Force: It is the force that enables us to do any work. To do anything, either we pull or push the object. Therefore, pull or push is called force.
Pdf is Loading Please Wait …….
Motion can be of different types depending upon the type of path by which the object is going through..
Pdf is Loading Please Wait …….
The method of arranging organisms into groups or sets on the basis of similarities and differences is called classification.
Pdf is Loading Please Wait …….
Pdf is Loading Please Wait …….
Cell is the structural and functional unit of life. It is the basic unit of life.
Pdf is Loading Please Wait …….
→ John Dalton considered atom to be an indivisible entity, but his concept had to be discarded at the end of nineteenth century, when scientists through experiments were able to find existence of charged (electrons and protons) and neutral particles (neutrons) in the atom. These particles were called the ‘Sub-atomic Particles’.
Pdf is Loading Please Wait …….
→ Around 500 B.C., Indian philosopher Maharishi Kanad, postulated the theory if we go on dividing matter (padarth), we will obtain smallest particle beyond which further division can’t be possible which is known as ‘parmanu’.
Pdf is Loading Please Wait …….
Introduction
Anything which occupies space and has mass is called matter. Matter can be divided in two categories.
(i) Pure Substance: It consists of single types of particles which are same in their chemical nature.
(ii) Mixtures: Mixture consists of two or more particles.
Pdf is Loading Please Wait …….
In This Post we are providing Chapter 15 Probability NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Probability Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Probability NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
| Number of girls in a family | 2 | 1 | 0 |
| Number of families | 475 | 814 | 211 |
| Outcome | 3 heads | 2 heads | 1 head | No head |
| Frequency | 23 | 72 | 77 | 28 |
| Monthly income (in ₹) | Vehicles per family | |||
| 0 | 1 | 2 | Above 2 | |
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000-10000 | 0 | 305 | 27 | 2 |
| 10000-13000 | 1 | 535 | 29 | 1 |
| 13000-16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
| Marks | Number of students |
| 0 – 20 | 7 |
| 20 – 30 | 10 |
| 30 – 40 | 10 |
| 40 – 50 | 20 |
| 50 – 60 | 20 |
| 60 – 70 | 15 |
| 70 – above | 8 |
| Total | 90 |
| Opinion | Number of students |
| like | 135 |
| dislike | 65 |
In This Post we are providing Chapter 14 Statistics NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Statistics Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Statistics NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise 14.1
1. Give five examples of data that you can collect from your day-to-day life.
Answer
Five examples from day-to-day life:
(i) Daily expenditures of household.
(ii) Amount of rainfall.
(iii) Bill of electricity.
(iv) Poll or survey results.
(v) Marks obtained by students.
2. Classify the data in Q.1 above as primary or secondary data.
Answer
Primary Data: (i) (iii) and (v)
Secondary Data: (ii) and (iv)
Exercise 14.2
1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Answer
The frequency means the number of students having same blood group. We will represent the data in table:
| Blood Group | Number of Students (Frequency) |
| A | 9 |
| B | 6 |
| O | 12 |
| AB | 3 |
| Total | 30 |
Most common Blood Group (Highest frequency): O
Rarest Blood Group (Lowest frequency): AB
2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Answer
The given data is very large. So, we construct a group frequency of class size 5. Therefore, class interval will be 0-5, 5-10, 10-15, 15-20 and so on. The data is represented in the table as:
The classes in the table are not overlapping. Also, 36 out of 40 engineers have their house below 20 km of distance.
3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Answer
(i) The given data is very large. So, we construct a group frequency of class size 2. Therefore, class interval will be 84-86, 86-88, 88-90, 90-92 and so on. The data is represented in the table as:
| Relative humidity (in %) | Frequency |
| 84-86 | 1 |
| 86-88 | 1 |
| 88-90 | 2 |
| 90-92 | 2 |
| 92-94 | 7 |
| 94-96 | 6 |
| 96-98 | 7 |
| 98-100 | 4 |
| Total | 30 |
(ii) The humidity is very high in the data which is observed during rainy season. So, it must be rainy season.
(iii) Range of data = Maximum value of data – Minimum = 99.2 − 84.9 = 14.3
4. The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table?
Answer
(i) The data with class interval 160-165, 165-170 and so on is represented in the table as:
| Height (in cm) | No. of Students (Frequency) |
| 150-155 | 12 |
| 155-160 | 9 |
| 160-165 | 14 |
| 165-170 | 10 |
| 170-175 | 5 |
| Total | 50 |
(ii) From the given data, it can be concluded that 35 students i.e. more than 50% are shorter than 165 cm.
5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Answer
(i) The data with class interval 0.00 – 0.04, 0.04 – 0.08 and so on is represented in the table as:
| Concentration of sulphur dioxide in air (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 |
| Total | 30 |
(ii) 2 + 4 + 2 = 8 days have the concentration of sulphur dioxide more than 0.11 parts per million.
Page No. 246
6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
Answer
The frequency distribution table for the data given above can be prepared as follow:
| Number of Heads | Frequency |
| 0 | 6 |
| 1 | 10 |
| 2 | 9 |
| 3 | 5 |
| Total | 30 |
7. The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Answer
(i)The frequency is given as follow:
| Digits | Frequency |
| 0 | 2 |
| 1 | 5 |
| 2 | 5 |
| 3 | 8 |
| 4 | 4 |
| 5 | 5 |
| 6 | 4 |
| 7 | 4 |
| 8 | 5 |
| 9 | 8 |
| Total | 30 |
(ii) The digit having the least frequency occurs the least and the digit with highest frequency occurs the most. 0 has frequency 2 and thus occurs least frequently while 3 and 9 have frequency 8 and thus occur most frequently.
8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?
Answer
(i) The distribution table for the given data, taking class width 5 and one of the class intervals as 5-10 is as follows:
| Number of Hours | Frequency |
| 0-5 | 10 |
| 5-10 | 13 |
| 10-15 | 5 |
| 15-20 | 2 |
| Total | 30 |
(ii) We observed from the given table that 2 children television for 15 or more hours a week.
9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Answer
A grouped frequency distribution table using class intervals of size 0.5 starting from the interval 2 – 2.5 is constructed.
| Lives of batteries (in years) | No. of batteries (Frequency) |
| 2-2.5 | 2 |
| 2.5-3 | 6 |
| 3-3.5 | 14 |
| 3.5-4 | 11 |
| 4-4.5 | 4 |
| 4.5-5 | 3 |
| Total | 40 |
| S.No. | Causes | Female fatality rate (%) |
| 1. | Reproductive health conditions | 31.8 |
| 2. | Neuropsychiatric conditions | 25.4 |
| 3. | Injuries | 12.4 |
| 4. | Cardiovascular conditions | 4.3 |
| 5. | Respiratory conditions | 4.1 |
| 6. | Other causes | 22.0 |
| S.No. | Section | Number of girls per thousand boys |
| 1. | Scheduled Caste (SC) | 940 |
| 2. | Scheduled Tribe (ST) | 970 |
| 3. | Non SC/ST | 920 |
| 4. | Backward districts | 950 |
| 5. | Non-backward districts | 920 |
| 6. | Rural | 930 |
| 7. | Urban | 910 |
| Political party | A | B | C | D | E | F |
| Seats won | 75 | 55 | 37 | 29 | 10 | 37 |
| S.No. | Length (in mm) | Number of leaves |
| 1. | 118 – 126 | 3 |
| 2. | 127 – 135 | 5 |
| 3. | 136 – 144 | 9 |
| 4. | 145 – 153 | 12 |
| 5. | 154 – 162 | 5 |
| 6. | 163 – 171 | 4 |
| 7. | 172 – 180 | 2 |
| S.No. | Length (in mm) | Number of leaves |
| 1. | 117.5 – 126.5 | 3 |
| 2. | 126.5 – 135.5 | 5 |
| 3. | 135.5 – 144.5 | 9 |
| 4. | 144.5 – 153.5 | 12 |
| 5. | 153.5 – 162.5 | 5 |
| 6. | 162.5 – 171.5 | 4 |
| 7. | 171.5 – 180.5 | 2 |
| Life Time (in hours) | Number of lamps |
| 300 – 400 | 14 |
| 400 – 500 | 56 |
| 500 – 600 | 60 |
| 600 – 700 | 86 |
| 700 – 800 | 74 |
| 800 – 900 | 62 |
| 900 – 1000 | 48 |
| Marks | Class Mark | Frequency |
| 0-10 | 5 | 3 |
| 10-20 | 15 | 9 |
| 20-30 | 25 | 17 |
| 30-40 | 35 | 12 |
| 40-50 | 45 | 9 |
| Marks | Class Mark | Frequency |
| 0-10 | 5 | 5 |
| 10-20 | 15 | 19 |
| 20-30 | 25 | 15 |
| 30-40 | 35 | 10 |
| 40-50 | 45 | 1 |
| Number of balls | Team A | Team B |
| 0.5-6.5 | 2 | 5 |
| 6.5-12.5 | 1 | 6 |
| 12.5-18.5 | 8 | 2 |
| 18.5-24.5 | 9 | 10 |
| 24.5-30.5 | 4 | 5 |
| 30.5-36.5 | 5 | 6 |
| 36.5-42.5 | 6 | 3 |
| 42.5-48.5 | 10 | 4 |
| 48.5-54.5 | 6 | 8 |
| 54.5-60.5 | 2 | 10 |
| Age (in years) | Number of children (frequency) | Width of class | Length of rectangle |
| 1-2 | 5 | 1 | (5/1)×1 = 5 |
| 2-3 | 3 | 1 | (3/1)×1 = 3 |
| 3-5 | 6 | 2 | (6/2)×1 = 3 |
| 5-7 | 12 | 2 | (12/2)×1 = 6 |
| 7-10 | 9 | 3 | (9/3)×1 = 3 |
| 10-15 | 10 | 5 | (10/5)×1 = 2 |
| 15-17 | 4 | 2 | (4/2)×1 = 2 |
| Number of letters | Number of surnames | Width of class | Length of rectangle |
| 1-4 | 6 | 3 | (6/3)×2 = 4 |
| 4-6 | 30 | 2 | (30/2)×2 = 30 |
| 6-8 | 44 | 2 | (44/2)×2 = 44 |
| 8-12 | 16 | 4 | (16/4)×2 = 8 |
| 12-20 | 4 | 8 | (4/8)×2 = 1 |
Exercise 14.4
1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer
Mean = Sum of all the observations/Total number of observations
= (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8
For Median, we will arrange the given data in ascending order,
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations (n) = 10
Number of observations are even so we will calculate median as,
2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer
Mean = Sum of all the observations/Total number of observations
= (41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/15 = 822/15 = 54.8
For Median, we will arrange the given data in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations (n) = 15
Number of observations are odd so we will calculate median as,
Salary (xi) | Number of workers (fi) | fixi |
| 3000 | 16 | 48000 |
| 4000 | 12 | 48000 |
| 5000 | 10 | 50000 |
| 6000 | 8 | 48000 |
| 7000 | 6 | 42000 |
| 8000 | 4 | 32000 |
| 9000 | 3 | 27000 |
| 10000 | 1 | 10000 |
| Total | Σfi = 60 | Σfixi = 305000 |
