NCERT Solutions 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.

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:

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.

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:

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 \[f\].

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:

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.

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:

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.

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

Ans:

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:

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.

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:

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:

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:

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.

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:

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.

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:

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:

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:

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:

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.

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:

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.

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:

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.

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:

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.

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:

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.

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 \[\mathbf{y}\] .

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.

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.

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.

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:

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.

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:

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.

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:

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.

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:

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

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.

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 :

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

Class 10 Maths Chapter 13 : Exercises Breakdown

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.