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Class 10 Mathematics Statistics Worksheets

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• Chapter 14 Statistics

NCERT Solutions For Class 10 Maths Chapter 14-Statistics

Ncert solutions for class 10 maths chapter 14 – cbse free pdf download.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here, which can be downloaded for free, in PDF format. The solutions are prepared by our subject experts who have mastery in Maths. All the solved questions of Statistics are with respect to the latest updates on the CBSE syllabus and guidelines, to help students solve each exercise question and effectively prepare for the CBSE exam.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 14 Statistics

Download most important questions for class 10 maths chapter – 14 statistics.

Using these solutions as a reference tool will be helpful for the students to score good marks. Students can also get the exercise-wise Solutions for Class 10 Maths in all chapters and practise solving the problems.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

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Access Answers to Maths NCERT Class 10 Chapter 14 – Statistics

Exercise 14.1 Page: 270

1. 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 for finding the mean, and why?

To find the mean value, we will use the direct method because the numerical value of f i and x i are small.

Find the midpoint of the given interval using the formula.

Midpoint (x i ) = (upper limit + lower limit)/2

The formula to find the mean is:

Mean = x̄ = ∑f i x i /∑f i

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

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 .

In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 550.

Class interval (h) = 20

So, u i = (x i – a)/h

u i = (x i – 550)/20

Substitute and find the values as follows:

So, the formula to find out the mean is:

Mean = x̄ = a + h(∑f i u i /∑f i ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20

Thus, mean daily wage of the workers = Rs. 545.20

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.

To find out the missing frequency, use the mean formula.

Given, mean x̄ = 18

The mean formula is

Mean = x̄ = ∑f i x i /∑f i = (752 + 20f)/ (44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/ (44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

So, the missing frequency, f = 20.

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

From the given data, let us assume the mean as a = 75.5

x i = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the u i and f i u i as follows:

Mean = x̄ = a + h(∑f i u i /∑f i )

= 75.5 + 3 × (4/30)

= 75.5 + (4/10)

= 75.5 + 0.4

Therefore, the mean heart beats per minute for these women is 75.9

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?

The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.

Here, assumed mean (a) = 57

Here, the step deviation is used because the frequency values are big.

The formula to find out the Mean is:

= 57 + 3(25/400)

= 57 + 0.1875

Therefore, the mean number of mangoes kept in a packing box is 57.19

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.

Let us assume the mean (a) = 225

Class size (h) = 50

= 225 + 50(-7/25)

Therefore, the mean daily expenditure on food is 211.

7. 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.

To find out the mean, first find the midpoint of the given frequencies as follows:

The formula to find out the mean is

= 0.099 ppm

Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.

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.

The mean formula is,

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

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

literacy rate.

In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 70.

Class interval (h) = 10

u i = (x i – 70)/10

So, Mean = x̄ = a + (∑f i u i /∑f i ) × h

= 70 + (-2/35) × 10

Therefore, the mean literacy part = 69.43%

Exercise 14.2 Page: 275

1. The following table shows the ages of the patients admitted to 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.

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 23, so the modal class = 35 – 45,

Lower limit of modal class = l = 35,

class width (h) = 10,

f 1 = 21 and f 2 = 14

The formula to find the mode is

Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

= 35 + (20/11)

= 36.8 years

So the mode of the given data = 36.8 years

Calculation of Mean:

First find the midpoint using the formula, x i = (upper limit +lower limit)/2

= 35.375 years

Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

Determine the modal lifetimes of the components.

From the given data the modal class is 60–80.

Lower limit of modal class = l = 60,

The frequencies are:

f m = 61, f 1 = 52, f 2 = 38 and h = 20

Mode = l+ [(f m – f 1 )/(2f m – f 1 – f 2 )] × h

Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20

Mode = 60 + [(9 × 20)/32]

Mode = 60 + (45/8) = 60 + 5.625

Therefore, modal lifetime of the components = 65.625 hours.

3. 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:

Given data:

Modal class = 1500-2000,

Frequencies:

f m = 40 f 1 = 24, f 2 = 33 and

Mode formula:

Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500

Mode = 1500 + [(16 × 500)/23]

Mode = 1500 + (8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, x i =(upper limit +lower limit)/2

Let us assume a mean, (a) be 2750.

The formula to calculate the mean,

Mean = x̄ = a +(∑f i u i /∑f i ) × h

Substitute the values in the given formula

= 2750 + (-35/200) × 500

= 2750 – 87.50

So, the mean monthly expenditure of the families = Rs. 2662.50

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

Modal class = 30 – 35,

Class width (h) = 5,

f m = 10, f 1 = 9 and f 2 = 3

Mode Formula:

Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5

= 30 + (5/8)

= 30 + 0.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x i =(upper limit +lower limit)/2

= 1022.5/35

= 29.2 (approx)

Therefore, mean = 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.

Modal class = 4000 – 5000,

class width (h) = 1000,

f m = 18, f 1 = 4 and f 2 = 9

Substitute the values

Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.7 (approximately)

Thus, 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 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, f m = 20, f 1 = 12 and f 2 = 11

Mode = l + [(f m – f 1 )/(2f m – f 1 – f 2 )] × h

Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10

= 40 + (80/17)

Thus, the mode of the given data is 44.7 cars.

Exercise 14.3 Page: 287

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

Find the cumulative frequency of the given data as follows:

From the table, it is observed that, N = 68 and hence N/2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, N = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

= 125 + [(34 − 22)/20] × 20

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f m or f 1 = 20, f 0 = 13, f 2 = 14 & h = 20

Mode = l+ [(f 1 – f 0 )/(2f 1 – f 0 – f 2 )] × h

Mode = 125 + [(20 – 13)/ (40 – 13 – 14)] × 20

= 125 + (140/13)

= 125 + 10.77

Therefore, mode = 135.77

Calculate the Mean:

x̄ = a + h (∑f i u i /∑f i ) = 135 + 20 (7/68)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5, find the value of x & y.

Given data, n = 60

Median of the given data = 28.5

Where, N/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

cf = 5 + x,

f = 20 & h = 10

28.5 = 20 + [(30 − 5 − x)/20] × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8.

Now, from cumulative frequency, we can identify the value of x + y as follows:

60 = 45 + x + y

Now, substitute the value of x, to find y

60 = 45 + 8 + y

y = 60 – 53

Therefore, the value of x = 8 and y = 7.

3. The life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Given data: N = 100 and N/2 = 50

Median class = 35-40

Then, l = 35, cf = 45, f = 33 & h = 5

Median = 35 + [(50 – 45)/33] × 5

= 35 + (25/33)

Therefore, the median age = 35.76 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as 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 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

So, the data obtained are:

N = 40 and N/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median = 144.5 + [(20 – 17)/ 12] × 9

= 144.5 + (9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a lifetime of 400 neon lamps.

Find the median lifetime of a lamp.

N = 400 & N/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130,

f = 86 & h = 500

Median = 3000 + [(200 – 130)/86] × 500

= 3000 + (35000/86)

= 3000 + 406.98

Therefore, the median lifetime of the lamps = 3406.98 hours

6. 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.

To calculate median:

N = 100 & N/2 = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

Median = 7 + [(50 – 36)/40] × 3

Median = 7 + (42/40)

Median = 8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f 1 = 40, f 0 = 30, f 2 = 16 & h = 3

Mode = 7 + [(40 – 30)/(2 × 40 – 30 – 16)] × 3

= 7 + (30/34)

Therefore mode = 7.88

Mean = 832/100 = 8.32

Therefore, mean = 8.32

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

Given: N = 30 and N/2= 15

Median class = 55-60

l = 55, C f = 13, f = 6 & h = 5

Median = 55 + [(15 – 13)/6] × 5

= 55 + (10/6)

= 55 + 1.666

Therefore, the median weight of the students = 56.67

Exercise 14.4 Page: 293

1. The following distribution gives the daily income of 50 workers in a factory.

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

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

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

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the median by making a table.

Here, N = 35 and N/2 = 35/2 = 17.5

Median class = 46 – 48

Here, l = 46, h = 2, cf = 14, f = 14

The mode formula is given as:

= 46 + [(17.5 – 14)/ 14] × 2

= 46 + 0.5 = 46.5

Thus, median is verified.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

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

Converting the given distribution to a more than type distribution, we get

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper. The graph obtained is known as more than type ogive curve.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Class 10 Maths Chapter 14, Statistics, is one of the most important chapters present in the textbook. The weightage of this chapter in the CBSE exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4 (it could vary as per question).

Topics covered in Chapter 14, Statistics are as follows:

• Mean of Grouped Data
• Mode of Grouped Data
• Median of Grouped Data
• Graphical Representation of Cumulative Frequency Distribution

List of Exercises in Class 10 Maths Chapter 14 : Exercise 14.1 Solutions 9 Questions ( 9 long) Exercise 14.2 Solutions 6 Questions ( 6 long) Exercise 14.3 Solutions 7 Questions ( 7 long) Exercise 14.4 Solutions 3 Questions ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14 – Statistics are made available for students aiming to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions , such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form, in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE exam.

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

• The solutions for the statistics chapter works as a reference for the students.
• It will help students to score marks against the questions asked from this chapter.
• Students can prepare and do the revision for Chapter 14 with this source.
• The questions of statistics have been solved by subject experts.
• The content of the material is as per the CBSE Syllabus (2023-24) and guidelines.

Statistics can also be understood in a more effective way by using the other solutions which are provided at BYJU’S. The solutions are prepared to help students perform well in the CBSE exams.

• RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics

Disclaimer – 

Dropped Topics –  14.5 Graphical representation of cumulative frequency distribution

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CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics (Published By CBSE)

Case study question bank for cbse class 10 maths chapter 14 - statistics is available here. practice this new format of questions to score good marks in your board exam..

CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics are published by the CBSE board itself. These questions are perfect to acquaint with the new format of the questions and make your board exam preparations. Questions are based on the real life situations. You can easily understand how concepts and logic are used in the case study questions. All the questions are provided with answers.  

Check Case Study Questions for Class 10 Maths Chapter 14 - Statistics

CASE STUDY 1:

COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.

The following tables shows the age distribution of case admitted during a day in two different hospitals

Refer to table 1

1. The average age for which maximum cases occurred is

Answer: c) 36.82

2. The upper limit of modal class is

Answer: d) 45

3. The mean of the given data is

Answer: d) 35.4

Refer to table 2

4. The mode of the given data is

Answer: a) 41.4

5. The median of the given data is

Answer: b) 40.2

Electricity Energy Consumption

CASE STUDY 2:

Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).

Refer to data received from Colony A

1. The median weekly consumption is

a) 12 units

b) 16 units

c) 20 units

d) None of these

Answer: c) 20 units

2. The mean weekly consumption is

a) 19.64 units

b) 22.5 units

c) 26 units

Answer: a) 19.64 units

3. The modal class of the above data is I

Answer: c) 20-30

Refer to data received from Colony B

4. The modal weekly consumption is

a) 38.2 units

b) 43.6 units

d) 32 units

Answer: b) 43.6 units

5. The mean weekly consumption is

a) 15.65 units

b) 32.8 units

c) 38.75 units

d) 48 units

Answer: c) 38.75 units

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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Assignment - Statistics, Class 10 Mathematics PDF Download

VERY SHORT ANSWER TYPE QUESTIONS

1. Which measure of central tendency is given by the x-coordinate of the point of intersection of the "more than ogive" and "less than ogive"?

2. Find the median class of the following data :

 2. The following table gives the distribution of expenditure of different families on education. Find the mean expenditure on education of a family :

Delhi-2004C

3. Find the mean of the following distribution:

4. If the mean of the following data is 18.75 find the value of p :

5. The Arithmetic Mean of the following frequency distribution is 50. Find the value of p :

6. If the mean of the following distribution is 50, find the value of f1 :

7. The mean of the following frequency distribution is 62.8. Find the missing frequency x.

LONG ANSWER TYPE QUESTIONS

1. A survery regarding the heights (in cm) of 50 girls of class x of a school was conducted and the following data was obtained :

Find the mean, median and mode of the above data.

2. Find the mean, mode and median of the following data :

3. Find the mean, median and mode of the following data :

Foreign-2008

4. The following table gives the daily income of 50 workers of a factory :

Find the mean, mode and median of the above data.

5. During the medical check-up of 35 students of a class their weights were recorded as follows:

6. Find the mode, median and mean for the following data :

Foreign-2009

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Statistics Assignment Worksheet Class 10 PDF with Answers

These Statistics Assignment worksheet PDF can be helpful for both teachers and students. Teachers can track their student’s performance in the chapter Statistics Assignment. Students can easily identify their strong points and weak points by solving questions from the worksheet. Accordingly, students can work on both weak points and strong points. 

All students studying in CBSE class 10th, need to practise a lot of questions for the chapter Statistics Assignment. Students can easily practise questions from the Statistics Assignment problems worksheet PDF. By practising a lot of questions, students can improve their confidence level. With the help of confidence level, students can easily cover all the concepts included in the chapter Statistics Assignment. 

Statistics Assignment Worksheets with Solutions

Solutions is the written reply for all questions included in the worksheet. With the help of Statistics Assignment worksheets with solutions, students can solve all doubts regarding questions. Students can have deep learning in the chapter Statistics Assignment by solving all their doubts. By solving doubts, students can also score well in the chapter Statistics Assignment. 

Statistics Assignment Worksheet PDF

Worksheet is a sheet which includes many questions to solve for class 10th students. The Statistics Assignment worksheet PDF provides an opportunity for students to enhance their learning skills. Through these skills, students can easily score well in the chapter Statistics Assignment. Students can solve the portable document format (PDF) of the worksheet from their own comfort zone. 

How to Download the Statistics Assignment Worksheet PDF? 

To solve questions from the Statistics Assignment worksheet PDF, students can easily go through the given steps. Those steps are- 

• Open Selfstudys website.

• Bring the arrow towards CBSE which can be seen in the navigation bar. 

• Drop down menu will appear, select KVS NCERT CBSE Worksheet. 

• A new page will appear, select class 10th from the given list of classes. 
• Select Mathematics from the given list of subjects. Now click the chapter’s name that is Statistics Assignment. 

Features of the Statistics Assignment Worksheet PDF

Before starting to solve questions from the Statistics Assignment problems worksheet PDF, students need to know everything about the worksheet. Those features are- 

• Variety of questions are included:  The Statistics Assignment Maths Worksheet for Class 10 includes varieties of questions. Those varieties of questions are- one mark questions, two mark questions, three mark questions, etc. 
• Solutions are provided:  Doubts regarding each question can be easily solved through the solutions given. Through solving questions, a student's comprehensive skill can be increased. 
• All concepts are covered:  By solving questions from the Statistics Assignment problems worksheet, students can easily cover all the concepts included in the chapter.  
• Created by Expert:  These worksheets are personally created by the subject experts. These Statistics Assignment worksheet pdf are created with proper research. 
• Provides plenty of questions:  The Statistics Assignment worksheet provides plenty of questions to practise. Through good practice, students can get engaged in the learning process. 

Benefits of the Statistics Assignment Worksheet PDF

With the help of Statistics Assignment problems worksheet PDF, students can easily track their performance. This is the most crucial benefit, other than this there are more benefits. Those benefits are- 

• Builds a strong foundation:  Regular solving questions from the worksheet can help students to build a strong foundation. Through the strong foundation, students can score well in the chapter Statistics Assignment. 
• Improves speed and accuracy:  While solving questions from the chapter Statistics Assignment, students need to maintain the speed and accuracy. Speed and accuracy can be easily maintained and improved by solving questions from the Statistics Assignment worksheet PDF. 
• Acts as a guide:  Statistics Assignment worksheets with solutions acts as guide for both the teachers and students. Through the worksheet, teachers can guide their students according to the answers given by them. Students can also analyse themselves with the help of answers and can improve accordingly. 
• Enhances the learning process:  Regular solving of questions from the worksheet can help students enhance their learning process. According to the learning skills, students can easily understand all topics and concepts included in the chapter Statistics Assignment.  
• Improvisation of grades:  Regular solving of questions from the worksheet can help students to improve their marks and grades. With the help of good marks and good grades, students can select their desired field further. 

Tips to Score Good Marks in Statistics Assignment Worksheet

Students are requested to follow some tips to score good marks in the Statistics Assignment worksheet. Those tips are-

• Complete all the concepts:  First and the most crucial step is to understand all the concepts included in the chapter Statistics Assignment.  
• Practise questions:  Next step is to practise questions from the Statistics Assignment problems worksheet. Through this students can identify all types of questions: easy, moderate, difficult, etc.  
• Note down the mistakes:  After practising questions, students need to note down the wrong sums that have been done earlier. 
• Rectify the mistakes:  After noting down the mistakes, students need to rectify all the mistakes made. 
• Maintain a positive attitude:  Students are requested to maintain a positive attitude while solving worksheets. By maintaining a positive attitude, students can improve speed and accuracy while solving the worksheets. 
• Remain focused:  Students need to remain focused while solving questions from the Statistics Assignment problems worksheet pdf. As it helps students to solve the questions as fast as possible. 

When should a student start solving the Statistics Assignment Worksheet PDF?

Students studying in class 10 should start solving worksheets after covering each and every concept included in the chapter. Regular solving questions from the Statistics Assignment worksheet PDF, can help students to have a better understanding of the chapter. Better understanding of the chapter Statistics Assignment can help students to score well in the class 10th board exam. 

Regular solving questions from the Statistics Assignment Worksheet PDF can help students to build a strong foundation for the chapter Statistics Assignment. Strong foundation of the chapter Statistics Assignment can help students to understand further chapters. 

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• Maths Notes Class 10
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• Chapter Wise NCERT Solutions for Class 10 Maths

Chapter 1: Real Numbers

• NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
• Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.1
• Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.2
• Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.3
• Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.4

Chapter 2: Polynomials

• NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
• Class 10 NCERT Solutions- Chapter 2 Polynomials - Exercise 2.1
• Class 10 NCERT Solutions - Chapter 2 Polynomials - Exercise 2.2
• Class 10 NCERT Solutions- Chapter 2 Polynomials - Exercise 2.3
• Class 10 NCERT Solutions - Chapter 2 Polynomials - Exercise 2.4

Chapter 3: Pair of Linear Equations in Two Variables

• NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
• Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1
• Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2
• Class 10 NCERT Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3
• Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4
• Class 10 NCERT Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5
• Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6
• Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7

Chapter 4: Quadratic Equations

• NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equation
• Class 10 NCERT Solutions- Chapter 4 Quadratic Equations - Exercise 4.1
• Class 10 NCERT Solutions- Chapter 4 Quadratic Equations - Exercise 4.2
• Class 10 NCERT Solutions- Chapter 4 Quadratic Equations - Exercise 4.3
• Class 10 NCERT Solutions- Chapter 4 Quadratic Equations - Exercise 4.4

Chapter 5: Arithmetic Progressions

• NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.1
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.2
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.3 | Set 1
• Class 10 NCERT Solutions - Chapter 5 Arithmetic Progressions - Exercise 5.3 | Set 2
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.4

Chapter 6: Triangles

• NCERT Solutions for Class 10 Maths Chapter 6 Triangles
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.1
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.3 | Set 1
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.3 | Set 2
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.4
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.5 | Set 1
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.5 | Set 2
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.6

Chapter 7: Coordinate Geometry

• NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
• Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.1
• Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.2
• Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.3
• Class 10 NCERT Solutions- Chapter 7 Coordinate Geometry - Exercise 7.4

Chapter 8: Introduction to Trigonometry

• NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
• Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.1
• Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.2
• Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.3
• Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry - Exercise 8.4

Chapter 9: Some Applications of Trigonometry

• NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry
• Class 10 NCERT Solutions- Chapter 9 Some Application of Trigonometry - Exercise 9.1 | Set 1
• Class 10 NCERT Solutions- Chapter 9 Some Application of Trigonometry - Exercise 9.1 | Set 2

Chapter 10: Circles

• NCERT Solutions for Class 10 Maths Chapter 10 Circles
• Class 10 NCERT Solutions - Chapter 10 Circles - Exercise 10.1
• Class 10 NCERT Solutions- Chapter 10 Circles - Exercise 10.2

Chapter 11: Constructions

• NCERT Solutions for Class 10 Maths Chapter 11 Constructions
• Class 10 NCERT Solutions- Chapter 11 Constructions - Exercise 11.1
• Class 10 NCERT Solutions - Chapter 11 Constructions - Exercise 11.2

Chapter 12: Areas Related to Circles

• NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
• Class 10 NCERT Solutions - Chapter 12 Areas Related to Circles - Exercise 12.1
• Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles - Exercise 12.2 | Set 1
• Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles - Exercise 12.2 | Set 2
• Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles - Exercise 12.3 | Set 1
• Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles - Exercise 12.3 | Set 2

Chapter 13: Surface Areas and Volumes

• Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.1
• Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.2
• Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.3
• Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.4
• Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.5

Chapter 14: Statistics

Ncert solutions for class 10 maths chapter 14 statistics.

• Class 10 NCERT Solutions - Chapter 14 Statistics - Exercise 14.1
• Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.2
• Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.3
• Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.4

Chapter 15: Probability

• NCERT Solutions for Class 10 Maths Chapter 15 Probability
• Class 10 NCERT Solutions- Chapter 15 Probability - Exercise 15.1 | Set 1
• Class 10 NCERT Solutions- Chapter 15 Probability - Exercise 15.1 | Set 2
• Class 10 NCERT Solutions - Chapter 15 Probability - Exercise 15.2

NCERT Solutions Class 10 Maths Chapter 14 Statistics – This article is a useful resource containing free NCERT Solutions for Class 10 Maths Chapter 14 Statistics. These NCERT solutions have been developed by the subject matter experts at GFG to assist students in easily solving questions related to Statistics from the NCERT textbook.

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics cover all four exercises of the NCERT Class 10 Maths Chapter 14 according to the latest CBSE syllabus 2023-24 and guidelines, which are as follows:

NCERT Class 10 Maths Chapter 14 Statistics will help the students learn important statistical concepts like mean, mode, standard deviation , and the graphical depiction of cumulative frequency distribution.

The solutions to all the problems in this Chapter 14 Statistics exercises from the NCERT textbook have been properly covered in the NCERT Solutions for Class 10 Maths .

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1

Question 1. 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 for finding the mean, and why.

Solution: 

Step 1: Let us find out the Class Mark (x i ) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. Step 3: Now we will apply the general formula to calculate the mean  

Now, Let’s see the detailed solution: 

Now, after creating this table we will be able to find the mean very easily –   = 16 = 8.1 Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.

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..

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20. u i = (x i – A)/h  => u i = (xi – 150)/20 Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

So, the formula to find out the mean is: Mean =             = 150 + (20 × -12/50)             = 150 – 4.8            = 145.20 Thus, mean daily wage of the workers = Rs. 145.20.

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.

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A). Step 3: Now we will apply the general formula to calculate the mean

The mean formula is Mean =            = (752 + 20f)/(44 + f) Now substitute the values and equate to find the missing frequency (f) ⇒ 18 = (752 + 20f)/(44 + f) ⇒ 18(44 + f) = (752 + 20f) ⇒ 792 + 18f = 752 + 20f ⇒ 792 + 18f = 752 + 20f ⇒ 792 – 752 = 20f – 18f ⇒ 40 = 2f ⇒ f = 20 So, the missing frequency, f = 20.

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

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3. d i = (x i – A)  => d i = (x i – 75.5) Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

Mean =  = 75.5 + (12/30) = 75.5 + 2/5 = 75.5 + 0.4 = 75.9 Therefore, the mean heartbeats per minute for these women is 75.9

Question 5. 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.

Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add  0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula  Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3. Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.

Now, Lets see the detailed solution: 

Mean =  = 57 + 3 * (25/400) = 57 + 0.1875 = 57.19 Therefore, the mean number of mangoes kept in a packing box is 57.19

Question 6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50. d i = (x i – A)  => d i = (x i – 225) u i = (x i – A)/h => u i = (x i – 225)/50 Step 3: Now we will apply the Step Deviation Formula to calculate the mean

Mean =  = 225 + 50 (-7/25) = 225 – 14 = 211 Therefore, the mean daily expenditure on food is ₹211

Question 7. 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..

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. Step 3: Now we will apply the general formula to calculate the mean

The formula to find out the mean is Mean =  = 2.96/30 = 0.099 ppm Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.

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.

The mean formula is, Mean =  = 499/40 = 12.48 days Therefore, the mean number of days a student was absent = 12.48.

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula Class Mark = (Upper class Limit + Lower Class Limit)/2 Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10. d i = (x i – A)  => d i = (x i – 70) u i = (x i – A)/h => u i = (x i – 70)/10 Step 3: Now we will apply the Step Deviation Formula to calculate the mean

So,  Mean =  = 70 + (-2/35) × 10 = 69.42 Therefore, the mean literacy rate = 69.42%.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2

Question 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..

The greatest frequency in the given table is 23, so the modal class = 35 – 45, l = 35, Class width = 10, and the frequencies are f m  = 23, f 1  = 21 and f 2  = 14 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  = 35 + (20/11) = 35 + 1.8 = 36.8 Hence, the mode of the given data is 36.8 year Now, we find the mean. So for that first we need to find the midpoint. x i = (upper limit + lower limit)/2 Class Interval Frequency (f i ) Mid-point (x i ) f i x i 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 23 40 920 45-55 14 50 700 55-65 5 60 300   Sum f i  = 80   Sum f i x i  = 2830 Mean =   = ∑f i x i  /∑f i = 2830/80 = 35.37 years

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components..

According to the given question The modal class is 60 – 80 l = 60, and the frequencies are f m  = 61, f 1  = 52, f 2  = 38 and h = 20 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  =  = 60 + 45/8 = 60 + 5.625 Hence, the modal lifetime of the components is 65.625 hours.

Question 3. 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:

According to the question Modal class = 1500-2000, l = 1500,and the frequencies are f m  = 40 f 1  = 24, f 2  = 33 and h = 500 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  =  = 1500 + 8000/23 = 1500 + 347.83 So, the modal monthly expenditure of the families is 1847.83 Rupees  Now, we find the mean. So for that first we need to find the midpoint. x i = (upper limit + lower limit)/2 Let us considered a mean, A be 2750 Class Interval f i x i d i = x i – a u i = d i /h f i u i 1000-1500 24 1250 -1500 -3 -72 1500-2000 40 1750 -1000 -2 -80 2000-2500 33 2250 -500 -1 -33 2500-3000 28 2750 0 0 0 3000-3500 30 3250 500 1 30 3500-4000 22 3750 1000 2 44 4000-4500 16 4250 1500 3 48 4500-5000 7 4750 2000 4 28   f i = 200       f i u i = -35 Mean =  On substituting the values in the given formula =  = 2750 – 87.50 = 2662.50 Hence, the mean monthly expenditure of the families is  2662.50 Rupees

Question 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

According to the question Modal class = 30 – 35, l = 30, Class width (h) = 5, and the frequencies are f m  = 10, f 1  = 9 and f 2  = 3 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  = 30 + 5/8 = 30 + 0.625 = 30.625 Hence, the mode of the given data is 30.625 Now, we find the mean. So for that first we need to find the midpoint. x i = (upper limit + lower limit)/2 Class Interval Frequency (f i ) Mid-point (x i ) f i x i 15-20 3 17.5 52.5 20-25 8 22.5 180.0 25-30 9 27.5 247.5 30-35 10 32.5 325.0 35-40 3 37.5 112.5 40-45 0 42.5 0 45-50 0 47.5 0 50-55 2 52.5 105.5   Sum f i  = 35   Sum f i x i  = 1022.5 Mean =  = 1022.5/35  = 29.2 Hence, the mean is 29.2

Question 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..

According to the question Modal class = 4000 – 5000, l = 4000, class width (h) = 1000, and the frequencies are f m  = 18, f 1  = 4 and f 2  = 9 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  Mode = 4000 + 14000/23 = 4000 + 608.695 = 4608.695 Hence, the mode of the given data is 4608.7 runs

Question 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

According to the question Modal class = 40 – 50, l = 40, Class width (h) = 10, and the frequencies are f m  = 20, f 1  = 12 and f 2  = 11 Now, we find the mode using the given formula Mode =  On substituting the values in the formula, we get Mode =  Mode = 40 + 80/17 = 40 + 4.7 = 44.7 Hence, the mode of the given data is 44.7 cars

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Question 1. the following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. find the median, mean, and mode of the data and compare them..

 Total number of consumer n = 68  n/2 =34 So, the median class is 125-145 with cumulative frequency = 42 Here, l = 125, n = 68, C f = 22, f = 20, h = 20 Now we find the median: Median =    = 125 + 12 = 137 Hence, the median is 137 Now we find the mode: Modal class = 125 – 145, Frequencies are f 1 = 20, f 0 = 13, f 2 = 14 & h = 20 Mode =  On substituting the values in the given formula, we get Mode =  = 125 + 140/13  = 125 + 10.77 = 135.77 Hence, the mode is 135.77 Now we find the mean: Class Interval f i x i d i = x i – a u i = d i /h f i u i 65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12   Sum f i = 68       Sum f i u i = 7 = 135 + 20(7/68) = 137.05 Hence, the mean is 137.05 Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

Question 2. If the median of a distribution given below is 28.5 then, find the value of x & y.

According to the question The total number of observations are n = 60 Median of the given data = 28.5 n/2 = 30   Median class is 20 – 30 with a cumulative frequency = 25 + x Lower limit of median class, l = 20, C f = 5 + x, f = 20 & h = 10 Now we find the median: Median =  On substituting the values in the given formula, we get 28.5 =  8.5 = (25 – x)/2 17 = 25 – x Therefore, x = 8 From the cumulative frequency, we can identify the value of x + y as follows: 60 = 5 + 20 + 15 + 5 + x + y On substituting the values of x, we will find the value of y 60 = 5 + 20 + 15 + 5 + 8 + y y = 60 – 53 y = 7 So the value of a is 8 and y is 7

Question 3. The 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 whose age is 18 years onwards but less than 60 years.

According to the given question the table is  Class interval Frequency Cumulative frequency 15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100 Given data: n = 100 and n/2 = 50 Median class = 35 – 45 Then, l = 35, c f = 45, f = 33 & h = 5 Now we find the median: Median =  On substituting the values in the given formula, we get Median =  = 35 + 5(5/33)  = 35.75 Hence, the median age is 35.75 years.

Question 4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Find the median length of leaves.             .

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit. We get a new table: Class Interval Frequency Cumulative frequency 117.5-126.5 3 3 126.5-135.5 5 8 135.5-144.5 9 17 144.5-153.5 12 29 153.5-162.5 5 34 162.5-171.5 4 38 171.5-180.5 2 40 From the given table n = 40 and n/2 = 20 Median class = 144.5 – 153.5 l = 144.5, cf = 17, f = 12 & h = 9 Now we find the median: Median =  On substituting the values in the given formula, we get Median =  = 144.5 + 9/4  = 146.75 mm Hence, the median length of the leaves is 146.75 mm.

Question 5. The following table gives the distribution of a lifetime of 400 neon lamps.

Find the median lifetime of a lamp..

According to the question Class Interval Frequency Cumulative 1500-2000 14 14 2000-2500 56 70 2500-3000 60 130 3000-3500 86 216 3500-4000 74 290 4000-4500 62 352 4500-5000 48 400 n = 400 and n/2 = 200 Median class = 3000 – 3500 l = 3000, C f = 130, f = 86 & h = 500 Now we find the median: Median =  On substituting the values in the given formula, we get Median =  = 3000 + 35000/86 = 3000 + 406.97 = 3406.97 Hence, the median lifetime of the lamps is 3406.97 hours

Question 6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Determine the number of median letters in the surnames. find the number of mean letters in the surnames and also, find the size of modal in the surnames..

According to the question Class Interval Frequency Cumulative Frequency 1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100 n = 100 and n/2 = 50 Median class = 7 – 10 Therefore, l = 7, C f = 36, f = 40 & h = 3 Now we find the median: Median =  On substituting the values in the given formula, we get Median =  Median = 7 + 42/40 = 8.05 Hence, the median is 8.05 Now we find the mode: Modal class = 7 – 10, Where, l = 7, f 1 = 40, f 0 = 30, f 2 = 16 & h = 3 Mode =  On substituting the values in the given formula, we get Mode =  = 7 + 30/34 = 7.88 Hence, the mode is 7.88 Now we find the mean: Class Interval f i x i f i x i 1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70   Sum f i  = 100   Sum f i x i  = 825 Mean =  = 825/100 = 8.25 Hence, the mean is 8.25

Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

According to the question Class Interval Frequency Cumulative frequency 40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30 n = 30 and n/2 = 15 Median class = 55 – 60 l = 55, C f = 13, f = 6 & h = 5 Now we find the median: Median =  On substituting the values in the given formula, we get Median =  = 55 + 10/6 = 55 + 1.666 = 56.67 Hence, the median weight of the students is 56.67

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Question 1. the following distribution gives the daily income of 50 workers if a factory. convert the distribution above to a less-than-type cumulative frequency distribution and draw its ogive..

According to the question, we convert the given distribution to a less than type cumulative frequency distribution,  Daily income Frequency Cumulative Frequency Less than 120 12 12 Less than 140 14 26 Less than 160 8 34 Less than 180 6 40 Less than 200 10 50 Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve. 

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. hence obtain the median weight from the graph and verify the result by using the formula..

According to the given table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive. Now, locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode: Class interval Number of students(Frequency) Cumulative Frequency Less than 38 0 0 Less than 40 3 – 0 = 3 3 Less than 42 5 – 3 = 2 8 Less than 44 9 – 5 = 4 9 Less than 46 14 – 9 = 5 14 Less than 48 28 – 14 = 14 28 Less than 50 32 – 28 = 4 32 Less than 52 35 – 22 = 3 35 The class 46 – 48 has the maximum frequency, hence, this is the modal class l  = 46, h = 2, f 1 = 14, f 0 = 5 and f 2  = 4 Now we find the mode: Mode =  On substituting the values in the given formula, we get = 46 + 0.95 = 46.95 Hence, the mode is verified.

Question 3. The following tables give the production yield per hectare of wheat of 100 farms of a village.

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

According to the question, we change the distribution to a more than type distribution. Production Yield (kg/ha) Number of farms More than or equal to 50 100 More than or equal to 55 100 – 2 = 98 More than or equal to 60 98 – 8 = 90 More than or equal to 65 90 – 12 = 78 More than or equal to 70 78 – 24 = 54 More than or equal to 75 54 – 38 = 16 Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.

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

• These NCERT solutions are developed by the GFG team, with a focus on students’ benefit.
• These solutions are entirely accurate and can be used by students to prepare for their board exams. 
• Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

Also Check:

NCERT Solutions for Class 10 Maths Chapter 14 Statistics – FAQs

Q1: why is it important to learn statistics in ncert class 10 maths chapter 14.

Statistics help people and groups in resolving issues regardless of the industry, market, economic sector, etc. Any issue where relevant data can be gathered, examined, analysed, and presented so as to work towards an efficient resolution can benefit from the knowledge and assistance provided by Statistics. The possibilities are endless.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics covers topics such as calculation of mean, median, mode, frequency distribution and standard deviation and variance etc.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 14 Statistics help me?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics can help you solve the NCERT exercises in an easy mannner. If you are stuck on a problem you can find its solution here and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 10 Maths Chapter 14 Statistics?

There are 4 exercises in the NCERT Class 10 Maths Chapter 14 Statistics which covers all the important topics and sub-topics.

Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

You can find the NCERT Solutions for Class 10 Maths Chapter 14 Statistics in this article created by our team of experts at GeeksforGeeks.

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Assignments For Class 10 Mathematics Statistics

Assignments for Class 10 Mathematics Statistics have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Statistics from our website as we have provided all Maths Assignment for Class 10 chapter wise free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Statistics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Statistics book and get good marks in class 10 exams.

Question . The median and mode of a frequency distribution are 26 and 29 respectively. Then, the mean is  (A) 27.5 (B) 24.5 (C) 28.4 (D) 25.8

Question . If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be    (A) 7.2 (B) 8.2 (C) 9.2 (D) 10.2

Question . Look at the frequency distribution table given below:      The median of the above distribution is

(A) 56.5 (B) 57.5 (C) 58.5 (D) 59

Question . Mode = ?       

Question . For a symmetrical frequency distribution, we have    (A) mean < mode < median (B) mean > mode > median (C) mean = mode = median (D) mode = 1/2 (mean median)

Question . Look at the cumulative frequency distribution table given below:    Number of families having income range 20000 to 25000 is

(A) 19 (B) 16 (C) 13 (D) 22

Measure of Central Tendency Mathematical Calculations

Where,l= lower limit of the model class, h=size of the class-interval,

Where, lower limit of the model class, size of the class-interval, f 1 = frequency of the model class,f 2 =frequency of the class, preceding the modal class,

f 2 =frequency the modal class,h=class size.

Where, l=lower limit of the median class n=number of observations, c.f=cumulative frequency if the class preceding the median class, f=frequency of the median class, h=class size.

Graphical Representation Cumulative Frequency Graph Less than Ogive More then Ogive All the formulae discussed are for grouped data calculation for ungrouped data have been done in previous class The three measures mean, mode and median are connected by the following relations. Mode = 3 median – 2 mean

Short / Long Answer types Questions :

Question. If the mean of the following data is 20.6, find the value of p.

Solution.      

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18. Find the missing frequency k. 

Solution.   k = 8

Question. Find the mean of the following data:   

Solution.    

Question. The mean of the following frequency distribution is 62.8 and sum of all frequencies is 50. Find the missing frequencies f 1 and f 2 .

Question. The table below shows the daily expenditure on grocery of 25 households in a locality.   

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

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18. Find the missing frequency f. 

Question. 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. 

Question. Find the class-mark of class 25–35. Solution.   Class-mark = 25 + 35/2 = 60/2 = 30

Question. Find the mean of first ten odd natural numbers. Solution.   10

Question. If the mean of the first n natural number is 15, then find n.    Solution.   15 = 1 + 2 + 3 + … + n/2

Question. Find the mean of the following distribution:

Solution.   

Question. Find the class-marks of the classes 10-25 and 35-55. Solution.   Class-mark of class 10 – 25  = 10 + 25/2 = 35/2 = 17.5 Class-mark of class 35–55 = 35 + 55/2 = 90/2 = 45

Question. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are as follows:   

Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km / l. Do you agree with this claim? Solution. 

Hence, mean mileage of car is 14.48 km/litre. So, the manufacturer’s statement is wrong that mileage is 16 km L –1 .     

Question. The arithmetic mean of the following frequency distribution is 53. Find the value of k.   

⇒ 3340 + 70k = 53 (72 + k) ⇒ 3340 + 70k = 3816 + 53k ⇒ 70k – 53k = 3816 – 3340 ⇒ 17k = 476 ⇒ k = 28

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given as follows: 

Determine the mean number of seats occupied over the flights. Solution.    

Question. If the mean of the following distribution is 6.4, then find the value of ‘p’.

Solution. 

Question. Find the arithmetic mean of 1, 2, 3, …, n          Solution.  We know that,   

Question. The following data gives the information on the observed life-times (in hours) of 225 electrical components.   

Determine the mean of the above data. Solution.

PRACTICE EXERCISE

Question. Calculate the median for the following data :   

Solution.  153.4

Question. Draw the ogive for the following frequency distribution :   

Also, find the median using ogive drawn. Solution.  37.5

Question. Find the mean of the following frequency distribution :   

Solution.  48.70

Question. Calculate the mode of the following data :   

Solution.  77.27

Question. Calculate the arithmetic mean of the following frequency distribution : 

Solution.  69.33

Question. The mean of the following frequency table is 50. But the frequencies f 1  and f 2  in class 20-40 and 60-80 are missing. Find the missing frequencies.   

Solution.  f 1  = 28, f 2  = 24

Question. The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:   

Draw the cumulative frequency curve and from it determine the median weight of the potatoes. Solution.  93 gm

Question. The marks of 200 students in a test were recorded as follows : 

Draw the cumulative frequency table and ogive. Use it to estimate the median. Solution.  52.8

Question. The following table gives the distribution of expenditure of different families on education. Find the mean expenditure on education in a family. 

Solution.  Rs. 2662.5

Question. Calculate the modal height from the following table   

Solution.  153.57 cm

Question. Find the value of p if the mean of the following distribution is 7.5.   

Solution.  p = 3

Question. Find the mode of the following distribution :     

Solution.  52

Solution.  25.2

Question. Find the average marks scored by students of a class in a test from the following data : 

Solution.  28 marks

Question. For the following frequency distribution, draw both the types of cumulative frequency curve on the same graph paper and hence find the median.   

Solution.  75

Question. Draw a ‘less than type’ cumulative frequency curve for the following data. 

Hence, estimate the median. Solution.  20

Question. Draw a cumulative frequency curve (less than type) for the following data and find the median from it.   

Solution.  32.6

Question. Find the mean of the following distribution :   

Solution.  19.92

Question. Calculate the mode of the following distribution :     

Solution.  107

Question. The marks obtained by 100 students of a class in an examination are given below :

Draw a cumulative frequency curved by using : (i) ‘less than type’ and (ii) ‘more than type’. Hence find the meadian. Solution.  28.8

Question. Compute the mode of the following data :   

Solution.  85.71

Question. Calculate the mode from the following data :   

Solution.  Rs. 7727.27

Question. Marks scored by 400 students in an examination are as follows :

Draw a ‘more than type’ cumulative frequency curve and from it determine the median. Solution.  57.5

Question. Calculate the mean, the median and the mode of the following distribution :   

Solution.  Mean = 14.96, Median = 15, Mode = 15

Question. The table given below shows the distribution of the daily wages, earned by 160 workers in a building site: 

Draw a cumulative frequency curve by using : (i) ‘less than type’ and (ii) ‘more than type’. Hence, estimate the median wages using graph. Solution.  Rs. 34.73

Question. Find the median of the following frequency distribution : 

Solution.  27

Solution.  55

Question. Calculate the mode for the following frequency distribution : 

Solution.  52.83 kg

Question. Find the mean of the following frequency distribution : 

Solution.  53

Question. Calculate the mean of the following frequency distribution: 

Solution.  196.8

Question. The arithmetic mean of the following frequency distribution is 50. Find the value of p. 

Solution.  p = 28

Solution.  50

Question. Calculate the median income :   

Solution.  Rs. 934.17

Solution.  148.61

Question. Find the median for the following frequency distribution :   

Solution.  35.76 years (approx).

Question. Find the mode for the following frequency distribution :   

Solution.  63.75

Question. Find the average height for the following frequency distribution : 

Solution.  153.45 cm

Solution.  153.64

Question. 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. Solution.  mode = 36.81 years; mean = 35.37 years

Question. The following table gives the marks obtained by 50 students in a class test :   

Find the median marks. Solution.  33

Question. Calculate the mean of the following distribution :   

Solution.  62.545

Question. The median of the following distribution is 35, find the value of a and b.   

Solution.  a = 35, b = 25

Solution.  35

Question. Find the mode for the following distribution table :   

Solution.  44.705

Solution.  12.89

Solution.  116.67

Question. Draw a cumulative frequency curve (more than type) for the following data and hence obtain the median.   

Solution.  35 (approx)

Question. If the median of the distribution given below is 28.5, find the value of x and y. 

Solution.  x = 8, y = 7

Question. If the mean of the distribution is 57.6 and the sum of its observations is 50, find the missing frequencies f 1 and f 2 .   

Solution.  f 1 = 8, f 2  = 10

Solution.  36.357

Question. Compute the mean for the following data :   

Solution.  24

Question. If the mean of the following distribution is 54, find the value of p :

Solution.  p = 11

Question. The wickets taken by a bowler in10 cricket matches are as follows:2,6,4,5,0,2,1,3,2,3 Find the mode of the data. Solution.  2

Question. How one can find median of a frequency distribution graphically. Solution.  OGIVE

Question. What is the mean of first ten prime numbers? Solution.  12.9

Question. What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect? Solution.  MEDIAN

Question. What important information one can get by the abscissa of the point of intersection of the less than type and the more than type cumulative frequency curve of a group data. Solution.  Median

Question. Using the assumed mean method find the mean of the following data.     

Solution.  27.2

Question. If the mode of a data is 45 and mean is 27, then median is .   Solution.  33

Question. Find the median of the following frequency distribution.   

Solution.  167

Question. Given below is the distribution of IQ of the 100 students .Find the median of IQ   

Solution.  106.1

Question. Find the median of the following distribution.   

Solution.  28.5l

Question. A class teacher has the following absentee records of 40 students of a class for the whole term 

Write the above distribution as less than type cumulative frequency distribution Solution.   

Question. Find the mode of the following     

Solution.  MODE =40

Question. Name the key terms used in central tendency Solution.  Mean, Median, Mode

Question. Draw cumulative frequency curve for the following data :   

Hence, determine the median. Solution.  32.5

Solution.  43

Question. Find the mode of the following data:  120, 110, 130, 110, 120, 140, 130, 120, 140, 120  Solution.  Let us first form the frequency table for the given data as given below: 

We observe that the value 120 has the maximum frequency. Hence, the mode or modal values is 120.

Question. Calculate the average daily income (in Rs) of the following data about men working in a company:   

Find the median for the above frequency distribution.    Solution.  n = 120 ⇒ n/2 = 60 Median is average of 60th and 61st observation Median = 5 + 5 /2 = 5 So, Median is 5.

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

Solution.  Take a = 75.5, h = 3

Using the step-deviation method,   

Hence, the mean heart beats per minute are 75.9.

Question. Calculate the median from the following data:   

Question. Find the unknown entries a, b, c, d in the following distribution of heights of students in a class :   

Question. On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table : 

Find mean and median of the above data. Solution. 

a= assumed mean = 12 we know that,   

Question. The data regarding marks obtained by 48 students of a class in a class test is given below. Calculate the modal marks of students.   

Solution.  Modal class is 30 – 35, l =30, f 1 = 25, f 0  = 10, 

= 30 + 2.27 or 32.27 approx.   

Question. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f 1 and f 2 .   

Solution.  Given,    sum of frequency = 50 5 + f 1  + 10 + f 2  + 7 + 8 = 50 f 1  + f 2  = 20 3f 1  + 3f 2  = 60 … (1) [multiply both side by 3] And mean = 62.8

30f 1  + 70f 2  = 3140 – 2060 30f 1  + 70f 2  = 1080 3f 1  + 7f 2  = 108 …(2) [divide it by 10] subtract equation (1) from equation (2) 3f 1  + 7f 2  – 3f 1  – 3f 2  =108 – 60 4f 2  = 48 f 2  =12 Put value of f 2  in equation (1) 3f 1  + 3(12) = 60 f 1  = 24/3 = 8 f 1  = 8, f 2  = 12

Question. From the following frequency distribution, find the median class : 

Question. The arithmetic mean of the following data is 14. Find the value of k. 

Question. Find the mode of the data , using a empirical formula , when it is given that median = 41.25 and mean = 33.75   Solution.  We know , Mode = 3(Median) – 2(Mean) Given, Median = 41.25 and Mean = 33.75 Therefore , Mode = 3 x 41.25 – 2 x 33.75 = 123. 75 – 67.50 = 56.25 So, mode is 56.25.

Question. The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs.18.00. Find out the missing frequency. 

Solution.  Given mean = 18, Let the missing frequency be ‘v’. 

Question. The following table gives the number of children of 150 families in a village 

Find the average number of children per family.    Solution.  Let the assumed mean (A) = 2 

Average number of children per family = 352/150 = 2.35(approx)

Question. Find the mode of the following data:  25,16,19, 48,19, 20,34,15, 19, 20, 21, 24,19, 16, 22, 16 ,18, 20,16, 19 ……. Solution.  Given, data 25,16,19, 48,19, 20,34,15, 19, 20, 21, 24,19, 16, 22, 16 ,18, 20,16,19

We observe that the value 19 has the maximum frequency i.e. 5.    Therefore, mode of the given data is 19.

Question. The arithmetic mean of the following data is 25, find the value of k.

Solution.  

Given mean = 25 sum /N = 25 15k + 390 = 25k + 350 25k – 15k = 40 10k = 40 k = 4

Question. Find the value of p for the following distribution whose mean is 16.6.

Question. Calculate the median from the following data:

Question. Draw an ogive to represent the following frequency distribution:

Solution.  The given frequency distribution is not continuous, so we will first make it continuous and then prepare the cumulative frequency:

Plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5,26) by taking the upper class limit over the x-axis and cumulative frequency over the y-axis.

Question. 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. Solution.  We may find class marks for each interval by using the relation

Class size of this data = 0.04 let a = 0.14

Question. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0,1,2,3,4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss

Solution.  Let the assumed mean (A) = 2

Mean number of per toss = 2 + 470/1000 = 2 + 0.47 = 2.47

Question. If ∑fi = 11, ∑fixi = 2p + 52 and the mean of any distribution is 6, find the value of p.  Solution.

Question. Compare the modal ages of two groups of students appearing for an entrance test:

For Group A: Here the maximum frequency is 78, then the corresponding class 18 – 20 is modal class.

For group B: Here the maximum frequency is 89, then the corresponding class 18 – 20 is the modal class.

Hence the modal age for the Group A is higher than that for Group B.

Question. The following distribution gives the daily income of 50 workers of a factory :

Convert the distribution to a ‘less than type’ cumulative frequency distribution and draw its ogive. Hence obtain the median of daily income.

Hence, Median of daily income = Rs 345

Question. The mean of a set of numbers is . If each number is multiplied by k, then find the mean of the new set.  Solution.

Question. Find the value of x, if the mode of the following data is 25: 15, 20, 25,18,14,15, 25,15,18,16, 20, 25, 20, x, 18 Solution.  The frequency table of the given data is as given below:

It is given that the mode of the given date is 25. So, it must have the maximum frequency. That is possible only when x =25. Hence, x = 25.

Question. The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below:

If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25. Solution.  The frequency table is as follows:

The frequency corresponding to the class 20 – 25 is 4. 12

Question. From the following information, construct less than and more than Ogive and find out median from it.

Question. A set of numbers consists of four 5’s, six 7’s, ten 9’s, eleven 12’s, three 13’s, two 14’s. Find the mode of this set of numbers. Solution.  Mode = 12

Since, It has highest frequency

Assignments for Class 10 Mathematics Statistics as per CBSE NCERT pattern

All students studying in Grade 10 Mathematics Statistics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Mathematics Statistics exam for standard 10. We have made sure that all topics given in your textbook for Mathematics Statistics which is suggested in Class 10 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

Some advantages of Free Assignments for Class 10 Mathematics Statistics

• Solving Assignments for Mathematics Statistics Class 10 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
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All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 10 Mathematics Statistics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 10 Mathematics Statistics Assignments PDF.

Free PDF download of Statics Class 10 Maths Assignment Chapter wise pdf created by master educators from the latest syllabus of CBSE Boards. By practicing these Maths Assignment for Class 10 Chapter Wise PDF will help you to score more marks in your CBSE Board Examinations. We also give free NCERT Solutions and other study materials for students to make their preparation better.

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Maths assignment class viii | quadrilateral, math assignment ch-14 class x | statistics.

Chapter: 14(Statistics) Formulas used to solve the Assignment on statistics

Range =  Highest value - Lowest value

Class size =  Upper limit - Lower limit

MODE of Data in Statistics

W here l = lower limit of the modal class,

 h = size of the class interval 

  f 1  = frequency of the modal class

f 0  = frequency of the class preceding the modal class,

f 2  = frequency of the class succeeding the modal class.

Median of Data in Statistics

Where l= lower limit of median class                           

N = Sum of all observations

f = frequency of the median class.       

Cf= cumulative frequency of class  preceding the median class

 h= height of  median class

Empirical Formula of Data in Statistics

      Mode= 3 Median  - 2 Mean

Empirical formula does not provide the same result in different situations. So, students should use it only when asked otherwise avoid it.

* If median class is the first class –interval then cumulative frequency (Cf) of the preceding class interval should be taken as zero.

* For finding Mean class interval need not to be continuous.

* For finding Mode and Median class interval should be continuous. 

* For making class interval continuous we should subtract 0.5 from the lower limits and add 0.5 to the upper limits of all class intervals. 

Assignment on Statistics Solve the following questions

Problems based on mean.

Do we use here step-deviation method ? Why or Why not   

No, we cannot use step deviation method here because height of the class intervals are not equal. So we can solve this question either by Direct Method or by Assumed mean method

PROBLEMS BASED ON MODE  

QUESTION 10. For the given data find mean & mode.              

Ans : Mean=26 & Mode=22.85

PROBLEMS BASED ON MEDIAN

QUESTION 15. Find median.         Ans: 40

Ans:  f 1 =35 &  f 2 = 25

MISCELLANEOUS PROBLEMS ON MEAN, MODE, MEDIAN  

QUESTION  21. 

If mode = 24.5 and mean is 29.75 then find median by using empirical formula.    [Ans: 28]

QUESTION 23. Calculate Mean, mode, median.     Ans : Mean = 51.63, Mode = 53.33, Median = 52

Marks no of students more than 0 80 more than 10 76 more than 20 71 more than 30 65 more than 40 57 more than 50 43 more than 60 28 more than 70 15 more than 80 10 more than 90 8 more than 100 0, higher order thinking skills (hots).

Answer:  f 1  = 28,  f 2  = 32 and  f 3  = 24

Solution Hint:   Use direct method to solve this question

For easy calculation use step deviation method and take assume mean (a) in front of f 1

QUESTIONS DELETED FROM CBSE SYLLABUS

QUESTION 25. Find median by using less than ogive & actual calculation.                               

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Statistics and probability

Unit 1: analyzing categorical data, unit 2: displaying and comparing quantitative data, unit 3: summarizing quantitative data, unit 4: modeling data distributions, unit 5: exploring bivariate numerical data, unit 6: study design, unit 7: probability, unit 8: counting, permutations, and combinations, unit 9: random variables, unit 10: sampling distributions, unit 11: confidence intervals, unit 12: significance tests (hypothesis testing), unit 13: two-sample inference for the difference between groups, unit 14: inference for categorical data (chi-square tests), unit 15: advanced regression (inference and transforming), unit 16: analysis of variance (anova).