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NCERT Solutions for Class 10 Maths Chapter Chapter 13 Statistics

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How Can You Use Statistics Class 10 Questions and Answers For Better Exam Preparation

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.

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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:

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

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

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

  • This article contains chapter notes, exercises, links  and important questions for Chapter 13 - Statistics which you can download as PDFs.

  • 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

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NCERT Solutions for Class 10 Maths Chapter Chapter 13 Statistics
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How Can You Use Statistics Class 10 Questions and Answers For Better Exam Preparation

Exercise 13.1: This exercise consists of 9 questions based on calculating the mean, median, and mode of grouped and ungrouped data. There are a total of seven questions in this exercise, and the solutions to each question provide step-by-step instructions on how to find the mean, median, and mode of the given data.


Exercise 13.2: This exercise involves 6 questions based on finding the cumulative frequency, quartiles, and interquartile range of given data. There are a total of six questions in this exercise, and the solutions to each question provide step-by-step instructions on how to find the cumulative frequency, quartiles, and interquartile range of the given data.


Exercise 13.3: This exercise contains 7 questions, this exercise focuses on constructing and interpreting various types of graphical representations of data, such as histograms, bar graphs, and pie charts. There are a total of six questions in this exercise, and the solutions to each question provide step-by-step instructions on how to construct and interpret different types of graphs.


Access NCERT Solutions for 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 \[\mathbf{50}\] 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.

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 $

$  \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:

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, 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) $

$  \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 of

Patients
\[{{f}_{i}}\] 

  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 of

students per

teacher

Number of

states /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:

Number

of 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:

Monthly

consumption

(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 $

 $ \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.

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 $ 

 $ 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 (in

hours)

Number of

lamps

$1500-2000$ 

\[14\]

$2000-2500$

\[56\]

$2500-3000$

\[60\]

$3000-3500$

\[86\]

$3500-4000$

\[74\]

$4000-4500$

\[62\]

$4500-5000$

\[48\]

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 (in

hours)

Number of

lamps

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:

Number

of Letters

\[1-4\] 

\[4-7\]  

\[7-10\] 

\[10-13\] 

\[13-16\] 

\[16-19\]

Number of

Surnames

\[6\]  

\[30\] 

\[40\] 

\[16\]  

\[4\] 

\[4\]

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.

Number

of Letters

Number of

Surnames

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:

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

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

$   \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

Number

of 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)

\[40-45\] 

\[45-50\]  

\[50-55\] 

\[55-60\] 

\[60-65\]

\[65-70\] 

\[70-75\]

Number

of

students

\[2\]  

\[3\] 

\[8\]

\[6\]

\[6\]

\[3\] 

\[2\]

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)

Number

of

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, 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


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.



Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter Chapter 13 Statistics

1. What is the most important concept students need to understand in statistics class 10?

The mean, median, and mode are the three fundamental measures of central tendency that form the foundation of statistics class 10. These measures help students understand how to find the typical or average value in a dataset, which is essential for analyzing real-world information like test scores, population data, or sales figures.

Why it matters: Central tendency measures appear in almost every statistical problem and form the basis for more advanced concepts like standard deviation and probability.

Example: In a class of 30 students with test scores ranging from 45 to 95, the mean gives the average performance, median shows the middle value, and mode reveals the most frequent score.

Tip: Always arrange data in ascending order before finding median, and remember that mean is affected by extreme values while median is not.

Mastering these three measures early helps students tackle complex statistical problems with confidence throughout the chapter.

2. How can students access NCERT Solutions for statistics problems effectively?

NCERT Solutions provide step-by-step explanations for all statistics problems, helping students understand the methodology behind each calculation rather than just memorizing formulas. These solutions cover both in-text questions and exercise problems, ensuring comprehensive coverage of mean, median, mode, and graphical representation topics.

Why it matters: Statistics problems often involve multiple steps and different approaches, making detailed solutions crucial for building problem-solving skills and avoiding common calculation errors.

Example: For finding the mean of grouped data, NCERT Solutions show how to create frequency tables, calculate midpoints, and apply the formula step-by-step with clear explanations.

Check: Vedantu's NCERT Solutions include alternative methods and common mistake alerts to help students verify their answers and understand different approaches to the same problem.

3. What are the key formulas students must memorize for statistics class 10 NCERT?

Students must memorize four essential formulas: Mean = Σ(f×x)/Σf for grouped data, Median = l + [(n/2-cf)/f] × h for grouped data, and the direct division method for ungrouped data calculations. These formulas are fundamental for solving all types of statistics problems in class 10.

Why it matters: Statistics problems in CBSE exams directly test formula application, and students lose marks when they confuse or incorrectly apply these basic formulas during calculations.

Example: For finding median of grouped data with class intervals 10-20, 20-30, 30-40, students need the median formula where l is the lower boundary of median class, n is total frequency, cf is cumulative frequency, f is frequency of median class, and h is class width.

Formula tip: Write down all given values clearly before substituting in formulas, and always double-check the median class identification as this is where most errors occur.

4. How should students prepare statistics questions and answers for better exam performance?

Students should practice a variety of statistics problems daily, focusing on both computational accuracy and understanding the reasoning behind each step to build confidence for CBSE board exams. Regular practice with different data sets helps students recognize patterns and choose the most efficient solution method.

Why it matters: Statistics chapter typically carries 8-10 marks in class 10 board exams, and consistent practice ensures students can handle both easy and challenging problems within time constraints.

Steps for effective preparation: Solve NCERT exercise problems first to build foundational understanding. Practice additional problems from previous year papers. Time yourself while solving to improve speed. Create formula sheets for quick revision. 

Check: After solving each problem, verify your answer using an alternative method when possible, and ensure your final answer makes logical sense in the given context.

5. Where can students download statistics class 10 NCERT PDF for offline study?

Students can download Free PDF materials for statistics from official educational platforms, ensuring they have access to complete chapter content, solved examples, and practice questions for offline preparation. These PDFs typically include the full NCERT textbook chapter along with additional practice materials and solutions.

Why it matters: Offline access allows students to study statistics concepts without internet dependency, enabling consistent learning even in areas with poor connectivity or during travel.

Example: A comprehensive statistics PDF includes the complete chapter 13 content, all exercise solutions, summary tables of formulas, and additional practice problems with step-by-step answers.

Instruction: Download PDFs from trusted educational sources only, and ensure the content aligns with the latest NCERT curriculum to avoid studying outdated material. Having offline resources ensures uninterrupted study sessions and helps students maintain consistent preparation schedules regardless of internet availability.

6. What common mistakes should students avoid while solving mean, median, and mode problems?

Students commonly make calculation errors when finding the median class in grouped data and often forget to arrange ungrouped data in ascending order before finding the median. These fundamental mistakes lead to incorrect answers even when students know the correct formulas and concepts.

Why it matters: Small computational errors can result in completely wrong answers, causing students to lose valuable marks in board exams despite having good conceptual understanding.

Common pitfalls to avoid: Not identifying the correct median class in grouped data. Confusing class boundaries with class limits. Forgetting to multiply frequency with midpoint when calculating mean. Using wrong cumulative frequency values in median formula.

Tip: Always double-check your median class identification by ensuring n/2 falls within the cumulative frequency range of your chosen class, and verify mean calculations by checking if the result lies within the data range.

7. How do graphical representations help in understanding statistics concepts better?

Graphical representations like histograms, frequency polygons, and cumulative frequency curves provide visual understanding of data distribution, making it easier for students to identify patterns and relationships in statistical data. These visual tools help students grasp abstract concepts like central tendency and data spread more intuitively.

Why it matters: Visual learning enhances comprehension for students who struggle with numerical calculations alone, and graphs often appear in board exam questions requiring interpretation skills. 

Example: A histogram showing students' test scores immediately reveals whether most students scored high, low, or average, while the same information might be difficult to interpret from a frequency table alone.

Steps for graph interpretation: Identify the highest and lowest frequency bars. Locate the modal class visually. Observe the overall shape and distribution pattern. Understanding graphs builds analytical thinking and helps students connect mathematical calculations with real-world data interpretation skills.

8. What is the best strategy for solving more than type questions in statistics class 10?

More than type questions require students to find cumulative frequencies and use them to determine how many observations exceed a particular value, typically solved using cumulative frequency tables or ogive curves. This question type tests understanding of cumulative data analysis and graph reading skills.

Why it matters: These questions combine computational skills with data interpretation, appearing frequently in board exams as they test multiple statistical concepts simultaneously.

Example: If asked \"How many students scored more than 75 marks?\", students need to find the cumulative frequency up to 75 marks and subtract from total frequency to get the answer.

Steps to solve: Create a cumulative frequency table (less than or more than type). Locate the given value in appropriate class interval. Use interpolation if exact value is not a class boundary. Calculate final answer using total frequency minus cumulative frequency