Table of Contents
Exercise Ex. 18A
Question 1
If the mean of 5 observation x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.Solution 1
Question 2
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?Solution 2
Question 3
Compute the mean of the following data:
Class | 1-3 | 3-5 | 5-7 | 7-9 |
Frequency | 12 | 22 | 27 | 19 |
Solution 3
Question 4
Find the mean, using direct method:
Class | Frequency |
0 – 1010 – 2020- 3030 – 4040 – 5050 – 60 | 7561282 |
Solution 4
We have
Class | Frequency | Mid Value | |
0-1010-2020-3030-4040-5050-60 | 7561282 | 51525354555 | 3575150420360110 |
Mean Question 5
Find the mean, using direct method:
Class | Frequency |
25 – 3535 – 4545 – 5555 – 6565 – 75 | 6108124 |
Solution 5
We have
Class | Frequency | Mid – value | |
25 – 3535 – 4545 – 5555 – 6565 – 75 | 6108124 | 3040506070 | 180400400720280 |
Mean, Question 6
Find the mean, using direct method:
Class | Frequency |
0 – 100100 – 200200 – 300300 – 400400 – 500 | 6915128 |
Solution 6
We have
Class | Frequency | Mid Value | |
0 – 100100 – 200200 – 300300 – 400400 – 500 | 6915128 | 50150250350450 | 3001350375042003600 |
= 50 |
Mean, Question 7
Using an appropriate method, find the mean of the following frequency distribution:
Class interval | 84-90 | 90-96 | 96-102 | 102-108 | 108-114 | 114-120 |
Frequency | 8 | 10 | 16 | 23 | 12 | 11 |
Which method did you use, and why?Solution 7
Question 8
If the mean of the following frequency distribution is 24, find the value of p.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 3 | 4 | p | 3 | 2 |
Solution 8
Question 9
The following distribution shows the daily pocket allowance of children of a locality. If 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 9
Question 10
If the mean of the following frequency distribution is 54, find the value of p.
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Frequency | 7 | p | 10 | 9 | 13 |
Solution 10
Question 11
The mean of the following data is 42. Find the missing frequencies x and y if the sum of frequencies is 100.
Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Solution 11
Question 12
The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is Rs.188.
Expenditure (in Rs.) | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
Number of families | 5 | 25 | f1 | f2 | 5 |
Solution 12
Question 13
The mean of the following frequency distribution is 57.6 and the sum of the observations is 50.
Class | Frequency |
0 – 20 | 7 |
20 – 40 | |
40 – 60 | 12 |
60 – 80 | |
80 – 100 | 8 |
100 – 120 | 5 |
Find and .Solution 13
We have
Class | Frequency | Mid Value | |
0 – 20 | 7 | 10 | 70 |
20 – 40 | 30 | 30 | |
40 – 60 | 12 | 50 | 600 |
60 – 80 | =18 – | 70 | 1260 – 70 |
80 – 100 | 8 | 90 | 720 |
100 – 120 | 5 | 110 | 550 |
= 50 |
Question 14
During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarized as follows:
Number of heart-beats per minute | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
Number of patients | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean heartbeats per minute for these patients, choosing a suitable method.Solution 14
Question 15
Find the mean, using assumed mean method:
Marks | No, of students |
0 – 1010 – 2020 -3030 – 4040 – 5050 – 60 | 12182720176 |
Solution 15
We have, Let A = 25 be the assumed mean
Marks | Frequency | Mid value | Deviation | |
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60 | 12182720176 | 51525 = A354555 | -20-100102030 | -240-1800200340180 |
= 100 | = 300 |
Hence mean = 28Question 16
Find the mean, using assumed mean method:
Class | Frequency |
100 – 120120 – 140140 – 160160 – 180180 – 200 | 102030155 |
Solution 16
Let the assumed mean be 150, h = 20
Marks | Frequency | Mid value | Deviationdi = – 150 | di |
100 – 120120 – 140140 – 160160 – 180180 – 200 | 102030155 | 110130150=A170190 | -40-2002040 | -400-4000300200 |
= 80 | di=-300 |
Hence, Mean = 146.25
Question 17
Find the mean, using assumed mean method:
Class | Frequency |
0 – 20 | 20 |
20 – 40 | 35 |
40 – 60 | 52 |
60 – 80 | 44 |
80 – 100 | 38 |
100 – 120 | 31 |
Solution 17
Let A = 50 be the assumed mean, we have
Marks | Frequency | Mid value | Deviation | |
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120 | 203552443831 | 103050 = A 7090110 | -40-200204060 | -800-700088015201860 |
= 220 |
Question 18
The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.
Literacy rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
Number of cities | 4 | 11 | 12 | 9 | 4 |
Solution 18
Question 19
Find the mean of the following frequency distribution using step-deviation method.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 7 | 10 | 15 | 8 | 10 |
Solution 19
Question 20
Find the mean of the following data, using step-deviation method:
Class | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 6 | 10 | 16 | 15 | 24 | 8 | 7 |
Solution 20
Question 21
The weights of tea in 70 packets are shown in the following table:
Weight (in grams) | 200-201 | 201-202 | 202-203 | 203-204 | 204-205 | 205-206 |
Number of packets | 13 | 27 | 18 | 10 | 1 | 1 |
Find the mean weight of packets using step-deviation method.Solution 21
Question 22
Find the mean of the following frequency distribution using a suitable method:
Class | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 25 | 40 | 42 | 33 | 10 |
Solution 22
Question 23
In a annual examination marks (out of 90) obtained by students of class X in mathematics are given below:
Marks obtained | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |
Number of students | 2 | 4 | 5 | 20 | 9 | 10 |
Find the mean marks.Solution 23
Question 24
Find the arithmetic mean of the following frequency distribution using step-deviation method:
Age (in year) | 18-24 | 24-30 | 30-36 | 36-42 | 42-48 | 48-54 |
Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Solution 24
Question 25
Find the arithmetic mean of each of the following frequency distribution using step-deviation method:
Class | Frequency |
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620 | 1495435 |
Solution 25
Let h = 20 and assume mean = 550, we prepare the table given below:
Age | Frequency | Mid value | ||
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620 | 1495435 | 510530550 = A570590610 | -2-10123 | -27-904615 |
= 40 |
Thus, A = 550, h = 20, and = 40,
Hence the mean of the frequency distribution is 544Question 26
Find the mean age from the following frequency distribution:
Age(in years) | No. of persons |
25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59 | 4142216653 |
Hint: change the given series to the exclusive seriesSolution 26
The given series is an inclusive series, making it an exclusive series, we have
Class | Frequency | Mid value | ||
24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5 | 4142216653 | 27323742 = A475257 | -3-2-10123 | -12-28-2206109 |
= 70 |
Thus, A = 42, h = 5, = 70 and
Hence, Mean = 39.36 years
Question 27
The following table shows the age distribution of patients of malaria in a village during a particular month:
Age(in years) | No. of cases |
5 – 1415 – 2424 – 3435 – 4445 – 5455 – 64 | 6112123145 |
Find the average age of the patients.Solution 27
The given series is an inclusive series making it an exclusive series,we get
class | Frequency | Mid value | ||
4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.5 | 6112123145 | 9.519.529.5=A39.549.559.5 | -2-10123 | -12-110232815 |
= 80 |
Thus, A = 29.5, h = 10, = 80 and
Hence, Mean = 34.87 years
Question 28
Weight of 60 eggs were recorded as given below:
Weight (in grams) | 75-79 | 80-84 | 85-89 | 90-94 | 95-99 | 100-104 | 105-109 |
Number of eggs | 4 | 9 | 13 | 17 | 12 | 3 | 2 |
Calculate their mean weight to the nearest gram.Solution 28
Question 29
The following table shows the marks scored by 80 students in an examination:
Marks | Less than 5 | Less than 10 | Less than 15 | Less than 20 | Less than 25 | Less than 30 | Less than 35 | Less than 40 |
Number of students | 3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
Calculate the mean marks correct to 2 decimal places.Solution 29
Exercise Ex. 18B
Question 1
In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age (in years) | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 |
Number of patients | 5 | 20 | 40 | 50 | 25 |
Solution 1
Question 2
Compute the median from the following data:
Marks | No. of students |
0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49 | 347110169 |
Solution 2
We prepare the frequency table, given below
Marks | No. of students | C.F. |
0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49 | 347110169 | 371425254150 |
N = = 50 |
Now,
The cumulative frequency is 25 and corresponding class is 21 – 28.
Thus, the median class is 21 – 28
l = 21, h = 7, f = 11, c = C.F. preceding class 21 – 28 is 14 and = 25
Hence the median is 28.Question 3
The following table shows the daily wages of workers in a factory:
Daily wages | No. of workers |
0 – 100100 – 200200 – 300300 – 400400 – 500 | 403248228 |
Find the median daily wage income of the workers.Solution 3
We prepare the frequency table given below:
Daily wages | Frequency | C.F. |
0 – 100100 – 200200 – 300300 – 400400 – 500 | 403248228 | 4072120142150 |
N = = 150 |
Now, N = 150, therefore
The cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.
Thus, the median class is 200 – 300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and
Hence the median of daily wages is Rs. 206.25.Hence the median is 28.Question 4
Calculate the median from the following frequency distribution:
Class | Frequency |
5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45 | 5615105422 |
Solution 4
We prepare the frequency table, given below:
Class | Frequency | C.F |
5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45 | 5615105422 | 511263641454749 |
= 49 |
Now, N = 49
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.
Thus, the median class is 15 – 20
l = 15, h = 5, f = 15
c = CF preceding median class = 11 and
Median of frequency distribution is 19.5Question 5
Given below is the number of units of electricity consumed in a week in a certain locality:
Consumption(in units) | No. of consumers |
65 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205 | 4513201474 |
Calculate the median.Solution 5
We prepare the cumulative frequency table as given below:
Consumption | Frequency | C.F |
65 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205 | 4513201474 | 492242566367 |
N = = 67 |
Now, N = 67
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.
Thus, the median class is 125 – 145
l = 125, h = 20, and c = CF preceding the median class = 22, = 33.5
Hence median of electricity consumed is 136.5Question 6
Calculate the median from the following data:
Height(in cm) | No. of boys |
135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175 | 6101822201563 |
Solution 6
Frequency table is given below:
Height | Frequency | C.F |
135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175 | 6101822201563 | 6163456769197100 |
N = =100 |
N = 100,
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155
Thus, the median class is 150 – 155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
Hence, Median = 153.64
Question 7
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Class | Frequency |
0 – 1010 – 2020 – 3030 – 4040 – 50 | 525×187 |
Solution 7
The frequency table is given below. Let the missing frequency be x.
Class | Frequency | C.F |
0 – 1010 – 2020 – 3030 – 4040 – 50 | 525×187 | 53030 + x48 + x55 + x |
Median = 24 Median class is 20 – 30
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
Hence, the missing frequency is 25.Question 8
The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequencies | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
Solution 8
Question 9
In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
Runs scored | 2500-3500 | 3500-4500 | 4500-5500 | 5500-6500 | 6500-7500 | 7500-8500 |
Number of batsmen | 5 | x | y | 12 | 6 | 2 |
Solution 9
Question 10
If the median of the following frequency distribution is 32.5, find the value of .
Solution 10
Let be the frequencies of class intervals 0 – 10 and 40 – 50
Median is 32.5 which lies in 30 – 40, so the median class is 30 – 40
l = 30, h = 10, f = 12, N = 40 and
Question 11
Calculate the median for the following data:
Age(in years) | Frequency |
19 – 2526 – 3233 – 3940 – 4647 – 5354 – 60 | 359668102354 |
Solution 11
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Class | Frequency | C.F |
18.5 – 25.525.5 – 32.532.5 – 39.539.5 – 46.546.5 – 53.553.5 – 60.5 | 359668102354 | 35131199301336340 |
fi = N = 340 |
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
Median class is 32.5 – 39.5
l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
Hence median is 36.5 yearsQuestion 12
Find the median wages for the following frequencies distribution:
Wages per day(in Rs) | Frequency |
61 – 7071 – 8081 – 9091 – 100101 – 110111 – 120 | 5152030208 |
Solution 12
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
Wages per day(in Rs) | Frequency | C.F |
60.5 – 70.570.5 – 80.580.5 – 90.590.5 – 100.5100.5 – 110.5110.5 – 120.5 | 5152030208 | 52040709098 |
fi = N =98 |
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.
median class is 90.5 – 100.5
l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
Hence, Median = Rs 93.50Question 13
Find the median from the following table:
Class | Frequency |
1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3535 – 4040 – 45 | 710163224161152 |
Solution 13
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks | Frequency | C.F |
0.5 – 5.55.5 – 10.510.5 – 15.515.5 – 20.520.5 – 25.525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.5 | 710163224161152 | 717336589105116121123 |
fi = N =123 |
The cumulative frequency just greater than 61.5 is 65.
The corresponding median class is 15.5 – 20.5.
Then the median class is 15.5 – 20.5
l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
Hence, Median = 19.95
Question 14
Find the median from the following data:
Marks | No. of students |
Below 10Below 20Below 30Below 40Below 50Below 60Below 70Below 80 | 1232578092116164200 |
Solution 14
Marks | Frequency | C.F |
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80 | 1220252312244836 | 1232578092116164200 |
N = |
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus the median class is 50 – 60
l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, = 100
Hence, Median = 53.33
Exercise Ex. 18C
Question 1
Find the mode of the following frequency distribution:
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 12 | 35 | 45 | 25 | 13 |
Solution 1
Question 2
Compute the mode of the following data:
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Frequency | 25 | 16 | 28 | 20 | 5 |
Solution 2
Question 3
Heights of students of Class X are given in the following frequency distribution:
Height (in cm) | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |
Number of students | 15 | 8 | 20 | 12 | 5 |
Find the modal height.
Also, find the mean height. Compare and interpret the two measures of central tendency.Solution 3
Question 4
Find the mode of the following distribution:
Class interval | Frequency |
10 – 1414 – 1818 – 2222 – 2626 – 3030 – 3434 – 3838 – 42 | 8611202522104 |
Solution 4
As the class 26 – 30 has maximum frequency so it is modal class
Hence, mode = 28.5Question 5
Given below is the distribution of total household expenditure of 200 manual workers in a city:
Expenditure | No. of manual workers |
1000 – 15001500 – 20002000 – 25002500 – 30003000 – 35003500 – 40004000 – 45004500 – 5000 | 244031283223175 |
Find the average expenditure done by maximum number of manual workers.Solution 5
As the class 1500 – 2000 has maximum frequency, so it os modal class
Hence the average expenditure done by maximum number of workers = Rs. 1820Question 6
Calculate the mode from the following data:
Monthly salary(in Rs) | No. of employees |
0 – 50005000- 1000010000 – 1500015000 – 2000020000 – 2500025000 – 30000 | 90150100807010 |
Solution 6
As the class 5000 – 10000 has maximum frequency, so it is modal class
Hence, mode = Rs. 7727.27Question 7
Compute the mode from the following data:
Age (in years) | No. of patients |
0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 35 | 611182417135 |
Solution 7
As the class 15 – 20 has maximum frequency so it is modal class.
Hence mode = 17.3 yearsQuestion 8
Compute the mode from the following series:
Size | Frequency |
45 – 5555 – 6565 – 7575 – 8585 – 9595 – 105105 – 115 | 712173032610 |
Solution 8
As the class 85 – 95 has the maximum frequency it is modal class
Hence, mode = 85.71Question 9
Compute the mode of the following data:
Class Interval | Frequency |
1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3536 – 4041 – 4546 – 50 | 381318282013864 |
Solution 9
The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get
Class | Frequency |
0.5 – 5.55.5- 10.510.5 – 15. 515.5 – 20.520.5 – 25. 525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.545.5 – 50.5 | 381318282013863 |
As the class 20.5 – 25.5 has maximum frequency, so it is modal class
Hence, mode = 23.28Question 10
The age wise participation of students in the Annual Function of a school is shown in the following distribution.
Age (in years) | 5-7 | 7-9 | 9-11 | 11-13 | 13-15 | 15-17 | 17-19 |
Number of students | x | 15 | 18 | 30 | 50 | 48 | x |
Find the missing frequencies when the sum of frequencies is 181. Also, find the mode of the data.Solution 10
Exercise Ex. 18D
Question 1
Find the mean, mode and median of the following data:
Class | Frequency |
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 70 | 510183020125 |
Solution 1
Let assumed mean be 35, h = 10, now we have
Class | Frequency | Mid-value | C.F | ||
0-1010-2020-3030-4040-5050-6060-70 | 510183020125 | 5152535 = A455565 | -3-2-10123 | 51533638395100 | -15-20-180202415 |
N = 100 |
(i)Mean
(ii)N = 100,
Cumulative frequency just after 50 is 63
Median class is 30 – 40
l = 30, h = 10, N = 100, c = 33, f = 30
(iii)Mode = 3 × median – 2 × mean
= 3 × 35.67 – 2 × 35.6 = 107.01 – 71.2
= 35.81
Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81Question 2
Find the mean, median and mode of the following data:
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | 120-140 |
Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
Solution 2
Question 3
Find the mean, median and mode of the following data:
Class | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
Frequency | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
Solution 3
Question 4
Find the mode, median and mean for the following data:
Marks obtained | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 |
Number of students | 7 | 31 | 33 | 17 | 11 | 1 |
Solution 4
Question 5
A survey regarding the heights of 50 girls of a class was conducted and the following data was obtained:
Height in cm | No. of girls |
120 – 130130 – 140140 – 150150 – 160160 – 170 | 2812208 |
Total | 50 |
Find the mean, Median and mode of the above data.Solution 5
Let the assumed mean A be 145.Class interval h = 10.
Class | Frequency | Mid-Value | C.F. | ||
120-130130-140140-150150-160160-170 | 2812208 | 125135145=A155165 | -2-1012 | -4-802016 | 210224250 |
N = 50 |
(i)Mean
(ii)N = 50,
Cumulative frequency just after 25 is 42
Corresponding median class is 150 – 160
Cumulative frequency before median class, c = 22
Median class frequency f = 20
(iii)Mode = 3 median – 2 mean
= 3 151.5 – 2 149.8 = 454.5 – 299.6
= 154.9
Thus, Mean = 149.8, Median = 151.5, Mode = 154.9Question 6
The following table gives the daily income of 50 workers of a factory:
Daily income(in Rs) | No. of workers |
100 – 120120 – 140140 – 160160 – 180180 – 200 | 12148610 |
Find the mean, mode and median of the above dataSolution 6
Class | Frequency | Mid-value | C.F. | ||
100-120120-140140-160160-180180-200 | 12148610 | 110130150= A170190 | -2-1012 | -24-140620 | 1226344050 |
N = 50 |
Let assumed mean A = 150 and h = 20
(i)Mean
(ii)
Cumulative frequency just after 25 is 26
Corresponding frequency median class is 120 – 140
So, l = 120, f = 14, h = 20, c = 12
(iii)Mode = 3 Median – 2 Mode
= 3 138.6 – 2 145.2
= 415.8 – 190.4
= 125.4
Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4Question 7
The table below shows the daily expenditure on food of 30 households in a locality:
Daily expenditure | No. of households |
100 – 150150 – 200200 – 250250 – 300300 – 350 | 671232 |
Find the mean and median daily expenditure on foodSolution 7
Class | Frequency | Mid-value | C.F. | ||
100-150150-200200-250250-300300-350 | 671232 | 125175225275325 | -2-1012 | -12-7034 | 613252830 |
N = 30 |
Let assumed mean = 225 and h = 50
(i)Mean =
(ii)
Cumulative frequency just after 15 is 25
corresponding class interval is 200 – 250
Median class is 200 – 250
Cumulative frequency c just before this class = 13
Hence, Mean = 205 and Median = 208.33
Exercise Ex. 18E
Question 1
Find the median of the following data by making a ‘less than ogive’.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Number of students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |
Solution 1
We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the ‘less than type’ ogive as follows:
At y = 26.5, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 68 units
Hence, median marks = 68Question 2
The given distribution shows the number of wickets taken by the bowlers in one -day international cricket matches:
Number of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
Draw a ‘less type’ ogive from the above data. Find the median.Solution 2
Number of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the ‘less than type’ ogive as follows:
At y = 40, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 78 units
Hence, median number of wickets = 78Question 3
Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Number of students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
Solution 3
We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the ‘more than type’ ogive as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 47 units
Hence, median marks = 47Question 4
The height of 50 girls of Class X of a school are recorded as follows:
Height (in cm) | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 |
Number of girls | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.Solution 4
We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the ‘more than type’ ogive as follows:
At y = 25, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 151 units
Hence, median height = 151 cmQuestion 5
The monthly consumption of electricity (in units) of some families of a locality is given in the following frequency distribution:
Monthly consumption (in units) | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 | 240-260 | 260-280 |
Number of families | 3 | 8 | 15 | 40 | 50 | 30 | 10 |
Prepare a ‘more than type’ ogive for the given frequency distribution.Solution 5
We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the ‘more than type’ ogive as follows:
At y = 78, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 226 units
Hence, median consumption of electricity = 226 unitsQuestion 6
The following table gives the production yield per hectare of wheat of 100 farms of a village.
Production yield (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. Using ogive, find the median of the given data.Solution 6
We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the ‘more than type’ ogive as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 70.5 units
Hence, median production yield = 70.5 kg/haQuestion 7
The table given below shows the weekly expenditures on food of some households in a locality.
Weekly expenditure (in Rs.) | Number of households |
100-200 | 5 |
200-300 | 6 |
300-400 | 11 |
400-500 | 13 |
500-600 | 5 |
600-700 | 4 |
700-800 | 3 |
800-900 | 2 |
Draw a ‘less tha type’ and a ‘more than type’ ogive for this distribution.Solution 7
Less Than Series:
Class interval | Cumulative Frequency |
Less than 200 | 5 |
Less than 300 | 11 |
Less than 400 | 22 |
Less than 500 | 35 |
Less than 600 | 40 |
Less than 700 | 44 |
Less than 800 | 47 |
Less than 900 | 49 |
We plot the points (200, 5), (300, 11), (400, 22), (500, 35), (600, 40), (700, 44), (800, 47) and (900, 49) to get ‘less than type’ ogive.
More Than Series:
Class interval | Frequency |
More than 100 | 49 |
More than 200 | 44 |
More than 300 | 38 |
More than 400 | 27 |
More than 500 | 14 |
More than 600 | 9 |
More than 700 | 5 |
More than 800 | 2 |
We plot the points (100, 49), (200, 44), (300, 38), (400, 27), (500, 14), (600, 9), (700, 5) and (800, 2) to get more than ogive.
Question 8
From the following frequency distribution, prepare the ‘More than Ogive’
Score | No. of candidates |
400 – 450450 – 500500 – 550550 – 600600- 650650 – 700700 – 750750 – 800 | 2035403224271834 |
Total | 230 |
Also find the medianSolution 8
More than series
We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)
Hence,
Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M
Then,OM = 590
Hence median = 590Question 9
The marks obtained by 100 students of a class in an examination are given below:
Marks | No. of students |
0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 4545 – 50 | 25681025201842 |
Draw cumulative frequency curves by using (1), less than series and (2) more than series
Hence, find the medianSolution 9
(i) Less than series:
Marks | No. of students |
Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35Less than 40Less than 45Less than 50 | 2713213156769498100 |
Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
Join these points free hand to get the curve representing “less than” cumulative curve.
(ii)From the given table we may prepare the ‘more than’ series as shown below
Marks | No. of students |
More than 45More than 40More than 35More than 30More than 25More than 20More than 15More than 10More than 5More than 0 | 2624446979879398100 |
Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)
Join these points free hand to get required curve
Here
Two curves intersect at point P(28, 50)
Hence, the median = 28
Question 10
From the following data, draw the two types of cumulative frequency curves and determine the median:
Height (in cm) | Frequency |
140 – 144144 – 148148 – 152152 – 156156 – 160160 – 164164 – 168168 – 172172 – 176176 – 180 | 392431426475828634 |
Solution 10
We may prepare less than series and more than series
(i)Less than series
Height in (cm) | Frequency |
Less than 140Less than 144Less than 148Less than 152Less than 156Less than 160Less than 164Less than 168Less than 172Less than 176Less than 180 | 03123667109173248330416450 |
Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)
(ii)More than series
Height in cm | C.F. |
More than 140More than 144More than 148More than 152More than 156More than 160More than 164More than 168More than 172More than 176More than 180 | 450447438414383341277202120340 |
Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)
The curves intersect at (167, 225).
Hence, 167 is the median.
Exercise Ex. 18F
Question 1
Write the median class of the following distribution:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 4 | 4 | 8 | 10 | 12 | 8 | 4 |
Solution 1
Question 2
What is the lower limit of the modal class of the following frequency distribution?
Age (in years) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number of patients | 16 | 13 | 6 | 11 | 27 | 18 |
Solution 2
Class having maximum frequency is the modal class.
Here, maximum frequency = 27
Hence, the modal class is 40 – 50.
Thus, the lower limit of the modal class is 40.Question 3
The monthly pocket money of 50 students of a class are given in the following distribution:
Monthly pocket money (in Rs.) | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
Number of students | 2 | 7 | 8 | 30 | 12 | 1 |
Find the modal class and also give class mark of the modal class.Solution 3
Question 4
A data has 25 observations arranged in a descending order. Which observation represents the median?Solution 4
Question 5
For a cetain distribution, mode and median were found to be 1000 and 1250 respectively. Find mean for this distribution using an empirical relation.Solution 5
Question 6
In a class test, 50 students obtained marks as follows:
Marks obtained | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Number of students | 4 | 6 | 25 | 10 | 5 |
Find the modal class and the median class.Solution 6
Question 7
Find the class marks of classes 10-25 and 35-55.Solution 7
Question 8
While calculating the mean of a given data by the assumed-mean method, the following values were obtained:
Find the mean.Solution 8
Question 9
The distributions X and Y with total number of observations 36 and 64, and mean 4 and 3 respectively are combined. What is the mean of the resulting distribution X + Y?Solution 9
Question 10
In a frequency distribution table with 12 classes, the class-width is 2.5 and the lowest class boundary is 8.1, then what is the upper class boundary of the highest class?Solution 10
Question 11
The observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 are arranged in ascending order. What is the value of x if the median of the data is 63?Solution 11
Question 12
The median of 19 observations is 30. Two more observations are made and the values of these are 8 and 32. Find the median of the 21 observations taken together.Solution 12
Question 13
Solution 13
Question 14
What is the cumulative frequency of the modal class of the following distribution?
Class | 3-6 | 6-9 | 9-12 | 12-15 | 15-18 | 18-21 | 21-24 |
Frequency | 7 | 13 | 10 | 23 | 4 | 21 | 16 |
Solution 14
Question 15
Find the mode of the given data:
Class interval | 0-20 | 20-40 | 40-60 | 60-80 |
Frequency | 15 | 6 | 18 | 10 |
Solution 15
Question 16
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in year) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.Solution 16
Question 17
In the following data, find the value of p and q. Also, find the median class and modal class.
Class | Frequency (f) | Cumulative frequency (cf) |
100-200 | 11 | 11 |
200-300 | 12 | p |
300-400 | 10 | 33 |
400-500 | q | 46 |
500-600 | 20 | 66 |
600-700 | 14 | 80 |
Solution 17
Question 18
The following frequency distribution gives the monthly consumption of electricity of 64 consumers of a locality.
Monthly consumption (in units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 |
Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 |
Form a ‘more than type’ cumulative frequency distribution.Solution 18
Question 19
The following table gives the life-time (in days) of 100 electric bulbs of a certain brand.
Life-time (in days) | Less than 50 | Less than 100 | Less than 150 | Less than 200 | Less than 250 | Less than 300 |
Number of bulbs | 7 | 21 | 52 | 79 | 91 | 100 |
From this table, construct the frequency distribution table.Solution 19
Question 20
The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
Marks obtained (in percent) | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
Number of students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark
(c) Find the modal class and write its cumulative frequency.Solution 20
Question 21
If the mean of the following distribution is 27, find the value of p.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 8 | p | 12 | 13 | 10 |
Solution 21
Question 22
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Age (in years) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Number of persons | 5 | 25 | ? | 18 | 7 |
Solution 22
Exercise MCQ
Question 1
Which of the following is not a measure of central tendency?
(a) Mean
(b) Mode
(c) Median
(d) RangeSolution 1
Correct option: (d)
Range is not a measure of central tendency.Question 2
Which of the following cannot be determined graphically? (a) Mean
(b) Median
(c) Mode
(d) None of theseSolution 2
Correct option: (a)
Mean cannot be determined graphically.Question 3
Which of the following measures of central tendency is influenced by extreme values?
(a) Mean
(b) Median
(c) Mode
(d) None of theseSolution 3
Correct option: (a)
Since mean is the average of all observations, it is influenced by extreme values.Question 4
The mode of a frequency distribution is obtained graphically from
(a) a frequency curve
(b) a frequency polygon
(c) a histogram
(d) an ogiveSolution 4
Correct option: (c)
Mode can be obtained graphically from a histogram.Question 5
The median of a frequency distribution is found graphically with the help of
(a) a histogram
(b) a frequency curve
(c) a frequency polygon
(d) ogivesSolution 5
Correct option: (d)
Ogives are used to determine the median of a frequency distribution.Question 6
The cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all of theseSolution 6
Correct option: (b)
The cumulative frequency table is useful in determining the median.Question 7
The abscissa of the point of intersection of the Less Than Type and of the More Than Type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) none of theseSolution 7
Correct option: (b)
Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.Question 8
(a) 1
(b) 0
(c) -1
(d) 2Solution 8
Question 9
Solution 9
Question 10
(a) Lower limits of the classes
(b) upper limits of the classes
(c) midpoints of the classes
(d) none of theseSolution 10
Correct option: (c)
di‘s are the deviations from A of midpoints of the classes.Question 11
While computing the mean of the grouped data, we assume that the frequencies are
(a) evenly distributed over the classes
(b) centred at the class marks of the classes
(c) centred at the lower limits of the classes
(d) centred at the upper limits of the classesSolution 11
Correct option: (b)
While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.Question 12
The relation between mean, mode and median is
(a) mode = (3 ⨯ mean) – (2 ⨯ median)
(b) mode = (3 ⨯ median) – (2 ⨯ mean)
(c) median = (3 ⨯ mean) – (2 ⨯ mode)
(d) mean = (3 ⨯ median) – (2 ⨯ mode)Solution 12
Correct option: (b)
Mode = (3 x median) – (2 x mean)Question 13
If the ‘less than type’ ogive and ‘more than type’ ogive intersect each other at (20.5, 15.5) then the median of the given data is
(a) 5.5
(b) 15.5
(c) 20.5
(d) 36.0Solution 13
Correct option: (c)
Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5Question 14
Consider the frequency distribution of the height of 60 students of a class:
Height (in cm) | No. of Students | Cumulative Frequency |
150-155 | 16 | 16 |
155-160 | 12 | 28 |
160-165 | 9 | 37 |
165-170 | 7 | 44 |
170-175 | 10 | 54 |
175-180 | 6 | 60 |
The sum of the lower limit of the modal class and the upper limit of the median class is
(a) 310
(b) 315
(c) 320
(d) 330Solution 14
Question 15
Consider the following frequency distribution :
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 3 | 9 | 15 | 30 | 18 | 5 |
The modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60Solution 15
Correct option: (c)
Class having maximum frequency is the modal class.
Here, maximum frequency = 30
Hence, the modal class is 30 – 40.Question 16
Mode =?
Solution 16
Question 17
Median = ?
Solution 17
Question 18
If the mean and median of a set of numbers are 8.9 and 9 respectively then the mode will be
(a) 7.2
(b) 8.2
(c) 9.2
(d) 10.2Solution 18
Correct option: (c)
Mean = 8.9
Median = 9
Mode = 3Median – 2Mean
= 3 x 9 – 2 x 8.9
= 27 – 17.8
= 9.2Question 19
Look at the frequency distribution table given below:
Class interval | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 8 | 12 | 20 | 10 |
The median of the above distribution is
(a) 56.5
(b) 57.5
(c) 58.5
(d) 59Solution 19
Question 20
Consider the following table :
Class interval | 10-14 | 14-18 | 18-22 | 22-26 | 26-30 |
Frequency | 5 | 11 | 16 | 25 | 19 |
The mode of the above data is
(a) 23.5
(b) 24
(c) 24.4
(d) 25Solution 20
Question 21
The mean and mode of a frequency distribution are 28 and 16 respectively. The median is
(a) 22
(b) 23.5
(c) 24
(d) 24.5Solution 21
Question 22
The median and mode of a frequency distribution are 26 and 29 respectively. The mean is
(a) 27.5
(b) 24.5
(c) 28.4
(d) 25.8Solution 22
Question 23
For a symmetrical frequency distribution, we have
(a) mean < mode < median
(b) mean > mode > median
(c) mean = mode = median
(d) Solution 23
Correct option: (c)
For a symmetrical distribution, we have
Mean = mode = medianQuestion 24
Look at the cumulative frequency distribution table given below :
Monthly income | Number of families |
More than Rs.10000 | 100 |
More than Rs.14000 | 85 |
More than Rs.18000 | 69 |
More than Rs.20000 | 50 |
More than Rs.25000 | 37 |
More than Rs.30000 | 15 |
Number of families having income range 20000 to 25000 is
(a) 19
(b) 16
(c) 13
(d) 22Solution 24
Correct option: (c)
Number of families having income more than Rs. 20000 = 50
Number of families having income more than Rs. 25000 = 37
Hence, number of families having income range 20000 to 25000 = 50 – 37 = 13Question 25
The median of first 8 prime numbers is
(a) 7
(b) 9
(c) 11
(d) 13Solution 25
Question 26
The mean of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
(a) 0
(b) 1
(c) 10
(d) 19Solution 26
Correct option: (d)
Mean of 20 numbers = 0
Hence, sum of 20 numbers = 0 x 20 = 0
Now, the mean can be zero if
sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),
sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),
…….
sum of 19 numbers is (S) and the 20th number is (-S), then their sum is zero.
So, at the most, 19 numbers can be greater than zero.Question 27
If the median of the data 4, 7, x – 1, x – 3, 16, 25, written in ascending order, is 13 then x is equal to
(a) 13
(b) 14
(c) 15
(d) 16Solution 27
Question 28
The mean of 2, 7, 6 and x is 5 and the mean of 18, 1, 6, x and y is 10.
What is the value of y?
(a) 5
(b) 10
(c) 20
(d) 30
Note: Question modifiedSolution 28
Question 29
Match the following columns :
Column I | Column II | ||
(a) | The most frequent value in a data is known as ……. | (p) | Standard deviation |
(b) | Which of the following cannot be determined graphically out of mean, mode and median? | (q) | Median |
(c) | An ogive is used to determine… | (r) | Mean |
(d) | Out of mean, mode, median and standard deviation, which is not a measure of central tendency? | (s) | Mode |
Solution 29
(a) – (s)
The most frequent value in a data is known as mode.
(b) – (r)
Mean cannot be determined graphically.
(c) – (q)
An ogive is used to determine median.
(d) – (p)
Standard deviation is not a measure of central tendency.Question 30
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
(a) Both Assertion (A) and Reason (R) are true and Reason (R.) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d)Assertion (A) is false and Reason (R) is true.
Assertion (A) | Reason (R) |
If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148. | Mean, median and mode of a frequency distribution are related as:mode = 3 median – 2 mean. |
The correct answer is: (a)/(b)/(c)/(d).Solution 30
Question 31
Assertion (A) | Reason (R) |
Consider the following frequency distribution :Class interval3-66-99-1212-1515-1818-21Frequency2521231012 The mode of the above data is 12.4 | The value of the variable which occurs most often is the mode. |
The correct answer is: (a)/(b)/(c)/(d).Solution 31
Exercise FA
Question 1
Which one of the following measures is determined only after the construction of cumulative frequency distribution?
(a) Mean
(b) Median
(c) Mode
(d) None of theseSolution 1
Correct option: (b)
The cumulative frequency table is useful in determining the median.Question 2
If the mean of a data is 27 and its median is 33 then the mode is
(a) 30
(b) 43
(c) 45
(d) 47Solution 2
Correct option: (c)
Mean = 27
Median = 33
Mode = 3Median – 2Mean
= 3 x 33 – 2 x 27
= 99 – 54
= 45Question 3
Consider the following distribution :
Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
Frequency | 10 | 15 | 12 | 20 | 9 |
The sum of the lower limits of the median class and the modal class is
(a) 15
(b) 25
(c) 30
(d) 35Solution 3
Question 4
Consider the following frequency distribution :
Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
Frequency | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is
(a) 16.5
(b) 18.5
(c) 18
(d) 17.5Solution 4
Question 5
If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.Solution 5
Question 6
In the table given below, the times taken by 120 athletes to run a 100-m-hurdle race are given.
Class | 13.8-14 | 14-14.2 | 14.2-14.4 | 14.4-14.6 | 14.6-14.8 | 14.8-15 |
Frequency | 2 | 4 | 15 | 54 | 25 | 20 |
Find the number of athletes who completed the race in less than 14.6 seconds.Solution 6
Number of athletes who completed the race in less than 14.6 seconds
= 2 + 4 + 15 + 54
= 75Question 7
Consider the following frequency distribution :
Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
Frequency | 13 | 10 | 15 | 8 | 11 |
Find the upper limit of the median class.Solution 7
Question 8
The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below :
Profit (in lakhs Rs.) | Number of shops |
More than or equal to 5 | 30 |
More than or equal to 10 | 28 |
More than or equal to 15 | 16 |
More than or equal to 20 | 14 |
More than or equal to 25 | 10 |
More than or equal to 30 | 7 |
More than or equal to 35 | 3 |
If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25Solution 8
The frequency table is as follows:
ClassesProfit (in lakhs Rs.) | FrequencyNumber of shops |
5 – 10 | 2 |
10 – 15 | 12 |
15 – 20 | 2 |
20 – 25 | 4 |
25 – 30 | 3 |
30 – 35 | 4 |
35 – 40 | 3 |
The frequency corresponding to the class 20 – 25 is 4.Question 9
Find the mean of the following frequency distribution :
Class | 1-3 | 3-5 | 5-7 | 7-9 |
Frequency | 9 | 22 | 27 | 18 |
Solution 9
Question 10
The maximum bowling speeds (in km/hr) of 33 players at a cricket coaching centre are given below :
Speed in km/hr | 85-100 | 100-115 | 115-130 | 130-145 |
No. of players | 10 | 4 | 7 | 9 |
Calculate the median bowling speed.Solution 10
Question 11
The arithmetic mean of the following frequency distribution is 25.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 16 | p | 30 | 32 | 14 |
Find the value of p.
Note: Question modifiedSolution 11
Question 12
Find the median of the following frequency distribution :
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Number of students | 6 | 16 | 30 | 9 | 4 |
Solution 12
Question 13
Following is the distribution of marks of 70 students in a periodical test :
Marks | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |
Number of students | 3 | 11 | 28 | 48 | 70 |
Draw a cumulative frequency curve for the above data.Solution 13
Marks | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |
Number of students | 3 | 11 | 28 | 48 | 70 |
We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:
Question 14
Find the median of the following data.
Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | Total |
Frequency | 8 | 16 | 36 | 34 | 6 | 100 |
Solution 14
Question 15
For the following distribution draw a ‘less than type’ ogive and from the curve find the median.
Marks obtained | Less than 20 | Less than 30 | Less than 40 | Less than 50 | Less than 60 | Less than 70 | Less than 80 | Less than 90 | Less than 100 |
Number of students | 2 | 7 | 17 | 40 | 60 | 82 | 85 | 90 | 100 |
Solution 15
Marks obtained | Less than 20 | Less than 30 | Less than 40 | Less than 50 | Less than 60 | Less than 70 | Less than 80 | Less than 90 | Less than 100 |
Number of students | 2 | 7 | 17 | 40 | 60 | 82 | 85 | 90 | 100 |
We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 56.
Hence, median = 56 Question 16
The median value for the following frequency distribution is 35 and the sum of all the frequencies is 170. Using the formula for median, find the missing frequencies.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 10 | 20 | ? | 40 | ? | 25 | 15 |
Solution 16
Question 17
Find the missing frequencies f1 and f2 in the table given below, it being given that the mean of the given frequency distribution is 50.
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | Total |
Frequency | 17 | f1 | 32 | f2 | 19 | 120 |
Solution 17
Question 18
Find the mean of the following frequency distribution using step-deviation method :
Class | 84-90 | 90-96 | 96-102 | 102-108 | 108-114 | 114-120 |
Frequency | 15 | 22 | 20 | 18 | 20 | 25 |
Solution 18
Question 19
Find the mean, median and mode of the following data :
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 6 | 8 | 10 | 15 | 5 | 4 | 2 |
Solution 19
Question 20
Draw ‘less than ogive’ and more than ogive’ on a single graph paper and hence find the median of the following data :
Class interval | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequency | 2 | 12 | 2 | 4 | 3 | 4 | 3 |
Solution 20
Less Than Series:
Class interval | Frequency |
Less than 10 | 2 |
Less than 15 | 14 |
Less than 20 | 16 |
Less than 25 | 20 |
Less than 30 | 23 |
Less than 35 | 27 |
Less than 40 | 30 |
We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.
More Than Series:
Class interval | Frequency |
More than 5 | 30 |
More than 10 | 28 |
More than 15 | 16 |
More than 20 | 14 |
More than 25 | 10 |
More than 30 | 7 |
More than 35 | 3 |
We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.
The two curves intersect at L. Draw LM ⊥ OX.
Thus, median = OM = 16 Question 21
The production yield per hectare of wheat of some farms of a village are given in the following table :
Production yield (in kg/ha) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 | 80-85 |
Number of farms | 1 | 9 | 15 | 18 | 40 | 26 | 16 | 14 | 10 |
Draw a less than type ogive and a more than type ogive for this dataSolution 21
Less Than Series:
Class interval | Frequency |
Less than 45 | 1 |
Less than 50 | 10 |
Less than 55 | 25 |
Less than 60 | 43 |
Less than 65 | 83 |
Less than 70 | 109 |
Less than 75 | 125 |
Less than 80 | 139 |
Less than 85 | 149 |
We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.
More Than Series:
Class interval | Frequency |
More than 40 | 149 |
More than 45 | 148 |
More than 50 | 139 |
More than 55 | 124 |
More than 60 | 106 |
More than 65 | 66 |
More than 70 | 40 |
More than 75 | 24 |
More than 80 | 10 |
We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.
Question 22
The following table gives the marks obtained by 50 students in a class test :
Marks | 11-15 | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 |
Number of students | 2 | 3 | 6 | 7 | 14 | 12 | 4 | 2 |
Calculate the mean, median and mode for the above data.Solution 22
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