Unit: 3: Data Handling
Find the range of heights of any ten students of your class.
S.No. 
Name of students 
Height (in feet) 
1. 
Gunjan 
$4.2$ 
2. 
Aditi 
$4.5$ 
3. 
Nikhil 
$5$ 
4. 
Akhil 
$5.1$ 
5. 
Riya 
$5.2$ 
6. 
Akshat 
$5.3$ 
7. 
Abhishek 
$5.1$ 
8. 
Mayank 
$4.7$ 
9. 
Rahul 
$4.9$ 
10. 
Ayush 
$4.5$ 
$\begin{array}{l}\text{Range}=\text{Highestheight}\text{Lowestheight}\\ \text{Range}=\text{5}\text{.3}\text{4}\text{.2}\\ \text{Range}=\text{1}\text{.1}\\ =5.34.2=1.1\text{feet}\text{.}\end{array}$ 
Organise 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 is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
S. No. 
Marks 
Tally marks 
Frequency 
1 
1 
I 
1 
2 
2 
II 
2 
3 
3 
I 
1 
4 
4 
III 
3 
5 
5 
IIIII 
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 $91=8$
(iv) Arithmetic mean $=$
$\frac{\begin{array}{l}4+6+7+5+3+5+4+5+2+6+2\\ +5+1+9+6+5+8+4+6+7\end{array}}{20}$
$=\frac{100}{20}=5$
Find the mean of the first five whole numbers.
The first five whole numbers are 0, 1, 2, 3 and 4. Therefore,
Mean of first five whole numbers
$=\frac{\text{Sumofnumbers}}{\text{Totalnumber}}$
$\begin{array}{l}=\frac{0+1+2+3+4}{5}\\ =\frac{10}{5}\\ =2\end{array}$
Thus, the mean of first five whole numbers is 5.
A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Number of innings $=8$
Mean of score
$=\frac{\text{Sumofscores}}{\text{Numberofinnings}}$
$=\frac{58+76+40+35+46+45+0+100}{8}$
$=\frac{400}{8}$
$=50$
Thus, the mean score is 50.
Following table shows the points of each player scored in four games:
Player 
Game 
Game 
Game 
Game 
A 
14 
16 
10 
10 
B 
0 
8 
6 
4 
C 
8 
11 
Did not Play 
13 
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 by $4$? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
(i) Average number of points of A
$=\frac{\text{SumofscoresbyA}}{\text{No}\text{.ofgamesplayedbyA}}\text{}$
$=\frac{14+16+10+10}{4}$
$=\frac{50}{4}\text{}$
$=12.5\text{}$
(ii) We need to divide the total points by 3 as player C played only three games.
(iii) Player B played in all the four games.
$\therefore \text{MeanscoreofplayerB}$
$=\frac{\text{SumofscoresbyB}}{\text{No}\text{.ofgamesplayedbyB}}$
$=\frac{0+8+6+4}{4}$
$=\frac{18}{4}$
$=4.5$
(iv) To find the best performer, we should know the mean of all the players played.
Mean score of player $\text{A}=12.5$
Mean score of player $\text{B}=4.5$
Mean score of player C
$=\frac{8+11+13}{3}$
$=\frac{32}{3}$
$=10.67$
Hence, on comparing mean scores of all the players, we found that player A is the best performer.
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) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
(i) Highest marks obtained by the student $=95$
Lowest marks obtained by the student $=39$
(ii) Range of marks
$=$ Highest marks $$ Lowest marks
$=9539=56$
(iii) Mean of obtained marks
$=\frac{\text{Sumofmarks}}{\text{Totalnumberofmarks}}$
$=\frac{85+76+90+85+39+48+56+95+81+75}{10}$
$=\frac{730}{10}$
$=73$
Thus, the mean marks obtained by the group of students is 73.
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.
Mean enrolment
$=\frac{\text{Sumofnumbersofenrolment}}{\text{Totalnumberofenrolment}}$
$=\frac{1555+1670+1750+2013+2540+2820}{6}$
$=\frac{12348}{6}$
$=2058$
Thus, the mean enrolment of the school is $2,058$.
The rainfall (in mm) in a city on $7$ days of a certain week was recorded as follows:
Day 
Mon 
Tue 
Wed 
Thurs 
Fri 
Sat 
Sun 
Rainfall (in mm) 
0.0 
12.2 
2.1 
0.0 
20.5 
5.5 
1.0 
(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.
(i) The range of the rainfall
$=\text{Highestrainfall}\text{Lowestrainfall}$
$=20.50.0=20.5\text{mm}$
(ii) Mean rainfall
$=\frac{\text{Sumofrainfallrecorded}}{\text{Totalnumberofdays}}$
$=\frac{0.0+12.2+2.1+2.2+20.5+5.5+1.0}{7}$
$=\frac{41.3}{7}$
$=5.9\text{\hspace{0.17em}}\text{mm}$
(iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday, rainfalls were less than the mean rainfall.
The heights 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 the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height.
(i) The height of the tallest girl $=151\text{cm}$
(ii) The height of the shortest girl $=128\text{\hspace{0.17em}}\text{cm}$
(iii) The range of the data $=\text{Highestheight}\text{Lowestheight}$
$=151128=23\text{cm}$
(iv) The mean height
$=\frac{\text{Sumofheightsofthegirsl}}{\text{Totalnumebrofgirls}}$
$=\frac{\begin{array}{l}135+150+139+128+151+132\\ +146+149+143+141\end{array}}{10}$
$\begin{array}{l}=\frac{1414}{10}\\ =141.4\text{cm}\end{array}$
(v) Five girls, i.e., $150,\text{}151,\text{}146,\text{}149,\text{}143$, have heights (in cm) more than the mean height.
The scores in mathematics test (out of 25) of 15 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?
Arranging the given data in ascending order,
$5,\text{}9,\text{}10,\text{}12,\text{}15,\text{}16,\text{}19,\text{}20,\text{}20,\text{}20,\text{}20,\text{}23,\text{}24,\text{}25,\text{}25$
Mode is the observation occurred the highest number of times. Therefore, Mode $=20$
Median is the middle observation $=20$
Yes, Mode and Median are same for given observation.
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?
Arranging the given data in ascending order,
$6,\text{}8,\text{}10,\text{}10,\text{}15,\text{}15,\text{}15,\text{}50,\text{}80,\text{}100,\text{}120$
$\text{Mean}=\text{}\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$=\text{}\frac{6+8+10+10+15+15+15+50+80+100+120}{11}$
$=\text{}\frac{429}{11}=39$
Mode is the observation occurred the highest number of times $=15$
Median is the middle observation $=15$
Therefore, Mode and Median is $15$.
No, the mean, median and mode are not same.
The weights (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?
Arrange the given data in ascending order.
$32,\text{}35,\text{}36,\text{}37,\text{}38,\text{}38,\text{}38,\text{}40,\text{}42,\text{}43,\text{}43,\text{}43,\text{}45,\text{}47,\text{}50$
Mode is the observation occurred the highest number of times.
Therefore, Mode $=38\text{and}43$
Median is the middle observation which is 40.
Yes, there are 2 modes
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Arrange the given data in ascending order,
$12,\text{}12,\text{}13,\text{}13,\text{}14,\text{}14,\text{}14,\text{}16,\text{}19$
Mode is the observation occurred the highest number of times. Thus, mode $=14$
Median is the middle observation which is 14.
Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean can be one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
(i) True
(ii) False
(iii) True
(iv) False
Use the bar graph (Fig 3.3) to answer the following questions.
(a) Which is the most popular pet?
Fig 3.3
(b) How many children have dog as a pet?
a. Cat is the most popular pet.
b. $8$ students have dogs as a pet.
Read the bar graph (Fig 3.4) which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
Fig 3.4
(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
According to the given bar graph,
(i) (a) In year $1989,\text{\hspace{0.17em}}180$ books were sold. (b) In year $1990,\text{\hspace{0.17em}}475$ books were sold. (c) In year $1992,\text{\hspace{0.17em}}225$ books were sold.
(ii) In year $1990$, about 475 books were sold and in year $1992$, about 225 books were sold.
(iii) In years $1989\text{and}1992$ fewer than 250 books were sold.
(iv) By reading the graph, we calculate that 180 books were sold in year 1989.
Number of children in six different classes are given below. Represent the data on a bar graph.
Class 
Fifth 
Sixth 
Seventh 
Eighth 
Ninth 
Tenth 
Number of Children 
135 
120 
95 
100 
90 
80 
a. How would you choose a scale.
b. Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of class eight.
Data represented by the bar graph is as follows:
a. Scale: $1\text{unit}=25\text{children}$
b. (i) Fifth class has the maximum number of children and Tenth class has the minimum number of children.
(ii) $\text{Ratio}=\frac{\text{Numberofstudentsinclasssixth}}{\text{numberofstudentsinclasseighth}}$
$=\frac{120}{100}=\frac{6}{5}=6:5$
The performance of students in 1st Term and 2nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:
Subject 
English 
Hindi 
Maths 
Science 
S. Science 
1st
Term 
67 
72 
88 
81 
73 
2nd Term 
70 
65 
95 
85 
75 
(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Data represented by bar graph is as follows:
Difference of marks of 1^{st} term and 2^{nd} term
English $=7067=3$
Hindi $=6572=7$
Maths $=9588=7$
Science $=8581=4$
S. Science $=7573=2$
(i) The child has improved most in Maths subject.
(ii) In S. Science subject, the child improvement is least.
(iii) Yes, in Hindi subject, the child performance has gone down.
Consider this data collected from a survey of a colony.
Favourite Sport 
Cricket 
Basket Ball 
Swimming 
Hockey 
Athletics 
Watching 
1240 
470 
510 
423 
250 
Participating 
620 
320 
320 
250 
105 
(i) Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Data represented by the double bar graph is as follows:
(i) This bar graph represents the number of persons who are watching and participating in their favourite sports.
(ii) Cricket is most popular.
(iii) Watching sports is more preferred.
Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this chapter (Table 3.1). Plot a double bar graph using the data and answer the following:
Table 3.1
Temperatures of cities as on 20.6.2006 


Max. 
Min. 
Ahmedabad 
38°C 
29°C 
Amritsar 
37°C 
26°C 
Bangalore 
28°C 
21°C 
Chennai 
36°C 
27°C 
Delhi 
38°C 
28°C 
Jaipur 
39°C 
29°C 
Jammu 
41°C 
26°C 
Mumbai 
32°C 
27°C 
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Data represented by double bar graph is as follows:
(i) Jammu has the largest difference in temperature i.e.,
Maximum temperature $=41\xb0\text{C}\text{\hspace{0.17em}}\text{and}$
Minimum temperature $=26\xb0\text{C}\text{.}$
$\therefore \text{Difference}=41\xb0\text{C}26\xb0\text{C}=15\xb0\text{C}$
(ii) Jammu is the hottest city due to maximum temperature is high and Bangalore is the coldest city due to maximum temperature is low.
(iii) Maximum temperature of Bangalore is $28\xb0\text{C}$
Minimum temperature of two cities whose minimum temperature is higher than the maximum temperature of Bangalore are Ahmedabad and Jaipur where the minimum temperature is $29\xb0\text{C}$
(iv) Mumbai has the least difference in temperature i.e.,
Maximum temperature $=32\xb0\text{C}\text{\hspace{0.17em}}\text{and}$
Minimum temperature $=27\xb0\text{C}\text{.}$
$\therefore \text{Difference}=32\xb0\text{C}27\xb0\text{C}=5\xb0\text{C}$
Tell whether the following is certain to happen, impossible, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
(i) It is certain to happen.
(ii) It can happen but not certain.
(iii) It is impossible.
(iv) It can happen but not certain.
(v) It can happen but not certain.
There are 6 marbles in a box with numbers from $1\text{to}6$ marked on each of them.
(i) What is the probability of drawing a marble with number $2$?
(ii) What is the probability of drawing a marble with number $5$?
Total marbles from $1\text{to}6$ marked in a box $=6$
(i) The probability of drawing a marble with number $2$
$\text{=P(drawingonemarble)}=\frac{1}{6}$
(ii) The probability of drawing a marble with number $5$
$\text{=P(drawingonemarble)}=\frac{1}{6}$
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
A coin has two possible outcomes Head and Tail.
Probability of getting Head or Tail is equal.
$\therefore \text{P(Startinggame)}=\frac{1}{2}$
A box contains pairs of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that it will make a pair?
It can be observed that while closing the eyes, one can draw either a black sock or a white sock. Therefore, there are two possible cases.
$\text{Probability}=\frac{\text{NumberofFavourableoutcome}}{\text{Numberofpossibleoutcomes}}$
Probability (a pair of white socks will be formed) $=\frac{1}{2}$