# NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Ex 14.3)

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**NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Ex 14.3) Exercise 14.3 **

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- It has been established that Mathematics fosters the growth of analytical and critical thinking.
- Students are given the ability to effectively assess, describe, and alter situations.
- Mathematics is sometimes regarded as the language of science.
- Every part of life can benefit from the use of Mathematics. It is regarded as a crucial form of knowledge because of this.

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**NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.3 Free PDF **

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Class 10 Maths Chapter 14 Exercise 14.3 is part of Chapter 14 of Class 10 Mathematics which covers the topic of Statistics. The chapter is divided into different exercises, and Statistics Class 10 Exercise 14.3 is just one of the multiple exercises from this chapter.

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**Class 10 Maths Chapter 14 Exercise 14.3 **

The study of data gathering, analysis, interpretation, presentation, and organisation is known as statistics. In other words, it is a mathematical discipline for gathering and summarising data. Additionally, statistics might be considered a subfield of applied Mathematics. However, uncertainty and variation are two crucial and fundamental concepts in statistics. Only statistical analysis can determine the uncertainty and variation in many sectors. Probability, which is a key concept in statistics, essentially determines these uncertainties. To understand the concepts of this chapter better, students can make use of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and many similar reference materials available on the Extramarks website. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 can help the student greatly in performing well in their examinations.

The chapter teaches some basics of Statistics. The measures of central tendency and dispersion are among the fundamental concepts in statistics. Mean, median, and mode are the core trends, whereas the variance and standard deviation are the dispersions. The mean represents the mean of the observations. When observations are sorted in order, the median value is in the middle. The most common observations in a data set are identified by their mode. The mean represents the mean of the observations. When observations are sorted in order, the median value is in the middle. The most common observations in a data set are identified by the mode. The degree of spread among the data collections is measured by variation. Data dispersion from the mean is measured by the standard deviation. The variance is equal to the square of the standard deviation. To understand all these concepts in-depth and practise the questions based on these, students can make use of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and all the other solutions for this chapter, which are all available on the Extramarks website.

The application of Mathematics in the form of Statistics was once envisioned as the science of the state. The gathering and analysis of data regarding a nation’s economy, military, population, and other factors — is known as mathematical statistics. Mathematical analysis, linear algebra, stochastic analysis, differential equations, and measure-theoretic probability theory are some of the mathematical methods utilised in various analytics. There are essentially two categories of Statistics which are referred to as Descriptive Statistics and Deductive Statistics. Students can take help from tools like the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and revision notes for the same chapter to understand the topics in-depth.

Descriptive Statistics are used to summarise and interpret given data. The summary is created using a population sample and numerous variables, including mean and standard deviation. A set of data can be organised, represented, and explained using Descriptive Statistics by employing graphs, charts, and summary measures. To summarise data and present it in tables or graphs, common methods include histograms, pie charts, bars, and scatter plots. Simply put, descriptive statistics are that. To understand this better, students can take help from the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and other helpful reference materials.

After the data has been gathered, assessed, and summarised, Inferential Statistics is used to explain what the data indicate. Inferential Statistics uses the probability principle to examine if patterns observed in a study sample can be extrapolated to the larger population from which the sample was taken. Inferential Statistics can be used to forecast population sizes in addition to testing hypotheses and examining connections between variables. Inferential statistics are used to draw conclusions and inferences from samples, or to make precise generalisations.Class 10 Maths 14.3 Solutions Question 1

The NCERT books for Mathematics are organised in such a way that the chapters include a wide variety of exercises. This makes it easier for students to focus on one sub-topic at a time and prepare for all the sub-topics separately. This can help the students not get confused between different formulas, equations, etc. The NCERT solutions and other similar tools available on the Extramarks website have also been designed in accordance with this pattern. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 for example, are the solutions only for exercise 14.3. The rest of the exercises from Chapter 14 are also available in a similar format. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and the solutions to all the other exercises can be accessed by students on the Extramarks website.

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**Class 10 Maths 14.3 Solutions Question 2 **

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**Class 10 Chapter 14 Exercise 14.3 Question 3 **

The CBSE is responsible for creating test requirements, organising tests at the end of Class 10, and offering. The Class 10 board examinations are among the most significant tests a student will ever take in their academic career. Students get to choose the subject they want to pursue further study in based on how well they perform on the Class 10 board exams. Students in Class 10 are advised to take their board exams seriously as a result. Students may benefit from using the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 to help them study for their board exams. The school requires Class 10 students to take pre-board tests before the boards in order to get them ready for their board exams. The purpose of the pre-board exams is to familiarise students with the format of the board exam and aid in their preparation. Students can use resources like the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and others to get ready for their pre-boards and boards.

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**Class 10th Exercise 14.3 Question Number 4 **

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**Ex 14.3 Class 10 Maths NCERT Solutions Question 5 **

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**Class 10 Maths Chapter 14 Exercise 14.3 Solutions Question 6 **

The importance of mathematics in human life cannot be overstated. Several lucrative career opportunities, as well as research opportunities, are closely associated with this academic field. There is no doubt that Mathematics is one of the most organised and vital scientific disciplines. India has a long history of brilliant mathematicians who captivated the world with their ingenuity and exceptional abilities. Mathematics is taught as a prominent scientific discipline in the early stages of school education, and students are encouraged to continually strive to improve their understanding of it. Mathematics holds great significance as an academic discipline. A significant amount of human life is influenced by mathematics. In addition to being an academic field, it is also closely associated with a wide range of lucrative career opportunities, as well as research opportunities.

**NCERT Solutions of Class 10 Maths Chapter 14 Question 7 **

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**Key Takeaways of NCERT Solutions Class 10 Maths Chapter 14 Exercise 14.3 free PDF **

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**Q.1** 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.

Monthly consumption (in units) | Number of consumers |

65 – 85
85– 105 105– 125 125–145 145–165 165–185 185–205 |
4
5 13 20 14 8 4 |

**Ans**

xmlns=”http://www.w3.org/1998/Math/MathML”>

Monthly consumption (in units) | Number of consumers (f)_{i} |
x_{i} |
d = _{i}x – 135_{i} |
xmlns=”http://www.w3.org/1998/Math/MathML”> |
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 |

Total | 68 | 7 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

We observe all these three measures are approximately same.

**Q.2 **If the median of the distribution given below is 28.5, find the values of x and y.

Class interval | Frequency |

0 – 10
10– 20 20– 30 30–40 40–50 50–60 |
5
x 20 15 y 5 |

Total | 60 |

**Ans**

We find cumulative frequency of the given data as below.

Class interval | Frequency | Cumulative frequency |

0 – 10 10– 20 20– 30 30–40 40–50 50–60 | 5 x 20 15 y 5 | 5 5 + x 25 + x 40 + x 40 + x + y 45 + x + y |

Total | 60 |

Here,

*n *= 60

or 45 + x + y = 60

or x + y = 15 …(1)

Median of data is given as 28.5 that lies in the interval 20 – 30.

So, median class = 20 – 30 xmlns=”http://www.w3.org/1998/Math/MathML”>

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

Class interval | Frequency |

Below 20
Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60 |
2
6 24 45 78 89 92 98 100 |

**Ans**

xmlns=”http://www.w3.org/1998/Math/MathML”>

Class interval | Number of policy holders (f)_{i} |
Cumulative frequency (cf) |

18 – 20 | 2 | 2 |

20 – 25 | 6 – 2 = 4 | 6 |

25 – 30 | 24 – 6 = 18 | 24 |

30 – 35 | 45 – 24 = 21 | 45 |

35 – 40 | 78 – 45 = 33 | 78 |

40 – 45 | 89 – 78 = 11 | 89 |

45 – 50 | 92 – 89 = 3 | 92 |

50 – 55 | 98 – 92 = 6 | 98 |

55 – 60 | 100 – 98 = 2 | 100 |

Total (n) |
100 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

xmlns=”http://www.w3.org/1998/Math/MathML”>

**Q.4 **The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm) | Number of leaves |

118 – 126
127– 135 136– 144 145–153 154–162 163–171 172–180 |
3
5 9 12 5 4 2 |

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)

**Ans**

$\begin{array}{l}\text{The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from each}\\ \text{of lower boundary and adding 0.5 in each of upper boundary.}\\ \\ \text{We write the class interval, number of leaves (}{\mathrm{f}}_{\mathrm{i}}),\text{and cumulative frequency (}\mathrm{cf}\text{) as below in the following table}\\ \text{on the basis of the given information.}\end{array}$

Class interval | Number of leaves (f)_{i} |
Cumulative frequency (cf) |

117.5 – 126.5 | 3 | 3 |

126.5 – 135.5 | 5 | 8 |

13.5 – 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 |

Total | 40 |

$\begin{array}{l}\text{We have,}\mathrm{n}=40.\\ \therefore \frac{\mathrm{n}}{2}=\frac{40}{2}=20\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=\text{20 is 29}\\ \text{which belongs to interval}144.5-153.5.\\ \therefore \text{Median class}=144.5-153.5\\ \text{Now, we have}\\ \mathrm{l}=144.5,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=17,\\ \text{Frequency of median class (}\mathrm{f})=12,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{Class size (}\mathrm{h})=9\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=144.5+\left(\frac{\frac{40}{2}-17}{12}\right)\times 9\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=146.75\\ \text{Hence, median length of leaves is}146.75\text{mm.}\end{array}$

**Q.5 **The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours) | Number of lamps |

1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000 |
14 56 60 86 74 62 48 |

Find the median life time of a lamp.

**Ans**

xmlns=”http://www.w3.org/1998/Math/MathML”>

Life time (in hours) | Number of lamps (f)_{i} |
Cumulative frequency (cf) |

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 |

Total (n) |
400 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

**Q.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:

Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |

Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |

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.

**Ans**

xmlns=”http://www.w3.org/1998/Math/MathML”>

Number of letters | Number of surnames (f)_{i} |
Cumulative frequency (cf) |

1 – 4 | 6 | 6 |

4 – 7 | 30 | 36 |

7 – 10 | 40 | 76 |

10 – 13 | 16 | 92 |

13 – 16 | 4 | 96 |

16 – 19 | 4 | 100 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

xmlns=”http://www.w3.org/1998/Math/MathML”>

We proceed as below to find *d _{i}, u_{i}, f_{i}u_{i}* .

Number of letters | Number of surnames (f)_{i} |
x_{i} |
d= _{i} x_{i} – a |
xmlns=”http://www.w3.org/1998/Math/MathML”> |
f_{i}u_{i} |

1 – 4 | 6 | 2.5 | -9 | -3 | -18 |

4 – 7 | 30 | 5.5 | -6 | -2 | -60 |

7 – 10 | 40 | 8.5 | -3 | -1 | -40 |

10 – 13 | 16 | 11.5 | 0 | 0 | 0 |

13 – 16 | 4 | 14.5 | 3 | 1 | 4 |

16 – 19 | 4 | 17.5 | 6 | 2 | 8 |

Total (n) | 100 | -106 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

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

Weight (in kg) |
40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |

Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |

**Ans**

xmlns=”http://www.w3.org/1998/Math/MathML”>

Weight (in kg) | Number of students (f)_{i} |
Cumulative frequency (cf ) |

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 |

Total (n) |
30 |

xmlns=”http://www.w3.org/1998/Math/MathML”>

## FAQs (Frequently Asked Questions)

### 1. What national level competitive examinations can Class 10 students appear for apart from the mandatory board examinations?

There are various competitive examinations that the students of Class 10 can appear for. Here is a list of few of the Olympiads Class 10 students can appear in for various subjects:

- Class 10 International Science Olympiad (ISO)
- Class 10 International Maths Olympiad (IMO)
- Class 10 English International Olympiad (EIO)
- Class 10 General Knowledge International Olympiad (GKIO)
- Class 10 International Computer Olympiad (ICO)

The helpful resources available at the Extramarks website like the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 can be of great help in these examinations. To prepare for these Olympiads, students should take advantage of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3 and other related resources.