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NCERT Solutions Class 10 Maths Chapter 13 Statistics

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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.

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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:

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

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

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

  • 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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Exercises Under NCERT Solutions for Class 10 Maths Chapter 13 – Statistics

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

\[40-45\] 

\[2\]

\[2\]

\[45-50\]

\[3\]

\[2+3=5\]

\[50-55\]

\[8\]

\[5+8=13\]

\[55-60\]

\[6\]

\[13+6=19\]

\[60-65\]

\[6\]

\[19+6=25\]

\[65-70\]

\[3\]

\[25+3=28\]

\[70-75\]