NCERT Solutions Class 10 Maths Chapter 14

NCERT Solutions for Class 10 Mathematics Chapter 14 Statistics

NCERT Class 10 Mathematics Chapter 14 is all about statistics i.e., analysing numerical data by using various approaches. It helps students understand the hows and whys of collecting & analysing numbers. The chapter has a high weightage in CBSE board exams, therefore, students must practise as many questions as possible to have a thorough understanding of the concepts related to statistics. 

When practising, students should never skip the questions given at the end of Chapter 14 in NCERT textbook. To help students solve all the questions accurately, Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 14. The solutions are prepared by subject matter experts in a simple and easy to understand manner. They have ensured that all the answers are written as per the latest guidelines by CBSE.

NCERT Solutions for Class 10 Mathematics Chapter 14 Statistics – 

Access NCERT Solutions for Class 10 Mathematics Chapter Number 14 – Statistics

NCERT Solutions for Class 10 Mathematics Chapter 14 Statistics

When it comes to subjects like Class 10 Mathematics Chapter 14 Statistics, practise is the key. No matter how bright a student is, they cannot secure high marks without practising. One of the sure-shot ways of ensuring enough practise is solving textbook questions by referring to NCERT Solutions for Class 10 Mathematics Chapter 14 by Extramarks. 

NCERT Solutions for Class 10 Mathematics Chapter 14 have solved answers to all textbook questions. Every solution is very elaborate and prepared  stepwise for better understanding.

NCERT Solutions for Class 10 Mathematics

To help students practise better for Mathematics exam, Extramarks offers NCERT Solutions for Class 10 Mathematics for all the chapters including:

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 14 – Statistics

Chapter 15 – Probability

NCERT Solutions For Class 10 Chapter 14 Mathematics Exercise Wise Marks Distribution

Chapter 14 Statistics is a vital chapter in CBSE Class 10 Board Examination. The entire chapter is divided into four different units or segments, making it easier for students to understand the chapter. 

Exercise 14.1 – Exercise 14.1 of NCERT Class 10 Mathematics Chapter 14 comes with a total of 9 questions explaining mean – the first measure of central tendency. Furthermore, Exercise 14.1 in NCERT Class 10 Mathematics chapter also shows the various methods to calculate the mean.

Once a student finishes the exercise, they will be able to solve any questions related to the calculation of the mean without any problem. For understanding how to solve or cross-checking if the answers of what they’ve solved is right, Class 10 students can tune in to NCERT Solutions Class 10 Mathematics Chapter 14.

Exercise 14.2 – Exercise 14.2 of NCERT Class 10 Mathematics Chapter 14 is a detailed exercise involving calculation of mode of data. In this, the student is required to solve 6 questions each and calculate the mode using data given in the question by applying the formula learned in NCERT Solutions Class 10 Mathematics Chapter 14.

Exercise 14.3 – Exercise 14.3 of NCERT Class 10 Mathematics Chapter 14 has 7 questions, all mandating mode calculation of grouped data.

Exercise 14.4 – – Exercise 14.4 of NCERT Class 10 Mathematics Chapter 14 is the last exercise of this chapter which has just three questions. It is also the exercise that students goof up the most while  – graph plotting! In this exercise, students can expect to learn how to demonstrate meaning, graphically. With NCERT Solutions Class 10 Mathematics Chapter 14 by Extramarks, students will be able to plot graphs with utmost precision and perfection.

Chapter 14: Statistics

NCERT Class 10 Mathematics Chapter 14 is a vital chapter not only in CBSE Class 10 Board examination but also in real life. Statistics is a branch of Mathematics that deals with numbers and their analysis using various methods. There will come a lot of instances in a student’s life when they will be required to use the basics of statistics in their life. As NCERT Class 10 Mathematics Chapter 14 also teaches how to interpret graphical representations of statistics like pie charts, graphics or tabular forms, it will prove to be very beneficial in their real life situation as well. . Students who aspire to enter fields like data mining, data research and analysis can strengthen their fundamentals with NCERT Class 10 Mathematics Chapter 14.

Benefits of NCERT Solutions For Class 10 Mathematics Chapter 14

One of the most obvious benefits of NCERT Solutions For Class 10 Mathematics Chapter 14 by Extramarks is gaining access to good study material prepared by experts. Other benefits are

  • NCERT Solutions Class 10 Mathematics Chapter 14 is a solved guide to textual questions which comes with an in-depth explanation of every statistical concept. This gives Class 10 students the much-needed confidence to solve sums in their Class 10 Board examination without worrying and they need not look for any help elsewhere
  • The content in NCERT Solutions Class 10 Mathematics Chapter 14 as per  the latest Class 10 CBSE syllabus.
  • Every problem in NCERT Class 10 Mathematics Chapter 14’s textbook is explained clearly and in a straightforward language.
  • With NCERT Solutions for Class 10 Mathematics Chapter 14, a student will not only learn the theoretical aspect, but they will also learn the practical aspects and apply them in their real-life situations. .
  • Class 10 Mathematics Chapter 14 NCERT Solutions by Extramarks not only help students with their initial preparation and class assignments, but also help them to revise the chapters right before their tests and examinations.

Related Questions

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

Solution:

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

xi = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the ui and fiui as follows:

Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiui
65-68 2 66.5 -3 -6
68-71 4 69.5 -2 -8
71-74 3 72.5 -1 -3
74-77 8 75.5 0 0
77-80 7 78.5 1 7
80-83 4 81.5 3 8
83-86 2 84.5 3 6
Sum fi= 30 Sum fiui = 4

Mean = x̄ = A + h∑fiui /∑f

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

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

For more such detailed answers to statistical questions, refer to NCERT Solutions Class 10 Mathematics Chapter 14.

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¯=fixifi=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=xi550h
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+(fiuifi)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+fidifior 18=18+2f4044+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=xi75.5h
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+(fiuifi)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=xi57h
 

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+(fiuifi)h=57+(25400)×3=57.187557.19Therefore, 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=xi57h
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+(fiuifi)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=xi0.14h

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+(fiuifi)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+fidifi                     =17+18140                    =12.47512.48Therefore, 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=xi70h
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+(fiuifi)h=70+(235)×10=69.43Therefore, mean literacy rate is 69.43.

Q.10 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.

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

                          Total=80                                        Total=430Mean=x¯=a+fidifi                     =30+43080                    =35.37535.38Therefore, mean of the given data is 35.38 years.Modal class is 3545.l=35, f1=23, f0=21,   f2=14, h=10

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.

While on average the age of a patient admitted to thehospital is 35.38 years

Q.11 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.

From the given data, we havel=60, f1=61, f0=52,   f2=38, h=20Mode=l+(f1f02f1f0f2)×h            =60+(61522×615238)×20            =65.625So, modal lifetime of electrical components is 65.625 hours.

Q.12 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.

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=2750We proceed as below to find di, ui,  fiui.

Expenditure (in ₹) Number of families (fi) xi di = xi – 2750 ui=xi2750h 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

Mean=x¯=a+fiuifi×h                     =2750+35200×500                    =2662.5Therefore, mean monthly expenditure is 2662.5.From the given data, we havel=1500, f1=40, f0=24,   f2=33, h=500Mode=l+(f1f02f1f0f2)×h            =1500+(40242×402433)×500            =1847.8261847.83So, modal monthly expenditure is 1847.83.

Q.13 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.

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  

ui=xi32.5h

b

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

Mean=x¯=a+fiuifi×h                     =32.5+2335×5                    =29.2Interpretation:Average teacher student ratio of the states/UT is 29.2.

Q.14 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.

From the given data, we havel=4000, f1=18, f0=4,   f2=9, h=1000Mode=l+(f1f02f1f0f2)×h            =4000+(1842×1849)×1000            =4608.695Mode of the given data is 4608.7 (approx.)

Q.15 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.

From the given data, we havel=40, f1=20, f0=12,   f2=11, h=10Mode=l+(f1f02f1f0f2)×h            =40+(20122×201211)×10            =44.7Mode of the given data is 44.7.

Q.16 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.

We find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=135We proceed as below to find di, ui,  fiui.

Monthly consumption (in units) Number of consumers (fi) xi di = xi – 135  

ui=xi135h
fiui
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

Mean=x¯=a+fiuifi×h                     =135+768×20                    =137.058Modal class=125145From the given data, we havel=125, f1=20, f0=13,   f2=14, h=20Mode=l+(f1f02f1f0f2)×h            =125+(20132×201314)×20            =135.76We know that 3 Median=Mode+ 2 Meanor Median=135.76+2×137.0583=136.625

We observe all these three measures are approximately same.

Q.17 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

From the given data, we havel=20,Cumulative frequency (cf) of the class preceding themedian class=5+x,Frequency of median class (f)=20,  Class size (h)=10     Median=l+(n2cff)×hor     28.5=20+(6025x20)×10or     x=8On putting this value of x in equation (1), we get        x+y=15or 8+y=15or        y=158=7Hence, values of x and y are 8 and 7 respectively.

Q.18 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.

We write the class interval, number of policy holders (fi),and cumulative frequency (cf) as below in the following tableon the basis of the given information.

Class interval Number of policy holders (fi) 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

We have, n=100.n2=1002=50Cumulative frequency (cf) just greater than  n2=50 is 78 which belongs to interval 3540. Median class=3540Now, we havel=35,Cumulative frequency (cf) of the class preceding themedian class=45,Frequency of median class (f)=33,  

Class size (h)=5We know that     Median=l+(n2cff)×h    Median =35+(10024533)×5or     x=35.76Hence, median age is 35.76 years.

Q.19 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.

The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from eachof lower boundary and adding 0.5 in each of upper boundary.We write the class interval, number of leaves (fi),and cumulative frequency (cf) as below in the following tableon the basis of the given information.

Class interval Number of leaves (fi) 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

We have, n=40.n2=402=20Cumulative frequency (cf) just greater than  n2=20 is 29 which belongs to interval 144.5153.5. Median class=144.5153.5Now, we havel=144.5,Cumulative frequency (cf) of the class preceding themedian class=17,Frequency of median class (f)=12,  Class size (h)=9We know that     Median=l+(n2cff)×h       Median =144.5+(4021712)×9or     Median =146.75Hence, median length of leaves is 146.75 mm.

Q.20 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.

We write the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Life time (in hours) Number of lamps (fi) 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

We have, n=400.n2=4002=200Cumulative frequency (cf) just greater than  n2=200 is 216 which belongs to interval 30003500. Median class=30003500Now, we havel=3000,Cumulative frequency (cf) of the class preceding themedian class=130,Frequency of median class (f)=86,  Class size (h)=500We know that     Median=l+(n2cff)×h       Median =3000+(400213086)×500or     Median =3406.98Hence, median life time of a lamp is 3406.98 hours.

Q.21 100 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.

We find the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Number of letters Number of surnames (fi) 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

Total(n)=100We have, n=100.n2=1002=50Cumulative frequency (cf) just greater than  n2=50 is 76 which belongs to interval 710. Median class=710

Now, we havel=7,Cumulative frequency (cf) of the class preceding themedian class=36,Frequency of median class (f)=40,  Class size (h)=3We know that     Median=l+(n2cff)×h       Median =7+(10023640)×3or     Median =8.05Median number of letters in the surnames is 8.05.Now, we find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=11.5

We proceed as below to find di, ui, fiui .

Number of letters Number of surnames (fi) xi di = xi – a  

ui=xiah
fiui
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

Mean=x¯=a+fiuifi×h                     =11.5+106100×3                    =8.32We know that 3 Median=Mode+2 Mean       3×8.05=Mode+2×8.32or Mode=7.51Hence, mean number of letters in the surnames is 8.32and modal size of the surnames is 7.51.

Q.22 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.

We find the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Weight (in kg) Number of students (fi) 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

We have, n=30.n2=302=15Cumulative frequency (cf) just greater than  n2=15 is 19 which belongs to interval 5560. Median class=5560Now, we havel=55,Cumulative frequency (cf) of the class preceding themedian class=13,Frequency of median class (f)=6,  Class size (h)=5We know that     Median=l+(n2cff)×h       Median =55+(302136)×5or     Median =56.67Hence, median weight of students is 56.67 kg.

Q.23 The following distribution gives the daily income of 50 workers of a factory.

Daily income

(in ₹)

100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10

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

Ans.

We find less than type cumulative frequency table as below.

Daily income (in ₹) Cumulative Frequency
Less than 120 12
Less than 140 12 + 14 = 26
Less than 160 26 + 8 = 34
Less than 180 34 + 6 = 40
Less than 200 40 + 10 = 50

Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw itsogive as following.

Q.24 During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg) Number of students
Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52

0

3

5

9

14

28

32

35

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

Ans.

Following is the given cumulative frequency distributions of less than type

 

Weight (in kg)upper class limits
 

Number of students(Cumulative frequency)
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Now locate 17.5 (half of 35) on y-axis and draw a line parallel to x-axis to cut the graph at a point.
Draw a perpendicular from this point to x-axis. The point where this perpendicular meets the x-axis
determines the median of the data. From the graph we see the median is 46.5.

Now, we mark the point A(17.5, 46.5). So median of the given data is 46.5. We find class intervals with their respective frequencies as the following.

Weight (in kg) Frequency (f) Cumulative Frequency
Less than 38 0 0
38 – 40 3 – 0 = 3 3
40 – 42 5 – 3 = 2 5
42 – 44 9 – 5 = 4 9
44 – 46 14 – 9 = 5 14
46 – 48 28 – 14 = 14 28
48 – 50 32 – 28 = 4 32
50 – 52 35 – 32 = 3 35
Total (n) 35

n2=352=17.5Cumulative frequency just greater that 17.5 is 28 which belongsto class interval 4648. Median class=4648Now, we havel=46,Cumulative frequency (cf) of the class preceding themedian class=14,Frequency of median class (f)=14,  Class size (h)=2

We know that     Median=l+(n2cff)×h       Median =46+(3521414)×2or     Median =46.5So median of the given data is 46.5.Hence, value of the median is verified.

Q.25 The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield

(in kg/ha)

50-55 55-60 60-65 65-70 70-75 75-80
Number of farms 2 8 12 24 38 16

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

Ans.

Following is the cumulative frequency distributions of more than type of the given data..

 

Production yield (in kg/ha)
 

Number of farms(Cumulative frequency)
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, draw the graph by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

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