NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1

Mathematics has always piqued the curiosity of students. Geometry is new to all students and may scare them even if Algebra, Trigonometry, and Statistics are all covered in Mathematics for Classes 9 and 10. While some kids struggle with it, others relish the challenges. Students may find it difficult to finish the NCERT Exercise because they may find it difficult to comprehend the themes on their first try. Despite the fact that learning something new is never simple, once students master it, they’ll feel tremendously satisfied.

Students may solve  Class 10 Maths Ex 14.1 with the aid of the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. Students always begin a new chapter or theory by going through the NCERT book. Students should practise the NCERT solutions because this is how they will grasp the concept of the chapter and its application. Mathematics places a high priority on problem-solving and practise. For students in Class 11 and Class 10, Statistics is a novel subject; it differs from other chapters that students have ever read. Students can use the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 to help them with the Class 10 Maths Chapter 14 Exercise 14.1 problems and doubts.

Notably, students who are interested in employment in engineering, data sciences, data analytics, economics, etc. must possess a specific level of mathematical competence. Students have a variety of possibilities if they want to pursue a more in-depth study of mathematics. To make a sensible choice, it is, however, always advisable to be aware of one’s interests and skills. Before making a decision, students should research a variety of career options.Although each decision students make opens up new possibilities for them,Students will learn about who they are through fresh encounters and discoveries. No matter what they choose to pursue, students need to recognise that trying is always valuable.

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The answers given in the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 have been provided by Extramarks experts to help students with their academics in this chapter of Mathematics in Class 10. The students of Class 10 may find the answers given in the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 difficult to comprehend as they have not studied those concepts before. Students must read through NCERT and prepare notes to make sure they grasp the concepts fully and make the least mistakes. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 are to help students find solutions to any queries they may have after they solve the exercises themselves.

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The information in this chapter has been broken down into a number of subtopics to make it more logically organised and simpler for students to understand. Mean of grouped data, mode of grouped data, median of grouped data and the graphical representation of cumulative frequency distribution are some of the subheadings for Mathematics Chapter 14 Class 10 as provided by the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. These subjects all offer organised and condensed information regarding the chapter’s headers and subheadings. Students will find it simpler to compile the chapter’s notes as a result. Students will learn the fundamentals of statistics through the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1, which will aid them in understanding how  data analytics work

The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 help students understand what the meaning of mean, median and modeand how the data is represented for cumulative frequency distribution. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 provide students with the knowledge of providing answers to various queries related to data analysis like mean. The mean of grouped data can be found using three formulas as mentioned in the chapter. The NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1 also help students to figure out a way to find the median of grouped data. The formula to find mode of grouped data has also been given in the chapter and the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. In addition to this, graphical representation for cumulative frequency distribution is also taught through the NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1. Students can learn a lot through the Chapter Statistics of Class 10. It will help them get a deeper understanding of Statistics and help choose their stream of choice in Class 11.

Access NCERT Solutions Maths Class 10 Chapter 14 – Statistics

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1

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Q.1 A survey was conducted by a group of students as a part of their environment awareness programme, 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.

Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses 1 2 1 5 6 2 3

 

 

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

Ans

Here, we observe that class marks and frequencies are smallquantities.So, we use direct method to compute the mean and proceedas below.

Number of plants Number of houses (fi) xi fixi
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
Total 20 162

 

 

 

 

 

 

 

Mean=x¯=∑fixi∑fi=16220=8.1Therefore, mean number of plants per house is 8.1.

Q.2 Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹) 500-120 520-140 540-160 560-180 580-200
Number of workers 12 14 8 6 10

 

 

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans

We find class mark for each interval by using the followingrelation.xi=Upper class limit + Lower class limit2Class size(h) of the given data=20Here, we take assumed mean(a)=550 and proceed as below.

Daily wages
(in ₹)		 Number of workers (fi) xi di = xi – 550 ui=xi−550h fiui
500 – 520 12 510 -40 -2 -24
520 – 540 14 530 -20 -1 -14
540 – 560 8 550 0 0 0
560 – 580 6 570 20 1 6
580 – 600 10 590 40 2 20
Total 50 -12

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=550+(−1250)×20=545.2

Therefore, mean daily wages of the workers is ₹545.20

Q.3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 18-21 21-23 23-25
Number of children 7 6 9 13 f 5 4

 

 

Ans

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=2Given mean(a)=18We proceed as below to find di, fidi.

Daily pocket allowance (in ₹) Number of children (fi) xi di = xi – 18 fidi
11 – 13 7 12 -6 -42
13 – 15 6 14 -4 -24
15 – 17 9 16 -2 -18
17 – 19 13 18 0 0
19 – 21 f 20 2 2f
21 – 23 5 22 4 20
23 – 25 4 24 6 24
Total ∑fi=44+f 2f – 40

 

 

 

 

 

 

 

Mean=x¯=a+∑fidi∑fior 18=18+2f−4044+for  f=20Therefore, missing frequency f is 20.

Q.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.

Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 7 4 2

 

 

Ans

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=3Here, we take assumed mean(a)=75.5 and proceed as below to find di, ui and fiui.

Number of heartbeats per minute Number of women (fi) xi di = xi – 75.5 ui=xi−75.5h fiui
65-68 2 66.5 -9 -3 -6
68-71 4 69.5 -6 -2 -8
71-74 3 72.5 -3 -1 -3
74-77 8 75.5 0 0 0
77-80 7 78.5 3 1 7
80-83 4 81.5 6 2 8
83-86 2 84.5 9 3 6
Total 30 4

 

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=75.5+(430)×3=75.9Therefore, mean heartbeats per minute is 75.9.

Q.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.

Number of mangoes 50-52 53-55 56-58 59-61 62-64
Number of boxes 15 110 135 115 25

 

 

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

Ans

The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limitof each interval. We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=3Here, we take assumed mean(a)=57 and proceed as below to find di, ui and fiui.

Class interval fi xi di = xi – 57 ui=xi−57h f i u i MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacaWGMbWaaSbaaSqaaiaadMgaaeqaaOGaamyDamaaBaaaleaacaWGPbaabeaaaaa@3D30@
49.5 – 52.5 15 51 -6 -2 -30
52.5 – 55.5 110 54 -3 -1 -110
55.5 – 58.5 135 57 0 0 0
58.5 – 61.5 115 60 3 1 115
61.5 – 64.5 25 63 6 2 50
Total 400 25

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=57+(25400)×3=57.1875≈57.19Therefore, mean number of mangoes is 57.19.We have chosen step deviation method as values of fi, di are bigand there is a common multiple between all di.

Q.6 The table below shows the daily expenditure on food of 25 households in a locality.

Daily Expenditure (in ₹) 100-150 150-200 200-250 250-300 300-350
Number of households 4 5 12 2 2

 

 

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

Ans

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=50Here, we take assumed mean(a)=225 and proceed as below to find di, ui and fiui.

Class interval fi xi di = xi – 225 ui=xi−57h fiui
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
Total -7

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=225+(−725)×50=211Therefore, mean daily expenditure is ₹211.

Q.7 To find out the concentration of SO­2 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.

Concentration of SO­2 (in ppm) Frequency
0.00 – 0.04

0.04– 0.08

0.08 – 0.12

0.12–0.16

0.16–0.20

0.20–0.24

 4

9

9

2

4

2

 

 

 

 

 

 

 

 

Find the mean concentration of SO2 in the air.

Ans

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2   Class size(h) of the given data=0.04Here, we take assumed mean(a)=0.14 and proceed as below to find di, ui and fiui.

Concentration of SO2 (in ppm) Frequency (fi) xi di = xi – 0.14 ui=xi−0.14h

b

 fiui
0.00 – 0.04 4 0.02 -0.12 -3 -12
0.04 – 0.08 9 0.06 -0.08 -2 -18
0.08 – 0.12 9 0.10 -0.04 -1 -9
0.12 – 0.16 2 0.14 0 0 0
0.16 – 0.20 4 0.18 0.04 1 4
0.20 – 0.24 2 0.22 0.08 2 4
Total 30 -31

 

 

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=0.14+(−3130)×0.04                                                     ≈0.099 ppmTherefore, mean concentration of SO2 in the air is 0.099 ppm.

Q.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.

Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40
Number of students 11 10 7 4 4 3 1

 

 

Ans

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Assumed mean(a)=17We proceed as below to find di, fidi.

Number of days Number of students fi xi di = xi – 17 fidi
0 – 6 11 3 -14 -154
6 – 10 10 8 -9 -90
10 – 14 7 12 -5 -35
14 – 20 4 17 0 0
20 – 28 4 24 7 28
28 – 38 3 33 16 48
38 – 40 1 39 22 22
Total 40 -181

 

 

 

 

 

 

 

Mean=x¯=a+∑fidi∑fi                     =17+−18140                    =12.475≈12.48Therefore, mean number of days for which a student was absent is 12.48.

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

Literacy rate (in %) 45-55 55-65 65-75 75-85 85-95
Number of cities 3 10 11 8 3

 

 

Ans

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=10Here, we take assumed mean(a)=70 and proceed as below to find di, ui and fiui.

Literacy rate (in %) Number of cities fi xi di = xi – 70 ui=xi−70h fiui
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
Total 35 -2

 

 

 

 

 

 

Mean=x¯=a+(∑fiui∑fi)h=70+(−235)×10=69.43Therefore, mean literacy rate is 69.43.

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