# NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.2

Class 10 plays a crucial role in the academic journey of every student. This is the year students have to sit for the board examinations. This matriculation mark sheet follows one throughout life. Thus, students tend to invest their endeavours to the fullest during this period so that they may  achievea good rank for themselves.

As a result, to support this learning journey, Extramarks comes to the forefront with its comprehensive and authentic NCERT Solutions for different subjects. These Extramarks Solutions are organised by chapter for Math, Science, English, and other subjects.For example, the NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 available on Extramarks’ website only provide solutions for Statistics Class 10 Exercise 14.2.Under this link, all the concepts and questions related to Exercise 14.2 have been elaborated. By working through these solutions, students may gain a better understanding of what statistics is all about.

## NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Ex 14.2) Exercise 14.2

Extramarks is a one-stop solution for Class 10 students who are looking for authentic solutions for Mathematics NCERT Solutions. For Chapter 14 Statistics, students may approach NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 on Extramarks. These NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 are available in a PDF format. The students may download the same through the Extramarks website.This makes the task of making notes easy for students.Students can print these NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 by downloading the PDF. With multiple revisions, they can polish their concepts related to Statistics. Thus, their chance to secure more marks in the examination increases.

## NCERT Class 10 Maths Exercise 14.2 Solutions – Free PDF Download

Statistics is a chapter that necessitates precision in calculations.If students are aware of the concepts and formulae, they can solve the questions related to Statistics within seconds. These concepts regarding Mean, Median and Mode can be learned in a well-explained manner using NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 on Extramarks. NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 explains the various definitions and methods for answering these questions.. Students can access the benefit of using these NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 by visiting the Extramarks website.

### Exercise 14.2 Class 10 NCERT Solutions

It is true that Math is not a subject that can be neglected during exams. Neglecting it can be detrimental to a student’s  overall rank. Obtaining a high rank in Math is not an easy task for everyone. It is an extremely challenging task and requires enormous dedication and guidance. Getting a proper solution to every question is required to score better in this subject.Even the smallest mistake can put students’ grades at stake. Thus, students who are not clear about even the basics of math can face a lot of problems in the exam..

In Maths, Chapter 14 entitled Statistics is considered one of the more difficult chapters and requires maximum concentration during exams. Statistics is a discipline that involves the study of various methods of collecting, organising, analysing and interpreting  given data. There are two basic types of Statistics: Descriptive statistics and Inferential statistics. Extramarks NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 can provide more detailed information about this chapter.

In Class 10 Chapter 14 Exercise 14.2 is one of the most challenging exercises in the chapter. However, all this confusion can be removed by going through NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2. Extramarks NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2 provide an explanative solution to each and every question in that exercise, making it easier for students to comprehend and master.

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Q.1 The following table shows the ages of the patients admitted in a hospital during a year.

 Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65 Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Ans

xmlns=”http://www.w3.org/1998/Math/MathML”>We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=30We proceed as below to find di, fidi.

 Age (in years) Number of patients (fi) Class mark (xi) di = xi – 30 fidi 5 -15 6 10 -20 -120 15 – 25 11 20 -10 -110 25 – 35 21 30 0 0 35 – 45 23 40 10 230 45 – 55 14 50 20 280 55 – 65 5 60 30 150

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

Total=80

Total=430Mean=x¯=a+fidifi

=30+43080

=35.37535.38

Therefore, mean of the given data is 35.38 years.

Modal class is 3545.l=35, f1=23, f0=21,   f2=14, h=10

xmlns=”http://www.w3.org/1998/Math/MathML”>Mode=l+(f1f02f1f0f2)×h

=35+(23212×232114)×10

=36.8Mode is 36.8

which represents that maximum number of patients admitted in hospital are of 36.8 years.

xmlns=”http://www.w3.org/1998/Math/MathML”>While on average the age of a patient admitted to the hospital is 35.38 years

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

 Lifetimes (in hours) 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Ans

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=60,\text{}{\mathrm{f}}_{1}=61,\text{}{\mathrm{f}}_{0}=52,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}_{2}=38,\text{}\mathrm{h}=20\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_{1}-{\mathrm{f}}_{0}}{2{\mathrm{f}}_{1}-{\mathrm{f}}_{0}-{\mathrm{f}}_{2}}\right)×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60+\left(\frac{61-52}{2×61-52-38}\right)×20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=65.625\\ \text{So, modal lifetime of electrical components is 65.625 hours.}\end{array}$

Q.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.

 Expenditure (in ₹) Number of families 1000 – 1500 1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000 24 40 33 28 30 22 16 7

Ans

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=2750\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

 Expenditure (in ₹) Number of families (fi) xi di = xi – 2750 ${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-2750}{\mathrm{h}}$ fiui 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 Total 200 -35

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2750+\frac{-35}{200}×500\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2662.5\\ \text{Therefore, mean monthly expenditure is}₹2662.5.\\ \text{From the given data, we have}\\ \mathrm{l}=1500,\text{}{\mathrm{f}}_{1}=40,\text{}{\mathrm{f}}_{0}=24,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}_{2}=33,\text{}\mathrm{h}=500\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_{1}-{\mathrm{f}}_{0}}{2{\mathrm{f}}_{1}-{\mathrm{f}}_{0}-{\mathrm{f}}_{2}}\right)×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1500+\left(\frac{40-24}{2×40-24-33}\right)×500\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1847.826\approx 1847.83\\ \text{So, modal monthly expenditure is}₹\text{\hspace{0.17em}}1847.83.\end{array}$

Q.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.

 Number of students per teacher Number of states / U .T. 15– 20 20– 25 25–30 30–35 35–40 40–45 45–50 50–55 3 8 9 10 3 0 0 2

Ans

xmlns=”http://www.w3.org/1998/Math/MathML”>From the given data, we havel=30, f1=10, f0=9,   f2=3, h=5Mode=l+(f1f02f1f0f2)×h

=30+(1092×1093)×5

=30.62530.6Interpretation:Most of the states/UT have a teacher student ratio as 30.6.

Now, we find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=32.5We proceed as below to find di, ui,  fiui.

 Number of students per teacher Number of states/UT (fi) xi di = xi – 32.5 xmlns=”http://www.w3.org/1998/Math/MathML”>ui=xi−32.5hb fiui 15 – 20 3 17.5 -15 -3 -9 20 – 25 8 22.5 -10 -2 -16 25 – 30 9 27.5 -5 -1 -9 30 – 35 10 32.5 0 0 0 35 – 40 3 37.5 5 1 3 40 – 45 0 42.5 10 2 0 45 – 50 0 47.5 15 3 0 50 – 55 2 52.5 20 4 8 Total 35 -23

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

Mean=x¯=a+fiuifi×h

=32.5+2335×5

=29.2Interpretation:Average teacher student ratio of the states/UT is 29.2.

Q.5 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

 Runs scored Number of batsmen 3000 – 4000 4000– 5000 5000– 6000 6000–7000 7000–8000 8000–9000 9000–10000 10000–11000 4 18 9 7 6 3 1 1

Find the mode of the data.

Ans

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=4000,\text{}{\mathrm{f}}_{1}=18,\text{}{\mathrm{f}}_{0}=4,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}_{2}=9,\text{}\mathrm{h}=1000\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_{1}-{\mathrm{f}}_{0}}{2{\mathrm{f}}_{1}-{\mathrm{f}}_{0}-{\mathrm{f}}_{2}}\right)×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4000+\left(\frac{18-4}{2×18-4-9}\right)×1000\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4608.695\\ \text{Mode of the given data is}4608.7 \left(\mathrm{approx}.\right)\end{array}$

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

 Number of cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 7 14 13 12 20 11 15 8

Ans

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=40,\text{}{\mathrm{f}}_{1}=20,\text{}{\mathrm{f}}_{0}=12,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{f}}_{2}=11,\text{}\mathrm{h}=10\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_{1}-{\mathrm{f}}_{0}}{2{\mathrm{f}}_{1}-{\mathrm{f}}_{0}-{\mathrm{f}}_{2}}\right)×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40+\left(\frac{20-12}{2×20-12-11}\right)×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=44.7\\ \therefore \text{Mode of the given data is}44.7.\end{array}$