NCERT Solutions for Class 10 Maths Chapter 14 - Statistics

Exercise 14.1: This exercise consists of 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 14.2: This exercise involves 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 14.3: 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.

Exercise 14.4: This exercise involves questions based on finding the probability of an event and its different properties. There are a total of nine questions in this exercise, and the solutions to each question provide step-by-step instructions on how to find the probability of an event and its different properties, such as complementary events, mutually exclusive events, and independent events.

Overall, the exercises in NCERT Solutions for Class 10 Maths Chapter 14 "Statistics" are designed to help students develop their understanding of the concepts and techniques used in statistics, and to enhance their problem-solving skills.

Access NCERT Solutions for Class 10 Maths  Chapter Number 14 - Statistics

Exercise 14.1

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 14.2

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 14.3

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.

Exercise 14.4

1. The following distribution gives the daily income of \[\mathbf{50}\]  workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Ans: The frequency distribution table of less than type is given as:

Plot the points $\left( 120,12 \right)$, $\left( 140,26 \right)$, $\left( 160,34 \right)$, $\left( 180,40 \right)$, $\left( 200,50 \right)$.

Taking upper class limits of class intervals on \[\text{x-}\]axis and their respective frequencies on \[\text{y-}\]axis, its ogive can be drawn as follows:

During the medical check-up of \[\mathbf{35}\] students of a class, their weights were recorded as follows:

2. Draw a less than type to give for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Ans: The given cumulative frequency distributions of less than type is given as:

Plot the points $\left( 38,0 \right)$, $\left( 40,3 \right)$, $\left( 42,5 \right)$, $\left( 44,9 \right)$, $\left( 46,14 \right)$, $\left( 48,28 \right)$, $\left( 50,32 \right)$, $\left( 52,35 \right)$

Taking upper class limits of class intervals on \[\text{x-}\]axis and their respective frequencies on \[\text{y-}\]axis, its ogive can be drawn as follows,

Here, \[n\text{ }=\text{ }35\] 

So,\[\frac{n}{2}=17.5\] 

Mark the point A whose ordinate is \[17.5\] and its \[\text{x-}\]coordinate is\[46.5\] .

Therefore, the median of this data is \[46.5\].

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

It can be observed that the difference between two consecutive upper-class limits is\[2\] . The class marks with their respective frequencies are obtained as below.

It can be observed from the given table

\[n\text{ }=35\] 

$\frac{\text{n}}{2}=17.5$

The cumulative frequency just greater than $\frac{n}{2}$ is\[28\] , Belonging to class interval \[46-48\] .

Median class = \[46-48\]

\[l=46\] 

\[f=14\] 

\[cf=14\] 

\[h=2\] 

Substituting these values in the formula of median we get

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

$m=46+\left( \frac{17.5-14}{14} \right)\times 2$ 

$m=46+\left( \frac{3.5}{7} \right)$ 

$m=46.5$

Hence, the median of this data is \[46.5\] .

Since, the value of median by graph and by formula are equal hence verified.

3. The following table gives production yield per hectare of wheat of \[\mathbf{100}\]  farms of a village.

Change the distribution to a more than type distribution and draw ogive.

Ans: The cumulative frequency distribution of more than type can be obtained as follows.

Plot the points $\left( 50,100 \right)$, $\left( 55,98 \right)$, $\left( 60,90 \right)$, $\left( 65,78 \right)$, $\left( 70,54 \right)$, $\left( 75,16 \right)$.

Taking upper class limits of class intervals on \[\text{x-}\]axis and their respective frequencies on \[\text{y-}\]axis, its give can be drawn as follows,

(image will be uploaded soon)

Importance of Maths Chapter 14 Statistics in Class 10

NCERT Class 10 Maths Chapter 14, Statistics, is one of the most important and scoring chapters of Class 10 CBSE. Since it carries around 11 to 12 marks in Term 2 of the Class 10 examination, it is very important for students to have their basics concepts clear to score higher marks. 

Practically, too, ‘Statistics’ is highly used in today’s data-oriented world, so, studying it is quite important. Statistics help you to represent relevant information easily in graphical and tabular form. Topics that you learn in Class 10 NCERT Chapter 14 such as step deviation methods, calculating mean, mode and median of grouped data, frequency distribution, etc. come in handy in many real-life situations.

The CBSE Class 10 Maths Chapter 14 Solutions are provided in PDF format on the mobile app and official website of Vedantu and students can download the PDF for free. The NCERT Solutions for Ch 14 Maths Class 10 PDF explains the chapter in a detailed manner to the students. Going through the Statistics Class 10 Solutions PDF will boost their confidence to appear in the exams without any fear. It will help them to clear all the doubts that they may face while solving the sums given in the exercise. The Ch 14 Class 10 Maths PDF are prepared to make the learning procedure easy for the students. 

The students of class 10, who have doubts and are facing difficulty in class 10 maths ch 14 can access the NCERT Solutions PDF on Vedantu to achieve good marks in the exams. These Solutions give an overview of the chapter and make it easier for students to understand the chapter thoroughly. Class 10 Maths Statistics Chapter is one of the most interesting chapters in the syllabus, so students will have a great time understanding the concepts of this chapter.

NCERT Solutions For Class 10 Chapter 14 Maths Exercise Wise Marks Weightage

Chapter 14: Statistics

The class 10 maths statistics gives a detailed explanation of the chapter to the students. The whole chapter is divided into four different segments or units. This division of exercises makes it easier for students to understand the chapter easily. When it comes to securing good marks in examinations, The NCERT solutions for ch14 maths class 10 is the answer to every problem. 

The Class 10th maths chapter 14 is divided according to the following:

Exercise 14.1 - Exercise 14.1 of class 10 ch 14 maths consists of 9 questions in total. This exercise explains the first measure of central tendency, which is mean. This exercise demonstrates and teaches the various methods to calculate the mean.

Once a student finishes the exercise, he/she will understand all the methods and will be able to solve any kind of problem. The NCERT solutions class 10 maths chapter 14 provides the solution to each of the 9 questions.

Exercise 14.2 - This exercise has a detailed study of the calculation of the Mode of given data. This exercise consists of 6 questions and each of which revolves around the calculation of mode. The NCERT solutions for class 10maths chapter 14 provides the answers to these questions. 

Exercise 14.3 - Exercise 14.3 talks about the calculation of the mode of grouped data. This exercise has a total of 7 questions. Here, we have provided the solutions to all these questions, so that it would act as a guide for the students.

Exercise 14.4 - This is the last exercise of the chapter, and in this exercise, you will study in detail the plotting of graphs and how to represent median graphically. The Class 10 maths chapter 14 solutions will help you to plot a graph perfectly. These solutions will help you grab good marks in your examinations.

Key Features for NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The CBSE 10th Board instructors strongly suggest the Class 10 Maths NCERT Solutions for Chapter 14 on Vedantu. These solutions are among the top resources for students in Class 10 who want to study independently. The important aspects of these solutions, as listed below, explain why they are a must-have for the CBSE Term 2 Class 10 Maths exam.

• These Class 10 Chapter 14 Statistics NCERT Solutions PDFs are simple to access online and free to download for offline practice.

• These NCERT Solution PDFs are available on the Vedantu mobile app, so students may download them and study whenever and wherever they choose.

• For a complete understanding of all students, each sum covered in the four exercises of Class 10 Ch-4 Quadratic Equations is answered in a step-by-step way.

• Our highly experienced Maths teachers have created these NCERT Solutions in accordance with the latest rules prescribed by CBSE for Class 10, so students can use them for exam preparation.

• These NCERT Solutions can be used by students if they are challenged with a problem. As a result, individuals can address their Statistics doubts during last-minute revisions without losing time or waiting to counsel from a buddy.

Conclusion 

Vedantu's subject experts have developed NCERT Solutions for Class 10 Maths Chapter 14 for students' Term 2 exam preparation and revision. Students need to have a solid understanding of other math concepts, such as mean, median, and mode, in order to solve the sums discussed in this chapter. As a result, obtaining and referring to the Maths NCERT Solutions Class 10 Chapter 14 will greatly assist students in their exam preparation.