NCERT Solutions Class 10 Math Chapter 14 Statistics

NCERT Solutions Class 10 Math Chapter 14 Statistics

NCERT Solutions Class 10 Math Chapter 14 Statistics: National Council of Educational Research and Training Class 10 Math Chapter 14 Solutions – Statistics. NCERT Solutions Class 10 Math Chapter 14 PDF Download.

 

NCERT Solutions Class 10 Math Chapter 14: Overview

Board NCERT
Class 10
Subject Math
Chapter 14
Chapter Name Statistics
Topic Exercise Solutions

 

NCERT Solutions Class 10 Math

Chapter 14 – Statistics

 

Exercise 14.1

 

(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?

Solution:

Number of plants Number or houses (fi) Class Mark (xi) Fi xi
0 – 2 1 0+2/2 = 1 1 × 1 = 1
2 – 4 2 2+4/2 = 3 2 × 3 = 6
4 – 6 1 4+6/2 = 5 1 × 5 = 5
6 – 8 5 6+8/2 = 7 5 × 7 = 35
8 – 10 6 8+10/2 = 9 6 × 9 = 54
10 – 12 2 10+12/2 = 11 2 × 11 = 22
12 – 14 3 12+14/2 = 13 3 × 13 = 39
  Σfi = 20   Σfi xi = 162

 

Mean (x̄) = Σfi xi/Σfi

x̄ = 162/20 = 8.1

∴ Mean = 8.1

We used the method as fi xi. Because it is easier to calculate and sum.

 

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

Daily wages (in ) 500 – 520 520 – 540 540 – 560 560 – 580 580 – 600
Number of workers 12 14 8 6 10

 

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

Solution:

Daily usage (in ₹) Number of workers (fi) xi di = xi – a fi di
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
  Σfi = 50   Σfi di = 240

 

Mean (x̄) = a + Σfi di/Σfi = 550 + (- 240/50)

= 550 – 4.8 = 545.2

∴ The mean daily wages of the factory is ₹545.2.

 

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

Daily Pocket allowance (in ₹) 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25
Number of Children 7 6 9 13 f 5 4

 

Solution:

Daily Pocket allowance (in ₹) Number of children (fi) Class Mark (xi) di = xi – a fi di
11 – 13 7 12 -6 -42
13 – 15 6 14 -4 -24
15 – 17 9 16 -2 -18
17 – 19 13 18 = a 0 0
19 – 21 F 20 2 2f
21 – 23 5 22 4 20
23 – 25 4 24 6 24
  Σfi = 44 + f   Σfidi = 2f – 40

 

Mean (x̄) = a + Σfi di/Σfi

= 18 + 2f – 40/44 + f

∴ Mean (x̄) = 18

18 + 2f-40/44+ f = 18

Or, 2f – 40/44 + f = 40

Or, 2f – 40 = 0 or, 2f = 40

Or, f = 20

 

(4) Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and 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 o women 2 4 3 8 7 4 2

 

Solution:

Number of heart beats per minute Number o women (fi) Class Mark (xi) Xi – a Ui = xi – a/h

(h = 3)

Fi ui
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 = a 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
  Σfi = 30   Σfi = 4

 

Mean (x̄) = a + h × Σfiui/Σfi

Where a = 75.5 h (Class interval) = 3

∴ x̄ = 75.5 + 3× 4/30

= 75.5 + 0.4

= 75.9

∴ The mean heart beats per minute for these women is 75.9

 

(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?

Solution:

No. of mangoes Number of boxes (fi) Number of mangoes Class mark (xi) Xi – a Ui = xi-a/h

(h = 3)

Fi ui
50 – 52 15 49.5 – 52.5 51 -6 -2 -30
53 – 55 110 52.5 – 55.5 54 -3 -1 -110
56 – 58 135 55.5 – 58.5 57 = a 0 0 0
59 – 61 115 58.5 – 61.5 60 3 1 115
62 – 64 25 61.5 – 64.5 63 6 2 50
  Σfi = 400   Σfiui = 25

 

Mean (x̄) = a + h × Σfiui/Σfi

Where, a = 57, h = 3

∴ x̄ = 57 + 3 × 255/400

= 57 + 3 × 1/16

= 57 + 0.1875

= 57.1875

= 57.1875

= 57.19

Mean number of boxes is 57.19

We used the method as fiui because it was easier to calculate and sum.

 

(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.

Solution:

Daily expenditure (in ₹) Number of hose holds (fi) Class Mark (xi) Xi – a Ui = xi-a/h

h = 50

fiui
100 – 150 5 125 -100 -2 -8
150 – 200 5 175 -50 -1 -5
200 – 250 12 225 = a 0 0 0
250 – 300 2 275 50 1 2
300 – 350 2 325 100 2 4
  Σfi = 25   Σfi ui = -7

 

Mean (x̄) = a + h Σfiui/Σfi

= 225 + 50 × -7/25

= 225 + (-14) = 225 – 14 = 211

The mean daily expenditure on food is ₹211.

 

(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.

Solution:

Concentration of SO2 (in ppm) Frequency (fi) Class mark (xi) fi xi
0.00 – 0.04 4 0.00+0.04/2 = 0.02 0.08
0.0.4 – 0.08 9 0.04+0.08/2 = 0.06 0.54
0.08 – 0.12 9 0.08+0.12/2 = 0.10 0.90
0.12 – 0.16 2 0.12+0.16/2 = 0.14 0.28
0.16 – 0.20 4 0.16+0.20/2 = 0.18 0.72
0.20 – 0.24 2 0.20+0.24/2 = 0.22 0.44
  Σfi = 30   Σfixi = 2.96

 

∴ Mean = Σfixi/Σfi = 2.96/30 = 0.099

∴ The mean concentration of So2 in the air is 0.99 ppm.

 

(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

 

Solution:

Number of days Number of students (fi) Class Mark (xi) fixi
0 – 6 11 0+6/2 = 3 33
6 – 10 10 6+10/2 = 8 80
10 – 14 2 10+14/2 = 12 84
14 – 20 4 14+20/2 = 17 68
20 – 28 4 20+28/2 = 24 96
28 – 38 3 28+38/2 = 33 99
38 – 40 1 38+40/2 = 39 39
  Σfi = 40   Σfixi = 499

 

Mean (x̄) = Σfixi/Σfi = 499/40 = 12.48

∴ The mean number of days a student was absent is 12.48 days.

 

(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

 

Solution:

Literacy rate (in %) Number of cities (fi) Class Mark (xi) Xi – a Ui = xi-a/h

(h = 10)

Fi ui
45 – 55 3 50 -20 -2 -6
55 – 65 10 60 -10 -1 -10
65 – 75 11 70 = a 0 0 0
75 – 85 8 80 10 1 8
85 – 95 3 90 20 2 6
  Σfi = 35   Σfiui = -2

 

Mean (x̄) = a + h Σfiui/Σfi

= 70 + 10 × -2/35

= 70 – 0.57

= 69.43

∴ The mean literacy rate is 69.43%

 

Exercise – 14.2

 

(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.

Solution:

Age (in years) Number of patients (fi) Class Mark (xi) fixi
5 – 15 6 5+15/2 = 10 6×10 = 60
15 – 25 11 15+25/2 = 20 220
25 – 35 21 25+35/2 = 30 630
35 – 45 23 35+45/2 = 40 920
45 – 55 14 45+55/2 = 50 700
55 – 65 5 55+65/2 = 60 300
  Σfi = 80   Σfixi = 2830

 

Mean (x̄) = Σfixi/Σfi = 2830/80 = 35.37

∴ The mean is 35.37

Age (in years) Number of patients
5 – 15 6
15 – 25 11
25 – 35 21 = f0
35 – 45 23 = f1
45 – 55 14 = f2
55 – 65 5

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

Where, lower limit of modal class (l) = 35

Class interval (h) = 15 – 5 = 10

Frequency of the modal class (f1) = 23

Frequency of the class before modal class (f0) = 21

Frequency of the class after modal class (f2) = 14

∴ Mode = 35 + 23 – 21 / 2 – (23) – 21 – 14 x 10

= 35 + 2 / 46 – 35 x 10

= 35 + 2 / 11 x 10

= 35 + 1.8

= 36.8

∴ Maximum number of patients admitted in the hospital are of the age 36.8 years (mode).

While, average age of a patients admitted to the hospital is 35.37 years (Mean).

 

(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.

Solution:

Life times (in hours) Frequency
0 – 20 10
20 – 40 35
40 – 60 52 = f0
60 – 80 61 = f1
80 – 100 38 = f2
100 – 120 29

 

Mode = L + f1 – f0/2f1 – f0 – f2 × L

Where, L = 60, f1 = 61, f0 = 52, f2 = 38, h = 20

∴ Mode = 60 + 61-52/2×(61) – 52.38 × 20

= 60 + 9/122-90 × 20

= 60 + 9/32 × 20

= 60+5.625 = 65.625

∴ The modal lifetime of components is 65.625 hours.

 

(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

 

Solution:

Expenditure (in ) Number of families
1000 – 1500 24 = f0
1500 – 2000 40 = f1
2000 – 2500 33 = f2
2500 – 3000 28
3000 – 3500 30
3500 – 4000 22
4000 – 4500 16
4500 – 5000 7

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

Where, L = 1500, h = 500, f1 = 40, f2 = 33, f0 = 24

Mode = 1500 + 40 – 24/2×40 – 24- 33 × 500

= 1500 + 16/23 × 500

= 1500 + 347.826

= 1847.83

∴ Modal monthly expenditure or families is ₹1847.83

 

Expenditure (in ) No. of families (fi) Class Mark (xi) Xi – a Ui = xi – a/h

(h = 500)

fi × ui
1000 – 1500 24 1250 – 2000 -4 – 96
1500 – 2000 40 1750 – 1500 -3 – 120
2000 – 2500 33 2250 – 1000 -2 – 66
2500 – 3000 28 2750 – 500 -1 – 28
3000 – 3500 30 3250 0 0 0
3500 – 4000 22 3750 500 1 22
4000 – 4500 16 4250 1000 2 32
4500 – 5000 7 4750 1500 3 21
  Σfi = 200   Σfiui = – 235

 

Mean (x̄) = a + h + Σfiui/Σfi

= 3250 + 500 × -235/200

= 2662.5

∴ Mean expenditure is ₹2662.5

 

(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

 

Solution:

Number of students per teacher Number of states/U.T.
15 – 20 3
20 – 25 8
25 – 30 9 = f0
30 – 35 10 = f1
35 – 40 3 = f2
40 – 45 0
45 – 50 0
50 – 55 2

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

Where = L = 30, f1 = 10, f0 = 9, f2 = 3, h = 5

∴ Mode = 30 + 10-9/2×10-9-3 × 5

= 30 + 1/20+12 × 5

= 30 + 5/8 = 30+0.625

= 30.625

Number of students per teacher Number of states/U.T. Class mark (xi) Xi – a Ui = xi-a/h

(h = 5)

Fi × ui
15 – 20 3 17.5 -20 -4 -12
20 – 25 8 22.5 -15 -3 -24
25 – 30 9 27.5 -10 -2 -18
30 – 35 10 32.5 -5 -1 10
35 – 40 3 37.5 = a 0 0 0
40 – 45 0 42.5 5 1 0
45 – 50 0 47.5 10 2 0
50 – 55 2 52.5 15 3 6
  Σfi = 35   Σfiui = -58

 

∴ Mean = a + h × Σfiui/Σfi

= 37.5 + 5× -58/35

= 37.5 – 8.286

= 29.214

∴ The most states have student teacher ratio = 30.625 (Mode) And average student – teacher ratio is 29.214 (Mean)

 

(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.

Solution:

Runs scored Number of batsmen (fi)
3000 – 4000 4 = f0
4000 – 5000 18 = f1
5000 – 6000 9 = f2
6000 – 7000 7
7000 – 8000 6
8000 – 9000 3
9000 – 10000 1
10000 – 11000 1

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

L = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000

∴ Mode = 4000 + 18-4/2×18-4-9 × 1000

= 4000 + 14/36- 13 × 1000

= 4000 + 14000/23

= 4000 + 608.69

= 4608.69

∴ The mode of the data is 4608.69

 

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

Numbers 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

 

Solution:

Numbers of cars Frequency (fi)
0 – 10 7
10 – 20 14
20 – 30 13
30 – 40 12 = t0
40 – 50 20 = t1
50 – 60 11 = t2
60 – 70 15

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

Where, L = 40, f1 = 20, f0 = 12, f2 = 11, h = 10

∴ Mode = 40 + 20-12/2×20-12-11 × 10

= 40 + 8/40-23 × 10

= 40 + 80/17

= 40 + 4.7 = 44.7

∴ The mode of the data is 44.7

 

Exercise – 14.3

 

(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

 

Solution:

Monthly consumption (in units) Number of consumers

 (fi)

65 – 85 4
85 – 105 5
105 – 125 13 = f0
125 – 145 20 = f1
145 – 165 14 = f2
165 – 185 8
185 – 205 4

 

Mode = L + f1-f0/2f1-f0-f2 × h

Where, L = 125, f1 = 20, f0 = 13, f2 = 14, h = 20

∴ Mode = 125 + 20-13/2×20-13-14 × 20

=125 + 7/40-27 × 20

= 125 + 7/13 × 20

= 125 + 10.77

= 135.77

∴ The data of the mode is 135.77

Monthly consumption (in units) Number of consumers

 (fi)

Cumulative  frequency
65 – 85 4 4
85 – 105 5 4+5 = 9
105 – 125 13 9+13 = 22
125 – 145 20 22+20 = 42
145 – 165 14 42+14 = 56
165 – 185 8 56+8= 64
185 – 205 4 64+4 = 68
  Σfi = 68  

 

Median = L + (N/2 – Cf)/f × h

Here, N/2 = 68/2 = 34

∴ Median class is = 125 – 145

L = 125, Cf = 22

f = 20, h = 20

∴ Median = 125 + 34-22/20 × 20

= 125+12

= 137

∴ The data of the median is 137

Monthly consumption (in units) Number of consumers

 (fi)

Class Marks Xi – a Ui = Xi – a/h

(h = 20)

Fi ui
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 = a 0 0 0
145 – 165 14 155 20 1 14
165 – 185 8 175 40 2 16
185 – 205 4 195 60 3 12
  Σfi = 68   Σfiui = 7

 

Mean = a + h × Σfi ui/Σfi

= 135 + 20 × 7/68

= 135 + 2.05 = 137.05

∴ The data of the mean is 137.05

∴ Mean = 137.05, median = 137, mode = 135.77

∴ Mean median and mode are approximately same.

 

(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

 

Solution:

Class Interval Frequency (fi) Cumulative frequency
0 – 10 5 5
10 – 20 x 5 + x
20 – 30 20 25 + x
30 – 40 15 40 + x
40 – 50 Y 40 + x + y
50 – 60 5 45 + x + y
Total Σfi = 60  

 

Median = l + (N/2 – Cf)/f × h

Here, N/2 = 60/2 = 30

∴ L = 20

h = 10

Cf = 5 + x

f = 20

Since median = 28.5

∴ Median class is = 20 – 30

∴ Median = 20 + 30 – (5+x)/20 × 10

Or, 28.5 = 20 + 30-5-x/2

Or, 28.5 = 20 + 25-x/2

Or, 25-x/2 = 8.5

Or, 25 – x = 8.5×2

Or, 25 – x = 17

Or, – x = 17 – 25

Or, – x = -8

Or, x = 8

Σfi = 45 + x + y

60 = 45 + 8 + y

Or, y = 60 – 53 = 7

∴ x = 8

y = 7

 

(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.

Age (in years) Number of policy holders
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

 

Solution:

Age (in years) Number of policy holders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

 

Age (in years) Number of policy holders
Below 20 2
Below 25 6 – 2 = 4
Below 30 24 – 6 = 18
Below 35 45 – 24 = 21
Below 40 78 – 45 = 33
Below 45 89 – 78 = 11
Below 50 92 – 89 = 3
Below 55 98 – 92 = 6
Below 60 100 – 98 = 2

 

Age (in years) Number of policy holders Cumulative Frequency
0 – 20 2 2
20 – 25 4 2 + 4 = 6
25 – 30 18 6 + 18 = 24
30 – 35 21 24 + 21 = 45
35 – 40 33 45 + 33 = 78
40 – 45 11 78 + 11 = 89
45 – 50 3 89 + 3 = 92
50 – 55 6 92 + 6 = 98
55 – 60 2 98 + 2 = 100
  Σfi = 100  

 

Median = L + (N/2 – cf)/f × h

Here, N/2 = 100/2 = 50

∴ Median class is 35 – 40

L = 35, h = 25-20 = 5, cf = 45, f = 33

Median = 35 + 50-45/33 × 5

= 35 + 5/33 x 5s

= 35 + 0.76 = 35.76

∴ The median age is 35.76 years.

 

(4) The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, 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.

Solution:

Length (in mm) Length (in mm) Number or leaves (fi) Cumulative frequency
118 – 126 117.5 – 126.5 3 3
127 – 135 126.5 – 135.5 5 3+5 = 8
136 – 144 135.5 – 144.5 9 8+9 = 17
145 – 153 144.5 – 153.5 12 17+12 = 29
154 – 162 153.5 – 162.5 5 29+5 = 34
163 – 171 162.5 – 171.5 4 34+4 = 38
172 – 180 171.5 – 180.5 2 38+2 = 4
  Σfi = 40  

 

Median = L + (N/2 – cf) × h

Here n/2 = 40/2 = 20

∴ The median class is 144.5 – 153.5

L = 144.5, cf = 15, f = 12, h = 9

∴ Median = 144.5 + 20-17/12 × 9

= 144.5 + 3/12 × 9

= 144.5 + 2.25

= 146.75

∴ The median length of the leaves is 146.75 mm

 

(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.

Solution:

Life Time (in hours) Number of lamps (fi) Cumulative frequency
1500 – 2000 14 14
2000 – 2500 56 14+56 = 70
2500 – 3000 60 70+60 = 130
3000 – 3500 86 130+86 = 216
3500 – 4000 74 216+74 = 290
4000 – 4500 62 290+62 = 352
4500 – 5000 48 352+48 = 400
  Σfi = 400  

 

∴ Median = L + (N/2 – cf) × h

Here, N/2 400/2 = 200

Median Class is 3000 – 3500

L = 3000, h = 2000 – 1500 = 500, cf = 130, f = 86

∴ Median = 3000 + 200-130/86 × 500

= 3000 + 70/86 × 500

= 3000 + 406.98

= 3406.98

∴ The Median life time of a lamp is 3406.98 hours.

 

(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.

Solution:

Number of letters Number of surnames (fi) Cumulative frequency
1 – 4 6 6
4 – 7 30 6 + 30 = 36
7 – 10 40 36 + 40 = 76
10 – 13 16 76 + 16 = 92
13 – 16 4 92 + 4 = 96
16 – 19 4 96 + 4 = 100
  Σfi = 100  

 

Median = L + (N/2 – cf)/f × h

Here, N/2 = 100/2 = 50

∴ Median class is 7 – 10

L = 7, h = 4-1 = 3, cf = 36, f = 40

Median = 7 + 50-36/40 × 3

= 7 + 14/40 × 3 = 7 + 1.05 = 8.05

∴ The median number of letters 8.05

Number of letters Number of surnames (fi) Class Mark (xi) Xi – a Xi = xi-a/h

(h = 3)

Fi ui
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 = a 0 0 0
13 – 16 4 14.5 3 1 4
16 – 19 4 17.5 6 2 8
  Σfi = 100   Σfi ui = -106

 

Mean = a + h × Σfiui/Σfi

Where, a = 11.5, h = 3, Σfiui = -106, Σfi = 100

∴ Mean = 11.5 + 3 × -106/100

= 11.5 – 3.18 = 8.32

∴ The mean is 8.32

Number of letters Number of surnames (fi)
1 – 4 6
4 – 7 30 = f0
7 – 10 40 = f1
10 – 13 16 = f2
13 – 16 4
16 – 19 4

 

Mode = L + f1 – f0/2f1 – f0 – f2 × h

Where, L = 7, h = 4 – 1 = 3 f0 = 30, f1 = 40, f2 = 16

∴ Mode = 7 + 40-30/2×40-30-16 × 3

= 7 + 10/80-46 × 3

= 7 + 10/34 × 3

= 7 + 0.88 = 7.88

∴ Modal number of letters is 7.88

 

(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

 

Solution:

Weight (in kg) Number of students (fi) Cumulative frequency
40 – 45 2 2
45 – 50 3 2+3 = 5
50 – 55 8 5+8 = 13
55 – 60 6 13+6 = 19
60 – 65 6 19+6 = 25
65 – 70 3 25+3 = 28
70 – 75 2 28+2 = 30
  Σfi = 30  

 

Median = L + (N/2 – cf)/f × h

Here, N/2 = 30/2 = 15

∴ Median class is 55 – 60

Here, L = 55, h = 45 – 40 = 5, cf = 13, f = 6

∴ Median = 55 + 15-13/6 × 5

= 15 + 2/6 × 5 = 55+1.67

= 56.67

∴ Median weight is 56.67 kg.

 

EXERCISE 14.4

 

(1) 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 ogive.

Solution:

Daily income (in ₹) Number of workers
100 – 120 12
120 – 140 14
140 – 160 8
160 – 180 6
180 – 200 10
  Σfi = 50

 

Daily income (in ₹) Number of workers
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

 

Daily income is plotted in x-axis and number of workers plotted in y-axis

 

(2) 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 ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula

Solution:

Weight (in kg) (x-axis) Number of students (x-axis)
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

 

Since, Σfi = 35

∴ N/2 = 35/2 = 17.5

We draw a line parallel to x-axis where number of students = 17.5

Where the line intersects are equation, is the median

∴ Median = 46.5

We convert Less than type table to normal form

Weight (in kg) Numbers of students
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

 

Weight (in kg) Number of students
0 – 38 0
38 – 40 3 – 0 = 3
40 – 42 5 – 3 = 2
42 – 44 9 – 5 = 4
44 – 46 14 – 9 = 5
46 – 48 28 – 14 = 14
48 – 50 32 – 28 = 4
50 – 52 35 – 32 = 3
  Σfi = 35

 

Weight (in kg) Number of students Cumulative frequency
0 – 38 0 0
38 – 40 3 0+3 = 3
40 – 42 2 3+2 = 5
42 – 44 4 5+4 = 9
44 – 46 5 9+5 = 14
46 – 48 14 14+14 = 28
48 – 50 4 28+4 = 32
50 – 52 3 32+3 = 35
  Σfi = 35  

 

Median = L + (N/2- cf)/f × h

Here, N/2 = 35/2 = 17.5

∴ Median class 46 – 48

L = 46, cf = 14, f = 14, h = 40 – 38 = 2

∴ Median = 46 + (35/2 – 14)/14 × 2

= 46 + 17.5-14/14 × 2

= 46 +3.5/14 × 2

= 46+0.5

= 46.5

∴ Median from the graph matches the median from the formula.

 

(3) 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 ogive.

Solution:

Production yield (in kg/ha) Number of farms
50 – 55 2
55 – 60 8
60 – 65 12
65 – 70 24
70 – 75 38
75 – 80 16
  Σfi = 100

 

Production of farms Number of farms (Years)
More than 50 100
More than 55 100 – 2 = 98
More than 60 98 – 8 = 90
More than 65 90 – 12 = 78
More than 70 78 – 24 = 54
More than 75 54 – 38 = 16

Updated: December 3, 2021 — 12:09 pm

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