Exercise 13.1 (Revised) - Chapter 13 - Statistics - Ncert Solutions class 10 - Maths

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Chapter 13 Statistics NCERT Solutions Class 10 Maths: Clear, Concise Answers

Ex 13.1 Question 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?
Answer.

Since, number of plants and houses are small in their values, so we use direct method.

Mean $(\bar{x})=\frac{\sum f_i x_i}{\sum f_i}=\frac{162}{20}=8.1$
Hence mean number of plants per house is 8.1.
Ex 13.1 Question 2.

Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer.

From given data, Assume mean $(a)=150$, Width of the class $(h)=20$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-12}{50}=-0.24
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=150+20(-0.24)=150-4.8=145.2$
Hence mean daily wages of the workers of factory is Rs. 145.20.
Ex 13.1 Question 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 $(f)$.

Answer

From given data, Assume mean $(a)=18$
$
\begin{aligned}
& \therefore(\bar{x})=a+\frac{\sum f_i d_i}{\sum f_i} \\
& \Rightarrow 18=18+\frac{2 f-40}{44+f} \\
& \Rightarrow \frac{2 f-40}{44+f}=0 \\
& \Rightarrow 2 f-40=0 \\
& \Rightarrow 2 f=40 \\
& \Rightarrow f=20
\end{aligned}
$

Hence missing frequency is 20 .
Ex 13.1 Question 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:

Answer:

( in the class interval 77-80, 78.4 changes to 78.5)
From given data, Assume mean $(a)=75.5$, Width of the class $(h)=3$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{4}{30}=0.13 \text { (approx.) }
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=75.5+3(0.13)=75.5+0.39=75.89$
Hence mean heart beat per minute for women is 75.89 .
Ex 13.1 Question 5.

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

\{change the frequency in above table as: $50-52$ (15) 53-55 (110) 56-58 (135) 59-61 (115) $62-$ $64(25)\}$

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

Answer.

Since value of number of mangoes and number of boxes are large numerically. So we use step-deviation method
we convert the class interval firstly into exclusive form given as

From given data, Assume mean $(a)=57$, Width of the class $(h)=3$
$\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{25}{400}=0.0625$ (approx.)
Using formula, Mean $(\bar{x})=a+h \bar{u}=57+3(0.0625)=57+0.1875=57.1875=57.19$ (approx.)
Hence mean number of mangoes kept in a packing box is 57.19.
Ex 13.1 Question 6.

The table below shows the daily expenditure on food of 25 households in a locality:

Find the mean daily expenditure on food by a suitable method.
Answer.

From given data, Assume mean $(a)=225$, Width of the class $(h)=50$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-7}{25}=-0.28
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=225+50(-0.28)=225-14=211$ Hence mean daily expenditure on food is Rs. 211.
Ex 13.1 Question 7.

To find out the concentration of $\mathrm{SO}_2$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of $\mathrm{SO}_2$ in the air.

Answer.

From given data, Assume mean $(a)=0.10$, Width of the class $(h)=0.04$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-1}{30}=-0.033 \text { (approx.) }
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=0.10+0.04(-0.033)=0.10-0.0013=0.0987$ (approx.) Hence mean concentration of $\mathrm{SO}_2$ in air is $0.0987 \mathrm{ppm}$.
Ex 13.1 Question 8.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Answer

From given data, Assume mean $(a)=17$
$
\therefore(\bar{x})=a+\frac{\sum f_i d_i}{\sum f_i}=17+\frac{(-181)}{40}=17-4.52=12.48
$

Hence mean 12.48 number of days a student was absent.
Ex 13.1 Question 9.

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer

From given data, Assume mean $(a)=70$, Width of the class $(h)=10$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-2}{35}=-0.057
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=70+10(-0.057)=70-0.57=69.43$
Hence mean literacy rate is $69.43 \%$.