10$ $-0.10$ $-1~$ $-9$ $0.12-0.16$ $2$ $0.14$ $0~$ $0~$ $0~$ $0.16-0.20$ $4$ $0.18$ $0.04$ $1$ $4$ $0.20-0.24$ $2$ $0.22$ $0.08$ $2$ $4$ Total $-31$

It can be observed that from the above table

$\sum{{{f}_{i}}=30}$

$\sum{{{f}_{i}}{{u}_{i}}}=-31$

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

The required mean:

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

$\overline{X}=0.14+\left( \frac{-31}{30} \right)\times (0.04)$

$\overline{X}=0.14-0.04133$

$\overline{X}=0.09867$

$\overline{X}=0.099$ ppm

Hence, mean concentration of $\text{S}{{\text{O}}_{\text{2}}}$ in the air is $0.099$ppm.

8. A class teacher has the following absentee record of  $\mathbf{40}$ students of a class for the whole term. Find the mean number of days a student was absent.

 Numberof Days $0-6$ $6-10$ $10-14$ $14-20$ $20-28$ $28-38$ $38-40$ NumberofStudents $11$ $10$ $7$ $4$ $4$ $3$ $1$

Ans: Let the class size of the data be $h$.

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 $17$.

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 calculated as follows:

 Number ofDays Number ofStudents${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-17$ ${{f}_{i}}{{d}_{i}}$ $0-6$ $11$ $3$ $-14$ $-154$ $6-10$ $10$ $8$ $-9$ $-90$ $10-14$ $7$ $12$ $-5$ $-35$ $14-20$ $4$ $17$ $0~$ $0~$ $20-28$ $4$ $24$ $7$ $28$ $28-38$ $3$ $33$ $16$ $48$ $38-40$ $1$ $39$ $22$ $22$ Total $40$ $-181$

It can be observed that from the above table

$\sum{{{f}_{i}}=40}$

$\sum{{{f}_{i}}{{u}_{i}}}=-181$

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

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

$\overline{X}=17+\left( \frac{-181}{40} \right)$

$\overline{X}=17-4.525$

$\overline{X}=12.475$

Hence, the mean number of days is $12.48$ days for which a student was absent.

9. The following table gives the literacy rate (in percentage) of $\mathbf{35}$ cities. Find the mean literacy rate.

 Literacy rate(in$%$) $45-55$ $55-65$ $65-75$ $75-85$ $85-95$ Number of  cities $3$ $10$ $11$ $8$ $3$

Ans: Let the class size of the data be $h$.

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 $70$.

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

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

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

$h=55-45$

$h=10$

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

 Literacy rate(in$%$) Number of cities ${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-70$ ${{u}_{i}}=\frac{{{d}_{i}}}{10}$ ${{f}_{i}}{{u}_{i}}$ $45-55$ $3$ $50$ $-20$ $-2$ $-6$ $55-65$ $10$ $60$ $-10$ $-1$ $-10$ $65-75$ $11$ $70$ $0~$ $0~$ $0~$ $75-85$ $8$ $80$ $10$ $1$ $8$ $85-95$ $3$ $90$ $20$ $2$ $6$ Total $35$ $-2$

It can be observed that from the above table

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

$\sum{{{f}_{i}}{{u}_{i}}}=-2$

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

The required mean:

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

$\overline{X}=70+\left( \frac{-2}{35} \right)\times 10$

$\overline{X}=70-\frac{20}{35}$

$\overline{X}=69.43$

Therefore, the mean literacy rate of cities is $69.43%$.

### 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.

### 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 ofConsumers${{f}_{i}}$ 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:

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

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

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

$\overline{X}=137.058$

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:

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

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

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

$M=135.76$

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.

 MonthlyConsumption(in units) Number ofConsumers 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:

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

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

$m=125+12$

$m=137$

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 the 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:

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

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

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

$17=25-x$

$x=8$

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 Policyholders 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 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 Policyholders $({{f}_{i}})$ 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:

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

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

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

$m=35.76$

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

5. The following table gives the distribution of the lifetime 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 lifetime 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, the median lifetime 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 ofsurnames${{f}_{i}}$ ${{x}_{i}}$ ${{d}_{i}}={{x}_{i}}-11.5$ ${{u}_{i}}=\frac{{{d}_{i}}}{20}$ ${{f}_{i}}{{u}_{i}}$ $1-4$ $6$ $2.5$ $-9$ $-3$ $-18$ $4-7$ $30$ $5.5$ $-6$ $-2$ $-60$ $7-10$ $40$ $8.5$ $-3$ $-1~$ $-40$ $10-13$ $16$ $11.5$ $0~$ $0~$ $0~$ $13-16$ $4$ $14.5$ $3$ $1$ $4$ $16-19$ $4$ $17.5$ $6$ $2$ $8$ 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:

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

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

$\overline{X}=11.5-3.18$

$\overline{X}=8.32$

Hence, the mean 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, the modal size of surnames 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) NumberofStudents 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, the median weight is $56.67$ kg.

## Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 13 Statistics

 Chapter Dropped Topics Statistics 13.5 - Graphical representation of cumulative frequency distribution Exercise - 13.4

## Class 10 Maths Chapter 13 : Exercises Breakdown

 Exercises Number of Questions Exercise 13.1 9 Questions and Solutions Exercise 13.2 6 Questions and Solutions Exercise 13.3 7 Questions and Solutions

## Conclusion

NCERT Solutions for maths statistics class 10 provide students with a thorough understanding and guidance to help them succeed in the subject.  It simplifies complex statistical concepts into easy-to-understand explanations,  which makes learning fun and efficient. It's important for students to focus on understanding the fundamental concepts such as mean, median, mode, and probability as they serve as the basis for more complex statistical subjects. For added confidence and great scores, practice with the supplied solutions and work through prior year's question papers. Students should expect around 4-5 questions from the statistics chapter.

## Other Study Material for CBSE Class 10 Maths Chapter 13

 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 5 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 Class 10 Maths Chapter 13 Statistics

1. How Do I Calculate the Mean of Grouped Data?

There are three different methods to calculate the mean.

(i) Direct method :

Mean or X= [Σfᵢxᵢ] / Σfᵢ

Where xᵢ is the class mark and fᵢ is the corresponding frequency of the class mark.

(ii) Assumed Mean method:

Mean = a + [Σfᵢdᵢ] / Σfᵢ

Where a is the assumed mean, dᵢ is the deviation from the class mark and fᵢ  is the

corresponding frequency of the class mark.

(iii) Step Deviation method:

Mean = a + { [Σfᵢuᵢ] / Σfᵢ } X h

Where a is assumed mean, h is the class size, fᵢ is the corresponding frequency of the class mark, and uᵢ = (xᵢ-a) / h.

2. Why Should I Prefer NCERT Solutions for Maths Chapter 14, Class 10?

The NCERT solutions of ch 14 class 10 maths provide a detailed overview of the chapter to the students. It guides the students throughout their learning process and revisions. These solutions are prepared by the most experienced teachers that assure the solutions of being top-grade and error-free. These Solution PDFs are designed as per the latest syllabus of CBSE, which helps the students to get acquainted with the examination process. If a student prefers these solutions, it will help him/her to secure good marks in examinations.

3. What topics do I have to read in Chapter 14 Statistics of Class 10 Maths?

The topics which you have to read in Chapter 14 Statistics of Class 10 Maths are as follows.

• Introduction to Statistics

• Methods for calculating the mean of grouped data

• Methods for the calculation of mode of grouped data

• Methods for calculating the median of grouped data

• Graphical Representation of Cumulative Frequency Distribution

• Summary

You must also check out the NCERT Solutions of Chapter 14 Statistics of Class 9 Maths to make its concepts crystal clear to you. You can avail these solutions on Vedantu website or app.

4. What do you mean by the term ‘Mean’ in Chapter 14 Statistics of Class 10 Maths?

The mean is the most common or the average value in a group of numbers. It is a measure of the central tendency of the distribution of probability along with mode and median. It is also termed as an expected value. There are two methods for calculating the mean. The first method is the Arithmetic mean. And the second method is known as the Geometric mean. Arithmetic mean is the sum of all the numbers divided by total numbers present in a collection of numbers whereas, the Geometric mean is the nth root of the numbers in a number collection.

5. What do you understand about ‘Median’ in Chapter 14 Statistics of Class 10 Maths?

In a list of numbers presented in ascending or descending format, the median is the middle number. It gives more information about numbers than the average. If there are odd numbers in the list then the median will be the middle number with the same numbers present above and below from that number. However, if there are even numbers in the number list then the median value is calculated by adding two numbers and then dividing the sum by two.

6. What is ‘Mode’ in Chapter 14 Statistics of Class 10 Maths?

In Statistics, the mode is the most frequent object that is repeated in the list. A set of numbers may have a single mode, more than one mode or no mode at all. In the normal distribution of numbers, the mode is the same value as the median and mean. In many cases, the modal value differs from the average value in the set of numbers.

7. What should I do to download the NCERT Solutions of Chapter 14 Statistics of Class 10 Maths?

If you are a Class 10 student and want to download the NCERT Solutions of Chapter 14 Statistics then go through the given steps.

• First, visit the page-NCERT Solutions of class 10 MAths Chapter 14 Statistics.

• Click on Chapter 14 Statistics.

• This link will take you to the webpage of the NCERT Solutions of Chapter 14 Statistics.

• On this webpage, you will see different types of study materials like revision notes, important questions, textbook questions and answers other than the NCERT Solutions.

• You can click on the icon of NCERT Solutions to download the content free of cost.

8. What are the three methods of statistics in Class 10 Maths Ch 13 ?

The three methods of Statistics are

• Mean: This is the most common method. It's basically the sum of all the data points divided by the total number of data points. Imagine you have a bunch of apples, and the mean tells you the average number of apples per person if you shared them all equally.

• Median: This is the middle value when you arrange the data points in order, from least to greatest. If you have an odd number of data points, the median is the exact middle number. With an even number, it's the average of the two middle numbers. Think of lining up your apples by size; the median is the one right in the center (or the average of the two in the center if you have an even number).

• Mode: This is the most frequent value in the data set. It tells you which number appears the most often. Imagine counting the colors of your apples; the mode is the color you have the most of.

9. What are the types of data in class 10 statistics solutions?

There are 4 types of Data in Statistics in statistics class 10 ncert solutions

1. Nominal Data

2. Ordinal Data

3. Discrete Data

4. Continuous Data