NCERT solutions for class 10 maths Chapter 13 Statistics

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

Below, you will find the NCERT Class 10 Maths Chapter 13 Statistics question answers explained step by step.

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,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{162}{20} = 8.1$
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,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 550 + \frac{-240}{50} = 550-4.8 = 545.20$
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,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$⇒18 = \frac{752+20f}{44+f}$
$⇒18(44+f) =( 752+20f)$
$⇒ 2f = 40 $
$⇒ f = 20$
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,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$
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,

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

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 225 + \frac{-7}{25}\times50 = 225 -14 = 211$
Therefore, the mean daily expenditure on food is Rs. 211.

Q7: To find out the concentration of $SO_2$ 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 $SO_2$ in the air.

Answer:

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2.96}{30} = 0.099$
Therefore, the mean concentration of $SO_2$ 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,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{499}{40} = 12.475$ $= \frac{499}{40} = 12.475\approx 12.48$
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,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43$
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.

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

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

$= 36.8$

Now,

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2830}{80} = 35.37$
The maximum number of patients is 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.

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

$\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right)\times20 \\ \\ = 60 + \frac{9}{32}\times20$

$= 65.62$

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.

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

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

$= 1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

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

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 2750 + \frac{-35}{200}\times500 = 2750 -87.5 = 2662.50$

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

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

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

$= 30.625$

Thus, the Mode of the data is 30.625

Now,

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

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 32.5 + \frac{-23}{35}\times5= 29.22$

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

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

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

$= 4608.70$

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

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

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

$= 44.70$

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

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

$= 137$

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.

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

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

$= 135.76$

Thus, the Mode of the data is 135.76

MEAN:

Mean,
$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 130 + \frac{7}{68}\times20 = 137.05$

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:

$N= 60$
$⇒\frac{N}{2} = 30$

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

Therefore, Median class = 20-30
The frequency corresponding to the 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

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8$

Also,

$\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7$

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:

$N= 100$
$⇒\frac{N}{2} = 50$
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$
$⇒\frac{N}{2} = 20$
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:

$N= 400$
$⇒\frac{N}{2} = 200$
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

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

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 $
$⇒\frac{N}{2} = 50$
$\therefore$ 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

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

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

$= 7.88$

Thus, the Mode of the data is 7.88

MEAN:

Mean,
$\overline x =\frac{\sum f_ix_i}{\sum f_i}$
$= \frac{825}{100} = 8.25$

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 $
$⇒\frac{N}{2} = 15$
$\therefore$ 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.

Statistics Class 10 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Statistics Class 10 Maths Chapter 13 are provided in the links below.

• Class 10 Maths NCERT Chapter 13 Exercise 13.1
• Class 10 Maths NCERT Chapter 13 Exercise 13.2
• Class 10 Maths NCERT Chapter 13 Exercise 13.3

Class 10 Maths NCERT Chapter 13: Extra Question

Question:

Calculate the mean from the following table.

Answer:

The formula for the mean ($X$) using the method of assumed mean is:
$X = A + \frac{\sum fd}{\sum f}$ where $A$ is assumed mean, $f$ is frequency, and $d$ is deviation from mean.
The assumed mean ($A$) is usually taken as the midpoint of the middle class. In this case, it's 35.
The midpoints of the score ranges are 5, 15, 25, 35, 45, 55, and 65.
The frequencies are 2, 4, 12, 21, 6, 3, and 2.

The formula for the mean ($X$) using the method of assumed mean is:
$X = A + \frac{\sum fd}{\sum f}$
⇒ $X = 35 + \frac{-80}{50} = 33.4$
Hence, the correct answer is 33.4.

Statistics Class 10 Chapter 13: Topics

Students will explore the following topics in NCERT Class 10 Maths Chapter 13 Statistics:

• Introduction
• Mean of Grouped Data
• Mode of Grouped Data
• Median of Grouped Data
• Summary

NCERT 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 the 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

$f_1$ = frequency of the modal class

$f_0$ = frequency of the class before the modal class

$f_2$ = frequency of the class after the modal class

$h$ = Class interval

Empirical formula

Mode = 3 Median - 2 Mean

Why are Class 10 Maths Chapter 13 Statistics question answers important?

Understanding statistics helps us make sense of numbers in real life, such as marks, surveys, or averages. It builds our skills to collect, organise, and study data properly. These Class 10 Maths chapter 13 Statistics question answers help us learn these ideas step by step. Here are some more points on why these question answers are important:

• It helps us find the mean, median, and mode easily, which are used to study data in many situations.
• Students get better at solving real-life problems that involve charts, tables, and grouped data.
• It prepares us for higher classes, where we will use statistics in subjects like economics and science.
• Knowing the Class 10 Maths chapter 13 Statistics question answers makes our basics strong for exams and daily data use.

NCERT Solutions for Class 10 Maths: Chapter Wise

We at Careers360 compiled all the NCERT class 10 Maths solutions in one place for easy student reference. The following links will allow you to access them.

Also, read,

• NCERT Exemplar Class 10 Maths Solutions Statistics
• NCERT Notes Class 10 Maths Statistics

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, which is very useful for students before strategising 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 for Class 10 Science
• NCERT Books Class 10
• NCERT Books for Class 10 English
• NCERT Syllabus Class 10

NCERT Exemplar solutions: Subject-wise

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

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