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2: Polynomials
3: Pair of Linear Equations in Two Variables
4: Quadratic Equations
5: Arithmetic Progressions
6: Co-ordinate Geometry
7: Triangles
8: Circles
9: Constructions
10: Trigonometric Ratios
11: Trigonometric Identities
12: Heights and Distances
13: Areas Related to Circles
14: Surface Areas and Volumes
▶ 15: Statistics
16: Probability
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Solutions for Chapter 15: Statistics
Below listed, you can find solutions for Chapter 15 of CBSE, Karnataka Board R.D. Sharma for Mathematics [English] Class 10.
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.1 [Pages 15.4 - 15.5]
BASIC
Calculate the mean for the following distribution:-
| x | 5 | 6 | 7 | 8 | 9 |
| f | 4 | 8 | 14 | 11 | 3 |
Find the missing value of p for the following distribution whose mean is 12.58
| x | 5 | 8 | 10 | 12 | P | 20 | 25 |
| f | 2 | 5 | 8 | 22 | 7 | 4 | 2 |
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
| Age (in years) | 15 | 16 | 17 | 18 | 19 | 20 |
| No. of students | 3 | 8 | 10 | 10 | 5 | 4 |
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
| No. of heads per toss | No. of tosses |
| 0 | 38 |
| 1 | 144 |
| 2 | 342 |
| 3 | 287 |
| 4 | 164 |
| 5 | 25 |
| Total | 1000 |
The arithmetic mean of the following data is 14. Find the value of k
| x1 | 5 | 10 | 15 | 20 | 25 |
| f1 | 7 | k | 8 | 4 | 5 |
The arithmetic mean of the following data is 25, find the value of k.
| x1 | 5 | 15 | 25 | 35 | 45 |
| f1 | 3 | k | 3 | 6 | 2 |
If the mean of the following data is 18.75. Find the value of p.
| xi | 10 | 15 | P | 25 | 30 |
| fi | 5 | 10 | 7 | 8 | 2 |
BASED ON LOTS
Find the value of p, if the mean of the following distribution is 20.
| x | 15 | 17 | 19 | 20+P | 23 |
| f | 2 | 3 | 4 | 5P | 6 |
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
| x | 10 | 30 | 50 | 70 | 90 | |
| f | 17 | f1 | 32 | f2 | 19 | Total 120 |
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.2 [Page 15.11]
BASIC
The number of telephone calls received at an exchange per interval for 250 successive one minute intervals are given in the following frequency table:
| No. of calls(x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of intervals (f) | 15 | 24 | 29 | 46 | 54 | 43 | 39 |
Compute the mean number of calls per interval.
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
| No. of heads per toss | No. of tosses |
| 0 | 38 |
| 1 | 144 |
| 2 | 342 |
| 3 | 287 |
| 4 | 164 |
| 5 | 25 |
| Total | 1000 |
The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
| Marks(x) | 15 | 20 | 22 | 24 | 25 | 30 | 33 | 38 | 45 |
| Frequency (f) | 5 | 8 | 11 | 20 | 23 | 18 | 13 | 3 | 1 |
Find the average number of marks.
The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:
| No. of students absent (x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| No. of days (f) | 1 | 4 | 10 | 50 | 34 | 15 | 4 | 2 |
Find the mean number of students absent per day.
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.3 [Pages 15.18 - 15.20]
BASIC
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?
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Fine the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean of the following distribution:
| Class: | Frequency: |
| 3 – 5 | 55 |
| 5 – 7 | 10 |
| 7 – 9 | 10 |
| 9 – 11 | 7 |
| 11 – 13 | 8 |
The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
| Class interval | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
| Frequency | 7 | 6 | 9 | 13 | - | 5 | 4 |
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 |
The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in Rs) | 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.
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 |
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 |
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 |
The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .
| Age below (in years): | 30 | 40 | 50 | 60 | 70 | 80 |
| Number of persons : | 100 | 220 | 350 | 750 | 950 | 1000 |
The marks obtained by 110 students in an examination are given below:
| Marks: | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 |
| Frequencу: | 14 | 16 | 28 | 23 | 18 | 8 | 3 |
Find the mean marks of the students.
BASED ON LOTS
Find the mean of the following data using assumed mean method:
| Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
| Frequency | 8 | 7 | 10 | 13 | 12 |
Find the mean of each of the following frequency distributions
| Classes | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
| Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
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 mangoe | 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?
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
| concentration of SO2 (in ppm) | Frequency |
| 0.00 − 0.04 | 4 |
| 0.04 − 0.08 | 9 |
| 0.08 − 0.12 | 9 |
| 0.12 − 0.16 | 2 |
| 0.16 − 0.20 | 4 |
| 0.20 − 0.24 | 2 |
Find the mean concentration of SO2 in the air.
If the mean of the following frequency distribution is 18, find the missing frequency.
| Class interval | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
| Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
The daily income of a sample of 50 employees are tabulated as follows:
| Income (in Rs.): | 1-1200 | 201 -400 | 401-600 | 601 - 800 |
| No.of employees : | 14 | 15 | 14 | 7 |
Find the mean daily income of employees.
The mean of the following frequency distribution is 25. Find the value of f.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 5 | 18 | 15 | f | 6 |
The following table shows the age of patients admitted in a hospital during a particular week:
| Age (in years): | 5 - 15 | 15 - 25 | 25 - 35 | 35 - 45 | 45 - 55 | 55 - 65 |
| Number of patients: | 5 | 12 | 20 | 24 | 15 | 4 |
Find the mean age of patients.
Find the mean of the following frequency distribution:
| Classes: | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 | 55 - 60 |
| Frequency: | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
In a test, the marks obtained by 100 students (out of 50) are given below:
| Marks obtained: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Number of students: | 12 | 23 | 34 | 25 | 6 |
Find the mean marks of the students.
The following distribution shows the weekly pocket allowance of 64 children of a locality.If the mean pocket allowance is ₹ 180, find the values of x and y.
| Pocket allowance (in ₹): |
110-130 | 130-150 | 150-170 | 170-190 | 190-210 | 210-230 | 230-250 |
| Number of children: |
7 | 6 | 9 | 13 | x | 5 | y |
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.4 [Pages 15.28 - 15.31]
BASIC
Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.
The table below shows the salaries of 280 persons:
| Salary (In thousand Rs) | No. of Persons |
| 5-10 | 49 |
| 10-15 | 133 |
| 15-20 | 63 |
| 20-25 | 15 |
| 25-30 | 6 |
| 30-35 | 7 |
| 35-40 | 4 |
| 40-45 | 2 |
| 45-50 | 1 |
Calculate the median salary of the data.
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 |
| No. of persons | 5 | 25 | ? | 18 | 7 |
Find the following table gives the distribution of the life time of 400 neon lamps:
| Life time (in hours) | Number of lamps |
| 1500 – 2000 | 14 |
| 2000 – 2500 | 56 |
| 2500 – 3000 | 60 |
| 3000 – 3500 | 86 |
| 3500 – 4000 | 74 |
| 4000 – 4500 | 62 |
| 4500 – 5000 | 48 |
Find the median life time of a lamp.
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 |
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
| Height in cm | Number of Girls |
| Less than 140 | 4 |
| Less than 145 | 11 |
| Less than 150 | 29 |
| Less than 155 | 40 |
| Less than 160 | 46 |
| Less than 165 | 51 |
Find the median height.
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 years.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| 20 - 25 | 4 |
| 25 - 30 | 18 |
| 30 - 35 | 21 |
| 35 - 40 | 33 |
| 40 - 45 | 11 |
| 45 - 50 | 3 |
| 50 - 55 | 6 |
| 55 - 60 | 2 |
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 | 3 |
| 127–135 | 5 |
| 136−144 | 9 |
| 145–153 | 12 |
| 154–162 | 5 |
| 163–171 | 4 |
| 172–180 | 2 |
Find the mean length of the leaves.
The frequency distribution given below shows the weight of 40 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 |
| No. of Students: | 9 | 5 | 8 | 9 | 6 | 3 |
BASED ON LOTS
An incomplete distribution is given below:
| Variable: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency: | 12 | 30 | - | 65 | - | 25 | 18 |
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
| Class interval : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency : | 4 | x | 5 | y | 1 |
The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.
| Marks: | 20 -30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
| Frequency: | P | 15 | 25 | 20 | q | 8 | 10 |
Heights of 50 students of class X of a school are recorded and following data is obtained:
| Height (in cm) | 130 – 135 | 135 – 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 |
| Number of students | 4 | 11 | 12 | 7 | 10 | 6 |
Find the median height of the students.
The monthly expenditure on milk in 200 families of a Housing Society is given below:
| Monthly Expenditure (in ₹) |
1000 – 1500 | 1500 – 2000 | 2000 – 2500 | 2500 – 3000 | 3000 – 3500 | 3500 – 4000 | 4000 – 4500 | 4500 – 5000 |
| Number of families | 24 | 40 | 33 | x | 30 | 22 | 16 | 7 |
Find the value of x and also, find the median and mean expenditure on milk.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
| Number of cars | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
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 | 3 |
| 127 – 135 | 5 |
| 136 − 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)
Find the mean and median for the following data:
| Classes: | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
| Frequencу: | 2 | 3 | 5 | 7 | 4 | 2 | 2 |
Medical check-up was carried out for 35 students of a class and their weights were recorded as follows:
| Weight (in kg): | 38-40 | 40-42 | 42-44 | 44-46 | 46-48 | 48-50 | 50-52 |
| Number of students: | 3 | 2 | 4 | 5 | 14 | 4 | 3 |
Find the difference between the mean weight and the median weight.
During a medical checkup, height of 35 students of a class were recorded as follows:
| Height (in cm): | 90-100 | 100-110 | 110-120 | 120-130 | 130-140 | 140-150 |
| Number of Students: | 3 | 2 | 4 | 5 | 14 | 7 |
Find the difference between the mean height and median height.
The following data shows the number of family members living in different bungalows of a locality:
| Number of Members | 0 − 2 | 2 − 4 | 4 − 6 | 6 − 8 | 8 − 10 | Total |
| Number of Bungalows | 10 | p | 60 | q | 5 | 120 |
If the median number of members is found to be 5, find the values of p and q.
The population of lions was noted in different regions across the world in the following table:
| Number of lions | Number of regions |
| 0 − 100 | 2 |
| 100 − 200 | 5 |
| 200 − 300 | 9 |
| 300 − 400 | 12 |
| 400 − 500 | x |
| 500 − 600 | 20 |
| 600 − 700 | 15 |
| 700 − 800 | 9 |
| 800 − 900 | y |
| 900 − 1000 | 2 |
| 100 |
If the median of the given data is 525, find the values of x and y.
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.5 [Pages 15.40 - 15.42]
BASIC
Find the mode of the following data:
3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Find the mode of the following data:
3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Find the mode of the following data:
15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:
| Shirt size: | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 |
| Number of persons: | 15 | 25 | 39 | 41 | 36 | 17 | 15 | 12 |
Find the model shirt size worn by the group.
Find the mode of the following distribution.
| Class-interval: | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 |
| Frequency: | 25 | 34 | 50 | 42 | 38 | 14 |
Find the mode of the following frequency distribution.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |
Compare the modal ages of two groups of students appearing for an entrance test:
| Age (in years): | 16-18 | 18-20 | 20-22 | 22-24 | 24-26 |
| Group A: | 50 | 78 | 46 | 28 | 23 |
| Group B: | 54 | 89 | 40 | 25 | 17 |
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.
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 |
| Number of workers: | 12 | 14 | 8 | 6 | 10 |
Find the mean, mode and median of the above data.
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 | 3 |
| 20 − 25 | 8 |
| 25 − 30 | 9 |
| 30 − 35 | 10 |
| 35 − 40 | 3 |
| 40 − 45 | 0 |
| 45 − 50 | 0 |
| 50 − 55 | 2 |
Find the mean, median and mode of the following data:
| Classes: | 0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
| Frequency: | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
| Number of cars | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Find the mean, median and mode of the following data:
| Classes: | 0-20 | 20-40 | 40-60 | 40-60 | 80-100 | 100-120 | 120-140 |
| Frequency: | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
The frequency distribution for agriculture holdings in a village is given below:
| Area of land (in hectares) | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 | 11 – 13 |
| Number of families | 20 | 45 | 80 | 55 | 40 | 12 |
Find the modal agriculture holdings of the village.
BASED ON LOTS
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 | 4 |
| 4000 − 5000 | 18 |
| 5000 − 6000 | 9 |
| 6000 − 7000 | 7 |
| 7000 − 8000 | 6 |
| 8000 − 9000 | 3 |
| 9000 − 10000 | 1 |
| 10000 − 11000 | 1 |
Find the mode of the data.
The monthly income of 100 families are given as below :
| Income in ( in ₹) | Number of families |
| 0-5000 | 8 |
| 5000-10000 | 26 |
| 10000-15000 | 41 |
| 15000-20000 | 16 |
| 20000-25000 | 3 |
| 25000-30000 | 3 |
| 30000-35000 | 2 |
| 35000-40000 | 1 |
Calculate the modal income.
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table:
| Mass (in grams) |
80 – 100 | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 |
| Number of apples |
20 | 60 | 70 | x | 60 |
Find the value of x and the mean mass of the apples.
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table:
| Mass (in grams) |
80 – 100 | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 |
| Number of apples |
20 | 60 | 70 | x | 60 |
Find the modal mass of the apples.
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.
Find the mode of the following data:
| Class: | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |
| Frequency: | 7 | 8 | 2 | 2 | 1 |
Find the mean and Mode of the following frequency distribution:
| Class: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequencу: | 8 | 7 | 15 | 20 | 12 | 8 | 10 |
Find the Mean and Mode of the following data:
| Class: | 4-8 | 8-12 | 12-16 | 16-20 | 20-24 | 24-28 | 28-32 | 32-36 |
| Frequency: | 2 | 12 | 15 | 25 | 18 | 12 | 13 | 3 |
The following table shows the number of patients of different age groups who were discharged from the hospital in a particular month:
| Age (in years) | Number of Patients Discharged |
| 5 − 15 | 6 |
| 15 − 25 | 11 |
| 25 − 35 | 21 |
| 35 − 45 | 23 |
| 45 − 55 | 14 |
| 55 − 65 | 5 |
| Total | 80 |
Find the ‘mean’ and the ‘mode’ of the above data.
The following table gives the daily income of 50 cab drivers of a particular city
| Income (₹) | 500 - 600 | 600 - 700 | 700 - 800 | 800 - 900 | 900 - 1000 |
| No. of Drivers | 12 | 14 | 8 | 6 | 10 |
Find the mean income and the modal income.
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics EXERCISE 15.6 [Page 15.53]
BASIC
Draw an ogive to represent the following frequency distribution:
| Class-interval: | 0 - 4 | 5 - 9 | 10 - 14 | 15 - 19 | 20 - 24 |
| Frequency: | 2 | 6 | 10 | 5 | 3 |
The monthly profits (in Rs.) of 100 shops are distributed as follows:
| Profits per shop: | 0 - 50 | 50 - 100 | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 |
| No. of shops: | 12 | 18 | 27 | 20 | 17 | 6 |
Draw the frequency polygon for it.
The following distribution gives the daily income of 50 workers of a factory.
| Daily income (in Rs | 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.
The following table gives production yield per hectare of wheat of 100 farms of a village:
| Production yield in kg per hectare: | 50 - 55 | 55 - 60 | 60 - 65 | 65 - 70 | 70 - 75 | 75 - 80 |
| Number of farms: | 2 | 8 | 12 | 24 | 38 | 16 |
Draw ‘less than’ ogive and ‘more than’ ogive.
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 | 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 |
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
The annual rainfall record of a city for 66 days is given in the following table :
| Rainfall (in cm ): | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Number of days : | 22 | 10 | 8 | 15 | 5 | 6 |
Calculate the median rainfall using ogives of more than type and less than type.
Change the following distribution to a 'more than type' distribution. Hence draw the 'more than type' ogive for this distribution.
| Class interval: | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | 80−90 |
| Frequency: | 10 | 8 | 12 | 24 | 6 | 25 | 15 |
The following distribution gives the daily income of 50 workers of a factory.
| Daily income (in ₹) | 200-220 | 220-240 | 240-260 | 260-280 | 280-300 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Convert the distribution above to a 'less than type' cumulative frequency distribution and draw its ogive.
BASED ON LOTS
The following table gives the height of trees:
| Height | No. of trees |
| Less than 7 Less than 14 Less than 21 Less than 28 Less than 35 Less than 42 Less than 49 Less than 56 |
26 57 92 134 216 287 341 360 |
Draw 'less than' ogive and 'more than' ogive.
The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:
| Profit (in lakhs in Rs) | Number of shops (frequency) |
| More than or equal to 5 More than or equal to 10 More than or equal to 15 More than or equal to 20 More than or equal to 25 More than or equal to 30 More than or equal to 35 |
30 28 16 14 10 7 3 |
Draw both ogives for the above data and hence obtain the median.
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics VERY SHORT ANSWER TYPE QUESTIONS (VSAQS) [Page 15.54]
BASIC
Define mean.
What is the algebraic sum of deviation of a frequency distribution about its mean?
Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?
Write the empirical relation between mean, mode and median.
Which measure of central tendency can be determine graphically?
Find the mode of the following frequency distribution:
| Classes: | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 |
| Frequencies: | 8 | 7 | 12 | 5 | 3 |
If mode of the following frequency distribution is 55, then find the value of x.
| Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
| Frequency | 10 | 7 | x | 15 | 10 | 12 |
Write the modal class for the following frequency distribution:
| Class-interval: | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
| Frequency: | 30 | 35 | 75 | 40 | 30 | 15 |
Write the median class for the following frequency distribution:
| Class-interval: | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
| Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
In the graphical representation of a frequency distribution, if the distance between mode and mean is ktimes the distance between median and mean, then write the value of k.
Find the class marks of classes 10−25 and 35−55.
Write the median class of the following distribution:
| Class-interval: | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
| Frequency: | 4 | 4 | 8 | 10 | 12 | 8 | 4 |
R.D. Sharma solutions for Mathematics [English] Class 10 15 Statistics FILL IN THE BLANK TYPE QUESTIONS (FBQS) [Page 15.55]
BASIC
If the mean of x, y, z is y, then mean of x and z is ______.
The mode of the data 2, x, 3, 4, 5, 2, 4, 6, where x > 2, is ______.
The mean of the first 673 natural numbers is ______.
The mean of the observations 425, 430, 435, ..., 495 is ______.
If the mean and median of a unimodal data are 34.5 and 32.5 respectively, then the mode of the data is ______.
The mean of the first n odd natural numbers is ______.
The mean of the observations 1, 3, 5, 7, 9, ..., 99 is ________.
If the mean of the first n natural numbers is 20, then n = ______.
If a mode exceeds a mean by 12, then the mode exceeds the median by ______.
BASED ON LOTS
If the median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m, then the median of x3, x4, x5, x6 (where x1 < x2 < x3 < x4 < x5 < x6 < x7 < x8) is ______.
If mode − median = 2, then median − mean = ______.
If the average of a, b, c, d is the average of b and c, then the value of a − b − c + d is ______.
Given that a, b, c, d are non-zero integers such that a < b < c < d. If the mean and median of a, b, c, d are equal to zero, then a = ______ and b = ______.
If a < b < 2a and the mean and median of a, b and 2a are 15 and 12 respectively, then a = ______.
If the mean of 26, 19, 15, 24 and x is x, then the median of the data is ______.
Solutions for 15: Statistics
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R.D. Sharma solutions for Mathematics [English] Class 10 chapter 15 - Statistics
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