NCERT solutions for class 10 maths Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 13 Statistics are provided here. This article is focused on the measurement of central tendencies like median, mode, and mean to analyze the grouped and ungrouped data. For students preparing for exams, the 10th Mathematics Statistics NCERT Answers is a comprehensive guide to solving exercise problems and understanding their applications. The Statistics Chapter Solutions Class 10 Math also includes different types of graphs such as pie charts, histograms, line graphs, bar graphs, etc to visualize data. Students should practice all the NCERT Solutions in class 10 chapter 13 thoroughly and clear all their doubts to strengthen their command of this chapter.

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1. NCERT Solutions for Class 10 Maths Chapter 13 Statistics PDF Free Download
2. NCERT Books and NCERT Syllabus
3. Importance of solving NCERT questions for class 10 Chapter 13 Statistics
4. NCERT solutions of class 10 subject-wise
5. NCERT Exemplar solutions: Subject-wise

These NCERT Solutions are created by the expert team at Careers360, keeping in mind the latest syllabus and pattern of CBSE. Students preparing for the Class 10 board exams must go through the Statistics Chapter 13 Class 10 Maths NCERT solutions. This is one of the most important chapters in NCERT books for Class 10 Maths. The NCERT solutions for Class 10 Maths Chapter 13 statistics cover each question of the exercise comprehensively and provide step-by-step solutions. Class 10 maths chapter 13 solutions pdf download from the below link. NCERT Chapter 13 Maths Class 10 Notes are useful for a deep understanding of topics. NCERT Exemplar Solutions for Class 10 Maths Chapter 13 Statistics can be used for deeper knowledge and practice.

NCERT Solutions for Class 10 Maths Chapter 13 Statistics PDF Free Download

Download PDFNCERT Solutions For Statistics Class 10 Chapter 13 - Important Formulae

Measures of Central Tendency - Mean, Median, and Mode:

Mean:

• Direct Method: The mean, represented as X, can be calculated directly using the formula:

$X=\frac{\sum (f_i{x}_i)}{\sum f_i}$

Where '$f_i$' denotes the frequency of the value '$x_i$'.

• Assumed Mean Method: Alternatively, the mean can be calculated using the Assumed Mean Method:

$X=a+\frac{\sum (f_i{d}_i)}{\sum f_i}$

Where '$a$' is an assumed mean and '$d_i$' is the deviation of each value '$x_i$' from the assumed mean.

• Step Deviation Method: Another approach is the Step Deviation Method:

$X=a+[\frac{\sum (f_i{u}_i)}{\sum f_i}]\times h$

Where '$u_i$' represents the step deviations and '$h$' is the class interval.

Median:

• The median is the central value in a set of observations, and its calculation depends on the number of observations.

• For an odd number of observations,

Median = Value of the $\frac{(n+1)}{2}$th term in the ordered set

• In the case of an even number of observations

Median = average of the values of two middle-terms

Median for grouped data

Median $=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).h$

Where,

l = lower limit of the median class

c.f = Cumulative frequency of preceding class

f = Frequency of median class

h = class interval

Mode:

• The mode represents the value that appears most frequently in a dataset.

• The formula to calculate the mode is:

Mode $=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

Where,

l = lower limit of the modal class

f₁ = frequency of the modal class

f₀ = frequency of the class before the modal class

f₂ = frequency of the class after the modal class

H = Class interval

Empirical formula

Mode = 3 Median - 2 Mean

NCERT Solutions for Class 10 Maths Chapter 13 Statistics (Exercises)

Q1. A survey was conducted by a group of students as a part of their environment awareness program, 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 to find the mean, and why?

Answer:

Mean,

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{162}{20}=8.1$
We used the direct method in this as the values of $x_i$ and $f_i$ are small.

Q2. 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:

Let the assumed mean be a = 550

Mean,

$\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$
$=550+\frac{-240}{50}=550-4.8=545.20$
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20.

Q3. 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:

Mean,

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$\Rightarrow 18=\frac{752+20f}{44+f}$
$\Rightarrow 18(44+f)=(752+20f)$
$\Rightarrow 2f=40$
$\Rightarrow f=20$
Therefore, the missing f = 20.

Q4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer:

Let the assumed mean be a = 75.5

Mean,

$\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$
$=75.5+\frac{12}{30}=75.5+0.4=75.9$
Therefore, the mean heartbeats per minute of these women are 75.9.

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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?

Answer:

Let the assumed mean be a = 57

Mean,

$\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$
$=57+\frac{75}{400}=57+0.1875=57.1875\approx 57.19$
Therefore, the mean number of mangoes kept in a packing box is approx 57.19.

Q6. 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:

Let the assumed mean be a = 225 and h = 50

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=225+\frac{-7}{25}\times 50=225-14=211$
Therefore, the mean daily expenditure on food is Rs. 211.

Q7. To find out the concentration of $S{O}_2$ 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 $S{O}_2$ in the air.

Answer:

Mean,

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{2.96}{30}=0.099$
Therefore, the mean concentration of $S{O}_2$ in the air is 0.099 ppm.

Q8. 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:

Mean,

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{499}{40}=12.475$ $=\frac{499}{40}=12.475\approx 12.48$
Therefore, the mean number of days a student was absent was 12.48 days.

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

Let the assumed mean be a = 75 and h = 10

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=70+\frac{-2}{35}\times 10=70-0.57=69.43$
Therefore, the mean literacy rate is 69.43%.

Q1. 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.

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 23, and hence the modal class = 35-45

Lower limit (l) of modal class = 35, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 23, frequency ( $f_0$ ) of class preceding the modal class = 21, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =35+\left(\frac{23-21}{2(23)-21-14}\right).10 \\ =35+\frac{2}{11}.10 \end{gathered}$$

$=36.8$

Now,

Mean,

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{2830}{80}=35.37$
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.

Q2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 61, and hence, the modal class = 60-80

Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( $f_1$ ) of the modal class = 61, frequency ( $f_0$ ) of class preceding the modal class = 52, frequency ( $f_2$ ) of class succeeding the modal class = 38.

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =60+\left(\frac{61-52}{2(61)-52-38}\right)\times 20 \\ =60+\frac{9}{32}\times 20 \end{gathered}$$

$=65.62$

Thus, the modal lifetime of 225 electrical components is 65.62 hours.

Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 40, and hence the modal class = 1500-2000

Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( $f_1$ ) of the modal class = 40, frequency ( $f_0$ ) of class preceding the modal class = 24, frequency ( $f_2$ ) of class succeeding the modal class = 33.

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =1500+\left(\frac{40-24}{2(40)-24-33}\right).500 \\ =1500+\frac{16}{23}.500 \end{gathered}$$

$=1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=2750+\frac{-35}{200}\times 500=2750-87.5=2662.50$

Thus, the Mean monthly expenditure is Rs. 2662.50.

Q4. 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.

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 10, and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( $f_1$ ) of the modal class = 10 frequency ( $f_0$ ) of class preceding the modal class = 9, frequency ( $f_2$ ) of class succeeding the modal class = 3

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =30+\left(\frac{10-9}{2(10)-9-3}\right).5 \\ =30+\frac{1}{8}.5 \end{gathered}$$

$=30.625$

Thus, the Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=32.5+\frac{-23}{35}\times 5=29.22$

Thus, the Mean of the data is 29.22.

Q5. 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.

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 18, and hence, the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( $f_1$ ) of the modal class = 18 frequency ( $f_0$ ) of class preceding the modal class = 4, frequency ( $f_2$ ) of class succeeding the modal class = 9

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =4000+\left(\frac{18-4}{2(18)-4-9}\right).1000 \\ =4000+\frac{14}{23}.1000 \end{gathered}$$

$=4608.70$

Thus, the Mode of the data is 4608.70.

Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Answer:

The class with the maximum frequency is the modal class.

The maximum frequency is 20, and hence, the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 20 frequency ( $f_0$ ) of class preceding the modal class = 12, frequency ( $f_2$ ) of class succeeding the modal class = 11

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =40+\left(\frac{20-12}{2(20)-12-11}\right).10 \\ =40+\frac{8}{17}.10 \end{gathered}$$

$=44.70$

Thus, the Mode of the data is 44.70.

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them.

Answer:

Let the assumed mean be a = 130 and h = 20

MEDIAN:
$N=68$
$\Rightarrow \frac{N}{2}=34$
$\therefore$ Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125;

c.f. = 22; f = 20; h = 20
$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$$\begin{gathered} =125+\left(\frac{34-22}{20}\right).20 \\ =125+12 \end{gathered}$$

$=137$

Thus, the median of the data is 137

MODE:

The class with the maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( $f_1$ ) of the modal class = 20; frequency ( $f_0$ ) of class preceding the modal class = 13, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =125+\left(\frac{20-13}{2(20)-13-14}\right).20 \\ =125+\frac{7}{13}.20 \end{gathered}$$

$=135.76$

Thus, the Mode of the data is 135.76

MEAN:

Mean,
$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$
$=130+\frac{7}{68}\times 20=137.05$

Thus, the Mean of the data is 137.05.

Q2. If the median of the distribution given below is 28.5, find the values of x and y.

Answer:

$N=60$
$\Rightarrow \frac{N}{2}=30$

Now,
Given median = 28.5, which lies in the class 20-30

Therefore, Median class = 20-30
The frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$$\begin{gathered} ⟹28.5=20+\left(\frac{30-5-x}{20}\right).10 \\ ⟹8.5=\frac{25-x}{2} \\ ⟹25-x=8.5(2) \\ ⟹x=25-17=8 \end{gathered}$$

Also,

$$\begin{gathered} 60=45+x+y \\ ⟹x+y=60-45=15 \\ ⟹y=15-x=15-8(\because x=8) \\ ⟹y=7 \end{gathered}$$

Therefore, the required values are: x = 8 and y = 7.

Q3. A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age if policies are given only to persons aged 18 years onwards but less than 60 years.

Answer:

$N=100$
$\Rightarrow \frac{N}{2}=50$
Therefore, Median class = 35-45
The frequency corresponding to the median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$=35+\left(\frac{50-45}{33}\right).5$

$=35.75$

Thus, the median age is 35.75 years.

Q4. The lengths of 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 to find the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Answer:

The data needs to be converted to continuous classes to find the median since the formula assumes continuous classes.

$N=40$
$\Rightarrow \frac{N}{2}=20$
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$=144.5+\left(\frac{20-17}{12}\right).9$

$=146.75$

Thus, the median length of the leaves is 146.75 mm.

Q5. The following table gives the distribution of the lifetime of 400 neon lamps :

Find the median lifetime of a lamp.

Answer:

$N=400$
$\Rightarrow \frac{N}{2}=200$
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$$\begin{gathered} =3000+\left(\frac{200-130}{86}\right).500 \\ =3000+406.97=3406.97 \end{gathered}$$

Thus, the median lifetime of a lamp is 3406.97 hours.

Q6. 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.

Answer:

MEDIAN:
$N=100$
$\Rightarrow \frac{N}{2}=50$
$\therefore$ Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$

$=7+\left(\frac{50-36}{40}\right)\cdot 3$

$=8.05$

Thus, the median of the data is 8.05

MODE:

The class with the maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( $f_1$ ) of the modal class = 40; frequency ( $f_0$ ) of class preceding the modal class = 30, frequency ( $f_2$ ) of class succeeding the modal class = 16

$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =7+\left(\frac{40-30}{2(40)-30-16}\right).3 \\ =125+\frac{10}{34}.3 \end{gathered}$$

$=7.88$

Thus, Mode of the data is 7.88

MEAN:

Mean,
$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{825}{100}=8.25$

Thus, the Mean of the data is 8.25.

Q7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

MEDIAN:
$N=30$
$\Rightarrow \frac{N}{2}=15$
$\therefore$ Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
$Median=l+\left(\frac{\frac{n}{2}-c.f}{f}\right).W$
$=55+\left(\frac{15-13}{6}\right).5$
$=55+\frac{2}{6}.5$

$=56.67$

Thus, the median weight of the students is 56.67 kg.

NCERT Solutions for Class 10 Maths: Chapter Wise

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, which is very useful for students before strategizing their study plans.
Also, links to some reference books which are important for further studies.

• NCERT Books Class 10 Maths
• NCERT Syllabus Class 10 Maths
• NCERT Books Class 10
• NCERT Syllabus Class 10

Importance of solving NCERT questions for class 10 Chapter 13 Statistics

Statistics is not only a useful chapter in class 10 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving Statistics questions in higher studies or competitive examinations.
Some important facts about solving statistics in class 10 are listed below.

• Students can study strategically at their own pace after accessing Class 10 Maths NCERT Solutions chapter 13. This will boost their confidence to attempt other questions from this chapter.
• Class 10 Maths Chapter 13 NCERT solutions are solved by subject-matter experts and are very reliable at the same time. The solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.
• These solutions will help students manage their time efficiently in this chapter and understand which questions are easier to approach and which are time-consuming. This will be helpful during the exam.
• NCERT solutions for class 10 Maths chapter 13 Statistics is designed to give the students step-by-step solutions for a particular question.

NCERT solutions of class 10 subject-wise

You can find NCERT Solutions for Maths as well as Science through the given links.

• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Science

NCERT Exemplar solutions: Subject-wise

You can find NCERT Exemplar Solutions for Maths as well as Science through the given links.

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science