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 (f_{i}) | Class Mark (x_{i}) | F_{i} x_{i} |
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 |
Σf_{i} = 20 | Σf_{i} x_{i} = 162 |
Mean (x̄) = Σf_{i} x_{i}/Σf_{i}
x̄ = 162/20 = 8.1
∴ Mean = 8.1
We used the method as f_{i} x_{i}. 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 (f_{i}) | x_{i} | d_{i} = x_{i} – a | f_{i} d_{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 |
Σf_{i} = 50 | Σf_{i }d_{i} = 240 |
Mean (x̄) = a + Σf_{i} d_{i}/Σf_{i} = 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 (f_{i}) | Class Mark (x_{i}) | d_{i} = x_{i} – a | f_{i} d_{i} |
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 |
Σf_{i} = 44 + f | Σf_{i}d_{i} = 2f – 40 |
Mean (x̄) = a + Σf_{i} d_{i}/Σf_{i}
= 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 (f_{i}) | Class Mark (x_{i}) | Xi – a | U_{i} = x_{i} – a/h
(h = 3) |
F_{i} u_{i} |
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 |
Σf_{i} = 30 | Σf_{i} = 4 |
Mean (x̄) = a + h × Σf_{i}u_{i}/Σf_{i}
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 (f_{i}) | Number of mangoes | Class mark (x_{i}) | X_{i} – a | U_{i} = x_{i}-a/h
(h = 3) |
F_{i} u_{i} |
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 |
Σf_{i }= 400 | Σf_{i}u_{i} = 25 |
Mean (x̄) = a + h × Σf_{i}u_{i}/Σf_{i}
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 f_{i}u_{i} 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 (f_{i}) | Class Mark (x_{i}) | X_{i} – a | U_{i} = x_{i}-a/h
h = 50 |
f_{i}u_{i} |
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 |
Σf_{i} = 25 | Σf_{i} u_{i} = -7 |
Mean (x̄) = a + h Σf_{i}u_{i}/Σf_{i}
= 225 + 50 × -7/25
= 225 + (-14) = 225 – 14 = 211
∴ The mean daily expenditure on food is ₹211.
(7) To find out the concentration of SO_{2} 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 SO_{2}
(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 SO_{2} in the air.
Solution:
Concentration of SO_{2} (in ppm) | Frequency (f_{i}) | Class mark (x_{i}) | f_{i} x_{i} |
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 |
Σf_{i} = 30 | Σf_{i}x_{i} = 2.96 |
∴ Mean = Σf_{i}x_{i}/Σf_{i} = 2.96/30 = 0.099
∴ The mean concentration of So_{2} 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 (f_{i}) | Class Mark (x_{i}) | f_{i}x_{i} |
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 |
Σf_{i} = 40 | Σf_{i}x_{i} = 499 |
Mean (x̄) = Σf_{i}x_{i}/Σf_{i} = 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 (f_{i}) | Class Mark (x_{i}) | X_{i} – a | U_{i} = x_{i}-a/h
(h = 10) |
F_{i} u_{i} |
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 |
Σf_{i} = 35 | Σf_{i}u_{i} = -2 |
Mean (x̄) = a + h Σf_{i}u_{i}/Σf_{i}
= 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 (f_{i}) | Class Mark (x_{i}) | f_{i}x_{i} |
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 |
Σf_{i} = 80 | Σf_{i}x_{i} = 2830 |
Mean (x̄) = Σf_{i}x_{i}/Σf_{i} = 2830/80 = 35.37
∴ The mean is 35.37
Age (in years) | Number of patients |
5 – 15 | 6 |
15 – 25 | 11 |
25 – 35 | 21 = f_{0} |
35 – 45 | 23 = f_{1} |
45 – 55 | 14 = f_{2} |
55 – 65 | 5 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
Where, lower limit of modal class (l) = 35
Class interval (h) = 15 – 5 = 10
Frequency of the modal class (f_{1}) = 23
Frequency of the class before modal class (f_{0}) = 21
Frequency of the class after modal class (f_{2}) = 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 = f_{0} |
60 – 80 | 61 = f_{1} |
80 – 100 | 38 = f_{2} |
100 – 120 | 29 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × L
Where, L = 60, f_{1} = 61, f_{0} = 52, f_{2} = 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 = f_{0} |
1500 – 2000 | 40 = f_{1} |
2000 – 2500 | 33 = f_{2} |
2500 – 3000 | 28 |
3000 – 3500 | 30 |
3500 – 4000 | 22 |
4000 – 4500 | 16 |
4500 – 5000 | 7 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
Where, L = 1500, h = 500, f_{1} = 40, f_{2} = 33, f_{0} = 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 (f_{i}) | Class Mark (x_{i}) | X_{i} – a | U_{i} = x_{i} – a/h
(h = 500) |
f_{i} × u_{i} |
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 |
Σf_{i} = 200 | Σf_{i}u_{i} = – 235 |
Mean (x̄) = a + h + Σf_{i}u_{i}/Σf_{i}
= 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 = f_{0} |
30 – 35 | 10 = f_{1} |
35 – 40 | 3 = f_{2} |
40 – 45 | 0 |
45 – 50 | 0 |
50 – 55 | 2 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
Where = L = 30, f_{1} = 10, f_{0} = 9, f_{2} = 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 (x_{i}) | X_{i} – a | U_{i} = x_{i}-a/h
(h = 5) |
F_{i} × u_{i} |
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 |
Σf_{i} = 35 | Σf_{i}u_{i} = -58 |
∴ Mean = a + h × Σf_{i}u_{i}/Σf_{i}
= 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 (f_{i}) |
3000 – 4000 | 4 = f_{0} |
4000 – 5000 | 18 = f_{1} |
5000 – 6000 | 9 = f_{2} |
6000 – 7000 | 7 |
7000 – 8000 | 6 |
8000 – 9000 | 3 |
9000 – 10000 | 1 |
10000 – 11000 | 1 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
L = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 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 (f_{i}) |
0 – 10 | 7 |
10 – 20 | 14 |
20 – 30 | 13 |
30 – 40 | 12 = t_{0} |
40 – 50 | 20 = t_{1} |
50 – 60 | 11 = t_{2} |
60 – 70 | 15 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
Where, L = 40, f_{1} = 20, f_{0} = 12, f_{2} = 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
(f_{i}) |
65 – 85 | 4 |
85 – 105 | 5 |
105 – 125 | 13 = f_{0} |
125 – 145 | 20 = f_{1} |
145 – 165 | 14 = f_{2} |
165 – 185 | 8 |
185 – 205 | 4 |
Mode = L + f_{1}-f_{0}/2f_{1}-f_{0}-f_{2} × h
Where, L = 125, f_{1} = 20, f_{0} = 13, f_{2} = 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
(f_{i}) |
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 |
Σf_{i} = 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
(f_{i}) |
Class Marks | X_{i} – a | U_{i} = X_{i} – a/h
(h = 20) |
F_{i} u_{i} |
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 |
Σf_{i} = 68 | Σf_{i}u_{i} = 7 |
Mean = a + h × Σf_{i }u_{i}/Σf_{i}
= 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 (f_{i}) | 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 | Σf_{i} = 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
Σf_{i} = 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 |
Σf_{i} = 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 (f_{i}) | 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 |
Σf_{i} = 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 (f_{i}) | 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 |
Σf_{i }= 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 (f_{i}) | 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 |
Σf_{i} = 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 (f_{i}) | Class Mark (x_{i}) | X_{i} – a | X_{i} = x_{i}-a/h
(h = 3) |
F_{i} u_{i} |
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 |
Σf_{i} = 100 | Σf_{i} u_{i} = -106 |
Mean = a + h × Σf_{i}u_{i}/Σf_{i}
Where, a = 11.5, h = 3, Σf_{i}u_{i} = -106, Σf_{i} = 100
∴ Mean = 11.5 + 3 × -106/100
= 11.5 – 3.18 = 8.32
∴ The mean is 8.32
Number of letters | Number of surnames (f_{i}) |
1 – 4 | 6 |
4 – 7 | 30 = f_{0} |
7 – 10 | 40 = f_{1} |
10 – 13 | 16 = f_{2} |
13 – 16 | 4 |
16 – 19 | 4 |
Mode = L + f_{1} – f_{0}/2f_{1} – f_{0} – f_{2} × h
Where, L = 7, h = 4 – 1 = 3 f_{0} = 30, f_{1} = 40, f_{2} = 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 (f_{i}) | 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 |
Σf_{i} = 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 |
Σf_{i} = 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, Σf_{i} = 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 |
Σf_{i} = 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 |
Σf_{i} = 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 |
Σf_{i} = 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 |