NCERT Solutions for Class 10 Maths Chapter 13 Statistics Ex 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:

$$\bar{X}=\frac{\sum f_i{x}_i}{\sum f_i}$$

Class mark $$\left(x_i\right)$$ for each interval is calculated as follows:$$xi=\frac{\text{Upper class limit+Lower class limit}}{2}$$ 

$$x_i$$ 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 $$\left(\bar{X}\right)$$:

$\bar{X}=\frac{\sum f_i{x}_i}{\sum f_i}$

$\bar{X}=\frac{162}{20}=8.1$        

Therefore, the mean number of plants per house is $$8.1$$.

In this case, we will use the direct method because the value of $$x_i$$ and $f_i$ .

2. Consider the following distribution of daily wages of $$\mathbf{50}$$ workers of a factory.

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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

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

$\Rightarrow \bar{X}=150-4.8$

$\bar{X}=145.2$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$$x_i=\frac{\text{Upper class limit+Lower class limit}}{2}$$ 

It is given that, mean pocket allowance, $$\bar{X}=Rs18$$

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:

$\bar{X}=a+\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)h$ 

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

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

$\Rightarrow 2f-40=0$ 

$\Rightarrow f=20$ 

Hence, the value of frequency $$f_i$$ is $$20$$.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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

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

$\Rightarrow \bar{X}=75.9$

Therefore, mean heart beats per minute for these women are $$75.9$$ beats per minute.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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

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

$\Rightarrow \bar{X}=57.1875$

Hence, mean number of mangoes kept in a packing box is $57.1875$.

In the above case, we used step deviation method as the values of $$f_i,d_i$$ 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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h$

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

 $\Rightarrow \bar{X}=221$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)h$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h$ 

$\bar{X}=0.14+\left(\frac{-31}{30}\right)\times (0.04)$

$\bar{X}=0.14-0.04133$

$\bar{X}=0.09867$

$\bar{X}=0.099$ ppm

Hence, the mean concentration of $$\text{S}{\text{O}}_{\text{2}}$$ in the air is $0.099$ppm.

8. A class teacher has the following absentee record of  $$\mathbf{40}$$ students of a class for the whole term. Find the mean number of days a student was absent.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$

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

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

$\Rightarrow \bar{X}=12.475$

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:

$\bar{X}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$

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

$$\sum f_i=35$$