NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

# NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

Edited By Ramraj Saini | Updated on Mar 26, 2024 10:45 AM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 Statistics are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2024-25. Class 10 Maths ex 14.1 covers the classification of ungrouped and grouped frequency distributions. In the prior class, students should have learned how to visualise data in a variety of graphical styles, such as bar graphs, histograms, and frequency polygons. They must understand concepts like numerical representations of ungrouped data and measures of central tendencies like mean, median, and mode. In NCERT solutions Class 10 Maths chapter 14.1, students will learn to transition from ungrouped to grouped data and study all three measures: mean, median, and mode.

Students will be introduced to new statistical concepts including cumulative frequency, cumulative frequency distribution, cumulative frequency curves, or 'ogives,' among others. We will also come across many statistical terms and their methods like class mark, direct method, assumed method for finding mean, etc. 10th class Maths exercise 14.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.1

Statistics C lass 10 Chapter 14 Statistics Exercise: 14.1

Which method did you use for finding the mean, and why?

 Number of plants Number of houses $f_i$ Classmark $x_i$ $f_ix_i$ 0-2 1 1 1 2-4 2 3 6 4-6 1 5 5 6-8 5 7 35 8-10 6 9 54 10-12 2 11 22 12-14 3 13 39 $\sum f_i$ =20 $\sum f_ix_i$ =162

Mean,

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

Let the assumed mean be a = 550

 Daily Wages Number of workers $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 500-520 12 510 -40 -480 520-540 14 530 -20 -280 540-560 8 550 0 0 560-580 6 570 20 120 580-600 10 590 40 400 $\sum f_i$ =50 $\sum f_ix_i$ =-240

Mean,

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

 Daily pocket allowance Number of children $f_i$ Classmark $x_i$ $f_ix_i$ 11-13 7 12 84 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 $\sum f_i$ =44+f $\sum f_ix_i$ =752+20f

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$\implies 18 = \frac{752+20f}{44+f}$

$\\ \implies 18(44+f) =( 752+20f) \\ \implies 2f = 40 \\ \implies f = 20$
Therefore the missing f = 20

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

Let the assumed mean be a = 75.5

 No. of heartbeats per minute Number of women $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 65-68 2 66.5 -9 -18 68-71 4 69.5 -6 -24 71-74 3 72.5 -3 -9 74-77 8 75.5 0 0 77-80 7 78.5 3 21 80-83 4 81.5 6 24 83-86 2 84.5 9 18 $\sum f_i$ =30 $\sum f_ix_i$ =12

Mean,

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

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Let the assumed mean be a = 57

 Number of mangoes Number of boxes $f_i$ Classmark $x_i$ $d_i = x_i -a$ $f_id_i$ 50-52 15 51 -6 -90 53-55 110 54 -3 -330 56-58 135 57 0 0 59-61 115 60 3 345 62-64 25 63 6 150 $\sum f_i$ =400 $\sum f_ix_i$ =75

Mean,

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

 Daily expenditure in rupees 100-150 150-200 200-250 250-300 300-350 Number of households 4 5 12 2 2

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

 Daily Expenditure Number of households $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 100-150 4 125 -100 -2 -8 150-200 5 175 -50 -1 -5 200-250 12 225 0 0 0 250-300 2 275 50 1 2 300-350 2 325 100 2 4 $\sum f_i$ =25 $\sum f_ix_i$ = -7

Mean,

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

Find the mean concentration of $SO_2$ in the air.

 Class Interval Frequency $f_i$ Classmark $x_i$ $f_ix_i$ 0.00-0.04 4 0.02 0.08 0.04-0.08 9 0.06 0.54 0.08-0.12 9 0.10 0.90 0.12-0.16 2 0.14 0.28 0.16-0.20 4 0.18 0.72 0.20-0.24 2 0.22 0.44 $\sum f_i$ =30 $\sum f_ix_i$ =2.96

Mean,

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

 Number of days Number of Students $f_i$ Classmark $x_i$ $f_ix_i$ 0-6 11 3 33 6-10 10 8 80 10-14 7 12 84 14-20 4 17 68 20-28 4 24 96 28-38 3 33 99 38-40 1 39 39 $\sum f_i$ =40 $\sum f_ix_i$ =499

Mean,

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

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

 Literacy rates Number of cities $f_i$ Classmark $x_i$ $d_i = x_i -a$ $u_i = \frac{d_i}{h}$ $f_iu_i$ 45-55 3 50 -20 -2 -6 55-65 10 60 -10 -1 -10 65-75 11 70 0 0 0 75-85 8 80 10 1 8 85-95 3 90 20 2 6 $\sum f_i$ = 35 $\sum f_ix_i$ = -2

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43$
Therefore, the mean mean literacy rate is 69.43%

## More About NCERT Solutions for Class 10 Maths Exercise 14.1 –

In this exercise we will now go through the definition of some basics terms like mean, median mode etc.

The mean (or average) of observations is the sum of all the observations' values divided by the total number of observations.

Where

- sum of the values of all the observations

- sum of the number of observations

In statistics, the median is the middle value of a set of data when ordered in a certain order. Data or observations might be arranged in ascending or descending order.

The mode is the observation with the greatest frequency in grouped data without class intervals.

Also Read| Statistics Class 10 Notes

## Benefits of NCERT Solutions for Class 10 Maths Exercise 14.1

• Exercise 14.1 Class 10 Maths, is based on Statistics and its uses.
• From Class 10 Maths chapter 14 exercise 14.1 we learn new formulas to calculate the mean and class mark.
• Understanding the concepts from Class 10 Maths chapter 14 exercise 14.1 will allow us to understand the concepts related to statistics of higher class.
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## Key Features For Class 10 Maths Chapter 14 Exercise 14.1

• Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 14.1 class 10, ensuring students understand the methodology.
• Conceptual Understanding: Explanation of statistical concepts involved in the exercises, aiding students in grasping the underlying principles are discussed in class 10 maths ex 14.1.
• Variety of Problems: In class 10 ex 14.1, diverse types of problems, offering a broad spectrum of challenges to reinforce understanding.
• Visual Aids, Diagrams, and Examples: Use of visual representations, diagrams, and real-life examples to simplify complex concepts and aid visual learners.
• Application-Based Problems: In10th class maths exercise 14.1 answers, application-oriented problems that relate statistics to real-world scenarios, fostering practical understanding and problem-solving skills.

Also, See

## NCERT Solutions for Class 10 Subject Wise

1. What is the main concept of Class 10 Maths chapter 14 exercise 14.1?

To find what is the basics of statistic. Definition of mean and its formula.  To learn about different methods of finding mean like Assumed mean method, step deviation method etc. Go through the ex 14.1 class 10 to command these concepts.

2. What do you understand by the mean of a set of data?

The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. Practice class 10 maths ex 14.1 to command these concepts.

3. What is class mark?

The class midpoint (or class mark) is a specific point in the centre of the bins (categories) in a frequency distribution table. Go through exercise 14.1 class 10 maths to get deeper understanding of concepts.

4. Find the mean of the data?

This class 10 ex 14.1 discusses the concept of mean in detail. to find mean of data use formula discussed in this exercise. mean = sum of total data / number of data

5. The centre of a bar in a histogram is known as?

The centre of a bar in a histogram is known as class mark.

6. What will happen to the mean of the data if every data set has increased by the value of 5?

The mean of the data will also increase by 5.

7. A data set has few intervals, one of such intervals is between 20 and 40. Find the class mark of this interval?

The mean of the data will also increase by 5.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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