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10: Trignometric Ratios
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15: Perimeter And Area of Plane Figures
16: Area of Circle, Sector and Segment
17: Volumes and Surface Areas of Solids
Statistics and Probability
▶ 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
19: Probability
Chapter 20: Additional Questions
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Solutions for Chapter 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
Below listed, you can find solutions for Chapter 18 of CBSE, Karnataka Board R.S. Aggarwal for Mathematics [English] Class 10.
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18A [Pages 859 - 863]
If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be new mean?
Compute the mean of the following frequency distribution:
| Class | 10 – 30 | 30 – 50 | 50 – 70 | 70 – 90 | 90 – 110 |
| Frequency | 15 | 18 | 25 | 10 | 2 |
Compute the mean of the following frequency distribution:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 | 120 – 140 |
| Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
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 | 6-12 | 12-18 | 18-24 | 24-30 | 30-36 | 36-42 |
| Number of students: | 10 | 11 | 7 | 4 | 4 | 3 | 1 |
The daily expenses of 40 families has been shown by the following frequency distribution:
| Expenses (in ₹) |
500 – 700 | 700 – 900 | 900 – 1100 | 1100 – 1300 | 1300 – 1500 |
| Number of families |
6 | 8 | 10 | 9 | 7 |
Find the mean daily expenses.
During a medical checkup, the number of heartbeats per minute of 40 patients were recorded and summarised as follows:
| Number of heartbeats per minute |
64 – 68 | 68 – 72 | 72 – 76 | 76 – 80 | 80 – 84 | 84 – 88 |
| Number of patients | 6 | 8 | 10 | 12 | 3 | 1 |
Find the mean heartbeats per minute for these patients.
The table given below shows the age distribution of 1000 persons who visited a marketing centre on a Sunday.
| Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Number of persons |
105 | 222 | 220 | 138 | 102 | 113 | 100 |
Find the mean age of the persons visiting the marketing centre on that day.
The arithmetic mean of the following frequency distribution is 53. Find the value of x.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 12 | 15 | 32 | x | 13 |
The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.
| Class interval | Frequency |
| 11 – 13 | 3 |
| 13 – 15 | 6 |
| 15 – 17 | 9 |
| 17 – 19 | 13 |
| 19 – 21 | f |
| 21 – 23 | 5 |
| 23 – 25 | 4 |
If the mean of the following distribution is 54, find the value of p.
| Class Interval | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 7 | p | 10 | 9 | 13 |
The mean of the following frequency 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 |
Find x and y.
The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹ 188.
| Expenditure (in ₹) |
140 – 160 | 160 – 180 | 180 – 200 | 200 – 220 | 220 – 240 |
| Number of families |
5 | 25 | f1 | f2 | 5 |
Find the mean of the following frequency distribution is 57.6 and the total number of observation is 50.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
| Frequency | 7 | `f_1` | 12 | `f_2` | 8 | 5 |
If the mean of the following frequency distribution is 188, find the missing frequencies x and y, if the sum of all frequencies is 100.
| Class | 0 – 80 | 80 – 160 | 160 – 240 | 240 – 320 | 320 – 400 |
| Frequency | 20 | 25 | x | y | 10 |
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 |
Find the mean marks per student, using assumed-mean method:
| Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of students |
12 | 18 | 27 | 20 | 17 | 6 |
Find the mean of the following frequency distribution, using the assumed-mean method:
| Class | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
| Frequency | 10 | 20 | 30 | 15 | 5 |
Find the mean of the following data, using the assumed-mean method:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
| Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
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 |
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 |
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.
Find the mean of the following frequency distribution using step-deviation method:
| Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Frequency | 25 | 40 | 42 | 33 | 10 |
Find the arithmetic mean of the following frequency distribution using step-deviation method:
| Age (in years) | 18 – 24 | 24 – 30 | 30 – 36 | 36 – 42 | 42 – 48 | 48 – 54 |
| Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Find the mean of the following data using step-deviation method:
| Class | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 | 600 – 620 |
| Frequency | 14 | 9 | 5 | 4 | 3 | 5 |
Find the mean age from the following frequency distribution:
| Age (in years) | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |
| Number of persons | 4 | 14 | 22 | 16 | 6 | 5 | 3 |
The following table shows the age distribution of patients of malaria in a village during a particular month:
| Age (in years) | 5 – 14 | 15 – 24 | 25 – 34 | 35 – 44 | 45 – 54 | 55 – 64 |
| No. of cases | 6 | 11 | 21 | 23 | 14 | 5 |
Find the average age of the patients.
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 |
| No. of eggs | 4 | 9 | 13 | 17 | 12 | 3 | 2 |
Calculate their mean weight to the nearest gram.
The following table shows the marks scored by 80 students in an examination:
| Marks | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
| No. of students |
3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18В [Pages 870 - 871]
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 |
| No. of patients | 5 | 20 | 40 | 50 | 25 |
Compute the mean from the following data:
| Marks | 0 – 7 | 7 – 14 | 14 – 21 | 21 – 28 | 28 – 35 | 35 – 42 | 42 – 49 |
| Number of students | 3 | 4 | 7 | 11 | 0 | 16 | 9 |
The following table shows the daily wages of workers in a factory:
| Daily wages in (Rs) | 0 – 100 | 100 – 200 | 200 – 300 | 300 – 400 | 400 – 500 |
| Number of workers | 40 | 32 | 48 | 22 | 8 |
Find the median daily wage income of the workers.
Calculate the median from the following frequency distribution table:
| Class | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |
Given below is the number of units of electricity consumed in a week in a certain locality:
| Class | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 – 200 |
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
Calculate the median.
Calculate the median from the following data:
| Height (in cm) |
135 – 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 |
| Frequency | 6 | 10 | 18 | 22 | 20 | 15 | 6 | 3 |
Calculate the missing frequency from the following distribution, it being given that the median of distribution is 24.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 5 | 25 | ? | 18 | 7 |
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 |
| Frequency | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
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 batsman |
5 | x | y | 12 | 6 | 2 |
If the median of the following frequency distribution is 32.5, find the values of `f_1 and f_2`.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | Total |
| Frequency | `f_1` | 5 | 9 | 12 | `f_2` | 3 | 2 | 40 |
Calculate the median for the following data:
| Class | 19 – 25 | 26 – 32 | 33 – 39 | 40 – 46 | 47 – 53 | 54 – 60 |
| Frequency | 35 | 96 | 68 | 102 | 35 | 4 |
Find the median wages for the following frequency distribution:
| Wages per day (in Rs) |
61 – 70 | 71 – 80 | 81 – 90 | 91 – 100 | 101 – 110 | 111 – 120 |
| No. of women workers |
5 | 15 | 20 | 30 | 20 | 8 |
Find the median from the following data:
| Class | 1 – 5 | 6 – 10 | 11 – 15 | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 7 | 10 | 16 | 32 | 24 | 16 | 11 | 5 | 2 |
Find the median from the following data:
| Marks | No of students |
| Below 10 | 12 |
| Below 20 | 32 |
| Below 30 | 57 |
| Below 40 | 80 |
| Below 50 | 92 |
| Below 60 | 116 |
| Below 70 | 164 |
| Below 80 | 200 |
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18C [Pages 877 - 879]
Find the mode of the following distribution:
| Marks | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 12 | 35 | 45 | 25 | 13 |
Compute the mode of the following data:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 25 | 16 | 28 | 20 | 5 |
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.
Find the mode of the following distribution:
| Class interval |
10 – 14 | 14 – 18 | 18 – 22 | 22 – 26 | 26 – 30 | 30 – 34 | 34 – 38 | 38 – 42 |
| Frequency | 8 | 6 | 11 | 20 | 25 | 22 | 10 | 4 |
Given below is the distribution of total household expenditure of 200 manual workers in a city:
| Expenditure (in Rs) | 1000 – 1500 | 1500 – 2000 | 2000 – 2500 | 2500 – 3000 | 3000 – 3500 | 3500 – 4000 | 4000 – 4500 | 4500 – 5000 |
| Number of manual workers |
24 | 40 | 31 | 28 | 32 | 23 | 17 | 5 |
Find the average expenditure done by maximum number of manual workers.
Calculate the mode from the following data:
| Monthly salary (in Rs) | No of employees |
| 0 – 5000 | 90 |
| 5000 – 10000 | 150 |
| 10000 – 15000 | 100 |
| 15000 – 20000 | 80 |
| 20000 – 25000 | 70 |
| 25000 – 30000 | 10 |
Compute the mode from the following data:
| Age (in years) | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
| No of patients | 6 | 11 | 18 | 24 | 17 | 13 | 5 |
Compute the mode from the following series:
| Size | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 | 95 – 105 | 105 – 115 |
| Frequency | 7 | 12 | 17 | 30 | 32 | 6 | 10 |
Compute the mode from the following data:
| Class interval | 1 – 5 | 6 – 10 | 11 – 15 | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 | 41 – 45 | 46 – 50 |
| Frequency | 3 | 8 | 13 | 18 | 28 | 20 | 13 | 8 | 6 | 4 |
The agewise 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.
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18D [Pages 881 - 882]
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 | 4 | 4 | 7 | 10 | 12 | 8 | 5 |
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 |
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 |
Find the mean, median and mode of the following data:
| Marks obtained | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 |
| No. of students | 7 | 31 | 33 | 17 | 11 | 1 |
A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:
| Height (in cm) | 120 – 130 | 130 – 140 | 140 – 150 | 150 – 160 | 160 – 170 | Total |
| No. of girls | 2 | 8 | 12 | 20 | 8 | 50 |
Find the mean, median and mode of the above data.
The following table gives the daily income of 50 workers of a factory:
| Daily income (in Rs) | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
| No. of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean, median and mode of the above data.
The table below shows the daily expenditure on food of 30 households in a locality:
| Daily expenditure (in Rs) | Number of households |
| 100 – 150 | 6 |
| 150 – 200 | 7 |
| 200 – 250 | 12 |
| 250 – 300 | 3 |
| 300 – 350 | 2 |
Find the mean and median daily expenditure on food.
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18E [Pages 892 - 895]
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 |
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 than type’ ogive from the above data. Find the median.
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 |
| No. of students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
The heights 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 |
| No of Students | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
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.
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 | 238 | 16 |
Change the distribution to a ‘more than type’ distribution and draw its ogive. Using ogive, find the median of the given data.
The following table gives production yield in kg per hectare of wheat of 100 farms of a village:
| Production yield |
40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 |
| Number of farms |
4 | 6 | 16 | 20 | 30 | 24 |
Change the distribution to 'a more than' type distribution, and draw its ogive.
From the following frequency distribution, prepare the ‘more than’ ogive.
| Score | Number of candidates |
| 400 – 450 | 20 |
| 450 – 500 | 35 |
| 500 – 550 | 40 |
| 550 – 600 | 32 |
| 600 – 650 | 24 |
| 650 – 700 | 27 |
| 700 – 750 | 18 |
| 750 – 800 | 34 |
| Total | 230 |
Also, find the median.
The marks obtained by 100 students of a class in an examination are given below:
| Marks | Number of students |
| 0 – 5 | 2 |
| 5 – 10 | 5 |
| 10 – 15 | 6 |
| 15 – 20 | 8 |
| 20 – 25 | 10 |
| 25 – 30 | 25 |
| 30 – 35 | 20 |
| 35 – 40 | 18 |
| 40 – 45 | 4 |
| 45 – 50 | 2 |
Draw cumulative frequency curves by using (i) ‘less than’ series and (ii) ‘more than’ series. Hence, find the median.
From the following data, draw the two types of cumulative frequency curves and determine the median:
| Marks | Frequency |
| 140 – 144 | 3 |
| 144 – 148 | 9 |
| 148 – 152 | 24 |
| 152 – 156 | 31 |
| 156 – 160 | 42 |
| 160 – 164 | 64 |
| 164 – 168 | 75 |
| 168 – 172 | 82 |
| 172 – 176 | 86 |
| 176 – 180 | 34 |
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive EXERCISE 18F [Pages 896 - 899]
Very-Short-Answer Questions
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 |
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 |
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 give class mark of the modal class.
A data has 25 observations arranged in a descending order. Which observation represents the median?
For a certain distribution, mode and median were found to be 1000 and 1250 respectively. Find mean for this distribution using an empirical relation.
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.
Find the class marks of classes 10 – 25 and 35 – 55.
While calculating the mean of a given data by the assumed-mean method, the following values were obtained.
`A = 25, sum f_i d_i = 110, sum f_i = 50`.
Find the mean.
The distribution 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?
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?
The observation 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?
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.
| HINT | Since 8 is less than 30 and 32 is more than 30, so the value of median (middle value) remains unchanged. |
If the median of `x/5, x/4, x/2, x and x/3`, where x > 0, is 8, find the value of x.
| HINT | Arranging the observations in ascending order, we have `x/5, x/4, x/3, x/2, x` Median `= x/3 = 8.` |
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 | 54 | 21 | 16 |
Short-Answer Questions
Find the mode of the given data:
| Class interval | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 |
| Frequency | 15 | 6 | 18 | 10 |
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
| Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Number of patients |
6 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.
In the following data, find the values 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 |
The following frequency distribution gives the monthly consumption of electricity of 64 consumers of 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.
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 | 9 | 91 | 100 |
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.
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 |
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 |
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive MULTIPLE-CHOICE QUESTIONS (MCQ) [Pages 900 - 904]
Choose the correct answer in each of the following questions:
Which of the following is not a measure of central tendency?
Mean
Mode
Median
Range
Which of the following cannot be determined graphically?
Mean
Median
Mode
None of these
Which of the following measures of central tendency is influenced by extreme values?
Mean
Medain
Mode
None of these
The mode of a frequency distribution is obtained graphically from ______.
a frequency curve
a frequency polygon
a histogram
an ogive
The median of a frequency distribution is found graphically with the help of ______.
a histogram
a frequency curve
a frequency polygon
ogives
The cumulative frequency table is useful in determining ______.
Mean
Median
Mode
All of these
The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its ______.
Mean
Median
Mode
All of the above
If xi's are the midpoints of the class intervals of a grouped data, fi's are the corresponding frequencies and `barx` is the mean then `sumf_i(x_i - barx) = ?`
1
0
–1
2
For finding the mean by using the formula, `barx = A + h((sumf_iu_i)/(sumf_i))`, we have ui = ?
`((A - x_i))/h`
`((x_i - A))/h`
`((A + x_i))/h`
h(xi – A)
In the formula, `barx = {A + (sumf_i d_i)/(sumf_i)}` for finding the mean of the grouped data, the di's are the deviations from A of ______.
lower limits of the classes
upper limits of the classes
midpoints of the classes
none of these
While computing mean of grouped data, we assume that the frequencies are ______.
evenly distributed over the classes
centred at the class marks of the classes
centred at the lower limit of the classes
centred at the upper limit of the classes
The relation between mean, mode and median is ______.
mode = (3 × mean) – (2 × median)
mode = (3 × median) – (2 × mean)
median = (3 × mean) – (2 × mode)
mean = (3 × median) – (2 × mode)
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 ______.
5.5
15.5
20.5
36.0
Consider the frequency distribution of the heights 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 ______.
310
315
320
330
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 ______.
10 – 20
20 – 30
30 – 40
50 – 60
Mode = ?
`x_k + h · {((f_(k - 1) - f_k))/((2f_k - f_(k - 1) - f_(k + 1)))}`
`x_k + h · {((f_k - f_(k - 1)))/((2f_k - f_(k - 1) - f_(k + 1)))}`
`x_k + h · {((f_k - f_(k - 1)))/((f_k - 2f_(k - 1) - f_(k + 1)))}`
`x_k + h · {((f_k - f_(k - 1)))/((f_k - f_(k - 1) - 2f_(k + 1)))}`
Median = ?
`l + {h xx ((N/2 - cf))/f}`
`l + {h xx ((cf - N/2 ))/f}`
`l - {h xx ((N/2 - cf))/f}`
None of these
If the mean and median of a set of numbers are 8.9 and 9 respectively then the mode will be ______.
7.2
8.2
9.2
10.2
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 ______.
56.5
57.5
58.5
59
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 ______.
23.5
24
24.4
25
The mean and mode of a frequency distribution are 28 and 16 respectively. The median is ______.
22
23.5
24
24.5
The median and mode respectively of a frequency distribution are 26 and 29. Then its mean is ______.
27.5
24.5
28.4
25.8
For a symmetrical frequency distribution, we have
mean < mode < median
mean > mode > median
mean = mode = median
mode = `1/2` (mean + median)
Look at the cumulative frequency distribution table given below:
| Monthly income | Number of families |
| More than ₹ 10000 | 100 |
| More than ₹ 14000 | 85 |
| More than ₹ 18000 | 69 |
| More than ₹ 20000 | 50 |
| More than ₹ 25000 | 37 |
| More than ₹ 30000 | 15 |
Number of families having income range 20000 to 25000 is ______.
19
16
13
22
The median of first 8 prime numbers is ______.
7
9
11
13
The mean of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
0
1
10
19
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 ______.
13
14
15
16
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?
5
10
20
30
Matching of columns
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 |
Assertion-and-Reason Type Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
| 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. |
Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
Assertion (A) is true and Reason (R) is false.
Assertion (A) is false and Reason (R) is true.
| Assertion (A) | Reason (R) | ||||||||||||||
Consider the following frequency distribution:
The mode of the above data is 12.4. |
The value of the variable which occurs most often is the mode. |
Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
Assertion (A) is true and Reason (R) is false.
Assertion (A) is false and Reason (R) is true.
R.S. Aggarwal solutions for Mathematics [English] Class 10 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive TEST YOURSELF [Pages 907 - 910]
MCQ
Which one of the following measures is determined only after the construction of cumulative frequency distribution?
Mean
Median
Mode
None of these
If the mean of a data is 27 and its median is 33 then the mode is ______.
30
43
45
47
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 ______.
15
25
30
35
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 ______.
16.5
18.5
18
17.5
Very-Short-Answer Questions
If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.
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 |
| Frequencу | 2 | 4 | 15 | 54 | 25 | 20 |
Find the number of athletes who completed the race in less than 14.6 seconds.
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.
The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below:
| Profit (in lakhs ₹) | 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 – 25.
Short-Answers Questions
Find the mean of the following frequency distribution:
| Class | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 |
| Frequency | 9 | 22 | 27 | 18 |
The maximum bowling speeds (in km/hr) of 30 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.
The arithmetic mean of the following frequency distribution is 50.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 16 | p | 30 | 32 | 14 |
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 |
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.
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 |
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 |
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 |
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 |
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 |
Long-Answer Questions
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 |
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 |
| Frequqncy | 2 | 12 | 2 | 4 | 3 | 4 | 3 |
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 data.
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.
Solutions for 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
R.S. Aggarwal solutions for Mathematics [English] Class 10 chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
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