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NCERT Solutions for Class 10 Maths Chapter 13 - Statistics Exercise 13.1

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NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 - FREE PDF Download

The NCERT Solutions for Maths Chapter 13 Exercise 13.1 Class 10 Statistics provides complete solutions to the problems in the Exercise. These NCERT Solutions are intended to assist students with the CBSE Class 10 board examination. The answers are formulated by following the updated CBSE curriculum; hence going through the solution will surely secure high marks for you in your maths exams. Students should thoroughly study this NCERT solution in order to solve all types of questions based on Statistics. By completing these practice questions with the NCERT Maths Solutions Chapter 10 Exercise 13.1 Class 10, students will be better prepared to understand all of the different types of questions that may be asked in the Class 10 board exams.

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Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 13 Exercise 13.1 Class 10 | Vedantu
3. Access NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1
4. Class 10 Maths Chapter 13: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 13 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 13 Exercise 13.1 Class 10 | Vedantu

  • This article focuses on the concept of class intervals, which group similar data points together.

  • This chapter teaches us how to construct a frequency table to effectively represent the data distribution. A frequency table summarizes how often each value (or range of values) appears in the data set.

  • Also learn how to classify data into appropriate class intervals based on the data range and desired level of detail and how frequency distribution helps visualize the distribution and patterns within a data set.

  • The ability to organize and represent data effectively using frequency tables.

  • This exercise lays the groundwork for applying statistical measures (like mean, median, mode) to analyze data further.

  • Exercise 13.1 class 10 NCERT solutions has overall 9 Questions & Solutions.

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Access NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1

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 \[\mathbf{50}\] workers 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 $

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

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

$ \Rightarrow  \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:

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

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

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

$\Rightarrow  2f-40=0$ 

$\Rightarrow  f=20$ 

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 $

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

$ \Rightarrow  \overline{X}=75.5+0.4 $

$ \Rightarrow  \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 a 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$

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

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

$ \Rightarrow \overline{X}=57.1875 $

Hence, 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.

Daily

Expenditure

(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 $

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

 $\Rightarrow  \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:

Concentration

of \[\text{S}{{\text{O}}_{\text{2}}}\] (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, the 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.

Number

of Days

\[0-6\] 

\[6-10\]  

\[10-14\] 

\[14-20\] 


\[20-28\]



\[28-38\] 

\[38-40\]

Number

of

Students


\[11\]  

\[10\] 

\[7\] 

\[4\]  

\[4\]

\[3\] 

$1$ 


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 of

Days

Number of Students \[{{f}_{i}}\] 

  \[{{x}_{i}}\]

${{d}_{i}}={{x}_{i}}-17$ 

\[{{f}_{i}}{{d}_{i}}\]

\[0-6\] 

\[11\] 

\[3\] 

\[-14\] 

\[-154\] 

\[6-10\] 

\[10\]  

\[8\]  

\[-9\] 

\[-90\] 

\[10-14\]

\[7\] 

\[12\] 

\[-5\] 

\[-35\]

\[14-20\]

\[4\]

\[17\] 

\[0~\] 

\[0~\]

\[20-28\]  

\[4\] 

\[24\] 

\[7\] 

\[28\]

\[28-38\]

\[3\]

\[33\]

\[16\]

\[48\]

\[38-40\]

\[1\]

\[39\]

\[22\]

\[22\]

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) $

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

$ \Rightarrow  \overline{X}=17-4.525 $

$ \Rightarrow  \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 $

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

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

 $\Rightarrow  \overline{X}=69.43 $

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


Conclusion

Class 10 Maths Chapter 13, exercise 13.1, focuses on the application of statistics to solve real-world problems. It's important to understand how to calculate measures of central tendency (mean, median, mode) and how to interpret these values in different contexts. Pay close attention to how data is organized and represented, as this will help in solving problems efficiently. Previous years' question papers have typically included 2-3 questions from this exercise, emphasizing practical application and interpretation of statistical data. Focus on understanding the concepts thoroughly and practicing different types of problems to build confidence and accuracy.


Class 10 Maths Chapter 13: Exercises Breakdown

Exercise

Number of Questions

Exercise 13.2

6 Questions & Solutions (6 Long Answers)

Exercise 13.3

7 Questions & Solutions (7 Long Answers)



CBSE Class 10 Maths Chapter 13 Other Study Materials



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 13.1

1. What are the topics covered in Class 10 Maths Chapter 14?

In the Class 10 Maths Chapter 14 named Statistics, the students will be introduced to various interesting topics and sub-topics such as cumulative frequency tables, graphical representations of data/ bar charts, histograms, and frequency polygons. Sometimes, we need to describe the data arithmetically, like describing the mean age of a class of students, mean height of a group of students, median score or model shoe size of a group. In this chapter, students will be introduced to three measures of central tendency i.e., mean, median, mode of ungrouped data and mean of grouped data.

  • 14: Statistics

  • 14.1: Introduction

  • 14.2: Mean of Grouped Data

  • 14.3: Mode of Grouped Data

  • 14.4: Median of Grouped Data

  • 14.5: Graphical Representation Of Cumulative Frequency Distribution

  • 14.6: Summary

2. Why should I refer to NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1?

NCERT Solutions for Class 10 Maths Chapter 14 is the best study material for all the CBSE students. Since these solutions are created by the best subject matter experts in the industry, these are the best quality resources available for exam preparation. You will get solved answers to the questions given in the exercises of your NCERT maths textbook. You will understand the question patterns and marks weightage as well. So you can prepare accordingly for the exam. There are a lot more benefits of using the NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1.

3. Can I download these NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 for free from Vedantu website?

Yes, these NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 are available on Vedantu website. You can avail these from Vedantu app as well at absolutely no cost. These are created by the best subject matter experts in the industry. If you want to score the highest possible marks in the Class 10 Maths exam, then you should definitely opt for Vedantu’s NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 as per your convenience.

4. What kind of questions should I expect from Class 10 Maths Chapter 14 Exercise 14.1?

In this exercise, Class 10th students will solve questions based on finding the mean number. A group of collection of data will be provided to the students to find the mean of them and the missing frequency as well. There are a few methods also involved in this i.e. Direct Method, Assumed Mean Method, and Step-deviation Method.


The questions in the Ex.14.1 are solved as per the latest CBSE syllabus and guidelines.  NCERT solutions are one of the best learning materials, where problems are solved in a detailed, concise manner following each and every necessary step and method.

5. How many exercises are there in Class 10 Maths Chapter 14 Statistics?

There are four exercises in Class 10 Maths Chapter 14 Statistics- 14.1, 14.2, 14.3 and 14.4. All four exercises are essential to attempt and significant for chapter understanding. For better assistance, Vedantu offers all solutions for free and also on its Vedantu Mobile app. You can download the same and start practising right away.

6. What are the key formulas in Chapter 14 of NCERT Solutions for Class 10 Maths?

The mean of grouped data using the direct technique, mean method and step deviation method are some of the key formulae in this chapter. For grouped data, there are further formulae such as median and mode. By systematically addressing questions, these formulae may be better comprehended. The NCERT Solutions Class 10 Maths Chapter 14 explains how to derive these formulae.

7. Why should I work on NCERT Solutions for Class 10 Maths and Statistics 14?

The NCERT Solutions Class 10 Maths Statistics 14 is a crucial chapter not only for examinations but also for gaining insights that may be applied in real-life circumstances. Students will learn how to organise and evaluate various types of data by placing it in a certain or necessary order in this chapter.

8. Give a brief on Class 10 Maths Chapter 14.

From ungrouped data to grouped data, learning will transition to all three measures such as mean, median, and mode in NCERT Solutions Class 10 Maths Chapter 14. Students will learn about new statistical concepts such as cumulative frequency, cumulative frequency distribution, cumulative frequency curves or "ogives," and so on.

9. How can I get full marks in Statistics Chapter 14 Maths Class 10?

You must write down all the formulas and revise them regularly. The procedure to solve a question is extremely important. Hence, you must practice all sorts of questions. Make sure you do all the NCERT questions thoroughly, and then you can practice from other books as well. Use Vedantu’s NCERT Solutions for free to get proper assistance and guidance.

10. What is the focus of Exercise 13.1 in Class 10 Maths Ch 13 Ex 13.1?

Class 10 Maths Chapter 13 Exercise 13.1 focuses on the introduction to statistics, specifically on understanding and calculating measures of central tendency such as mean, median, and mode for a given set of data.

11. How do you calculate the mean for a given set of data in Class 10 Chapter 13 Exercise 13.1?

To calculate the mean (average) of a given set of data:

Step 1: Sum all the data values.

Step 2: Divide the sum by the total number of data values.

12. What is mode, and how do you find it in Class 10 Maths Ex 13.1?

The mode is the value that appears most frequently in a data set. To find the mode:

  • Identify the value(s) that occur most often in the data set.

  • The value with the highest frequency is the mode.

13. Why is it important to understand measures of central tendency in class 10th maths Chapter 13 exercise 13.1?

In Class 10 Maths Chapter Statistics Exercise 13.1, understanding measures of central tendency (mean, median, and mode) is crucial because they provide a summary of a data set, representing its central point. These measures help in analyzing and interpreting data, making it easier to compare different data sets and draw meaningful conclusions.

14. Are there any specific examples in Class 10 Ch 13 Ex 13.1 to illustrate these concepts?

Yes, Cass 10 Maths Statistics Exercise 13.1 provides various examples and problems that help students practice calculating the mean, median, and mode for different sets of data. These examples are designed to reinforce the understanding of these statistical measures and their applications.