NCERT Solutions: Data Handling- 1 Notes | Study Mathematics (Maths) Class 7 - Class 7

Exercise 3.1

Q1. Find the range of heights of any ten students of your class. 
Ans:

Range = Highest height – Lowest height
= 5.3 – 4.2
= 1.1 feet

Q2. Organize the following marks in a class assessment, in a tabular form:
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7. 
(i) Which number is the highest? 
(ii) Which number of the lowest? 
(iii) What is the range of the numbers? 
(iv) Find the arithmetic mean
Ans:

S. No.	Marks	Tally marks	Frequency (No. of students)
1.	1	I	1
2.	2	II	2
3.	3	I	1
4.	4	III	3
5.	5		5
6.	6	IIII	4
7.	7	II	2
8.	8	I	1
9.	9	I	1

(i) The highest number is 9.
(ii) The lowest number is 1.
(iii) The range of the data is 9 - 1 = 8
(iv) Arithmetic mean
=   
= 100/20 = 5

Q3. Find the mean of the first five whole numbers. 
Ans: The first five whole numbers are 0, 1, 2, 3 and 4.
Therefore,
Mean of first five whole numbers

Thus, the mean of first five whole numbers is 2.

Q4. A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100 Find the mean score. 
Ans: Number of innings = 8


= 400 / 8 = 50
Thus, the mean score is 50.

Q5. Following table shows the points of each player scored in four games: 

Now answer the following questions: 
(i) Find the mean to determine A’s average number of points scored per game. 
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or 47? Why? 
(iii) B played in all the four games. How would you find the mean? 
(iv) Who is the best performer? 
Ans: 
(i) A’s average number of points scored per game
= Total points scored by A in 4 games / Total number of games
= (14 + 16 + 10 + 10) / 4
= 50 / 4
= 12.5 points
(ii) We should divide the total points by 3 because player C played only three games.
(iii) Player B played in all the four games.


(iv) To find the best performer, we should know the mean of all players.

Therefore, on comparing means of all players, player A is the best performer.

Q6. The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. 
Find the: 
(i) The highest and the lowest marks obtained by the students. 
(ii) Range of the marks obtained. 
(iii) Mean marks obtained by the group. 
Ans: (i) Highest marks obtained by the student = 95
Lowest marks obtained by the student = 39
(ii) Range of marks = Highest marks – Lowest marks
= 95 – 39
= 56
(iii)

Thus, the mean marks obtained by the group of students is 73.

Q7. The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2820 Find the mean enrolment of the school for this period. 
Ans: 

Thus, the mean enrolment of the school is 2,058.

Q8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows: 

(i) Find the range of the rainfall in the above data. 
(ii) Find the mean rainfall for the week. 
(iii) On how many days was the rainfall less than the mean rainfall? 
Ans: (i) The range of the rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0 = 20.5 mm
(ii)  

(iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls were less than the mean rainfall.

Q9. The height of 10 girls were measured in cm and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141 
(i) What is the height of the tallest girl? 
(ii) What is the height of the shortest girl? 
(iii) What is the range of data? 
(iv) What is the mean height of the girls? 
(v) How many girls have heights more than the mean height? 
Ans: First we have to arrange the given data in an ascending order,
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) The height of the tallest girl = 151 cm
(ii) The height of the shortest girl = 128 cm
(iii) The range of the data = Highest height – Lowest height
= 151 – 128 = 23 cm
(iv) 


(v) Five girls, i.e., 150, 151, 146, 149, 143 have heights (in cm) more than the mean height.

Exercise 3.2 

Q1. The scores in mathematics test (out of 25) of students is as follows: 
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 
Find the mode and median of this data. Are they same? 
Ans: Arranging the given scores in an ascending order, we get
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode,
Mode is the value of the variable which occurs most frequently.
Clearly, 20 occurs maximum number of times.
Hence, mode of the given sores is 20
Median,
The value of the middle-most observation is called the median of the data.
Here n = 15, which is odd.
Where, n is the number of the students.
∴ median = value of ½ (n + 1)^th observation.
= ½ (15 + 1)
= ½ (16)
= 16 / 2
= 8
Then, value of 8^th term = 20
Hence, the median is 20. Yes, both the values are same.

Formulas for Mean, Median and Mode Q2. The runs scored in a cricket match by 11 players is as follows: 
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 
Find the mean, mode and median of this data. Are the three same? 
Ans: Arranging the runs scored in a cricket match by 11 players in an ascending order, we get
6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mean,
Mean of the given data = Sum of all observations / Total number of observations
= (6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120) / 11
= 429 / 11
= 39
Mode, Mode is the value of the variable which occurs most frequently.
Clearly, 15 occurs maximum number of times.
Hence, mode of the given sores is 15
Median,
The value of the middle-most observation is called the median of the data.
Here n = 11, which is odd.
Where, n is the number of players.
∴ median = value of ½ (n + 1)th observation.
= ½ (11 + 1)
= ½ (12)
= 12 / 2
= 6
Then, value of 6^th term = 15
Hence, the median is 15.
No, these three are not same.

Q3. The weight (in kg) of 15 students of a class are: 
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 
(i) Find the mode and median of this data. 
(ii) Is there more than one mode? 
Ans: Arranging the given data in ascending order,
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Mode and Median
Mode,
Mode is the value of the variable which occurs most frequently.
Clearly, 38 and 43 both occurs 3 times.
Hence, mode of the given weights are 38 and 43.
Median,
The value of the middle-most observation is called the median of the data.
Here n = 15, which is odd.
Where, n is the number of the students.
∴ median = value of ½ (n + 1)^th observation.
= ½ (15 + 1) = ½ (16)
= 16 / 2
= 8
Then, value of 8^th term = 40
Hence, the median is 40.
(ii) Yes, there are 2 modes for the given weights of the students.

Q4. Find the mode and median of the data: 
13, 16, 12, 14, 19, 12, 14, 13, 14. 
Ans: Arranging the given data in an ascending order, we get
= 12, 12, 13, 13, 14, 14, 14, 16, 19
Mode,
Mode is the value of the variable which occurs most frequently.
Clearly, 14 occurs maximum number of times.
Hence, mode of the given data is 14.
Median,
The value of the middle-most observation is called the median of the data.
Here n = 9, which is odd.
Where, n is the number of the students.
∴ median = value of ½ (9 + 1)^th observation.
= ½ (9 + 1)
= ½ (10)
= 10 / 2
= 5
Then, value of 5th term = 14
Hence, the median is 14.

Q5. Tell whether the statement is true or false: 

(i) The mode is always one of the numbers in a data.
Ans: True
Solution: Because, Mode is the value of the variable which occurs most frequently in the given data.
Hence, mode is always one of the numbers in a data.

(ii) The mean is one of the numbers in a data. 
Ans: False
Solution: Because, mean is may be or may not be one of the number in a data.

(iii) The median is always one of the numbers in a data.
Ans: True
Solution: Because, median is the value of the middle-most observation in the given data while arranged in ascending or descending order.
Hence, median is always one of the numbers in a data

(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
Ans: False
Solution: Mean = Sum of all given observations / number of observations
= (6 + 4 + 3 + 8 + 9 + 12 + 13 + 9) / 8
= (64 / 8) = 8
Hence, the given statement is false.