- Find the range of heights of any ten students of your class.
Solution:
Let us taken the height of 10 students in cm
150, 122, 135, 166, 117, 118, 113, 100, 112, 132
Now, By observing this data we find the highest value =166
By observing this data we find the lowest value = 100
Then,
Range of heights = Highest value – Lowest value
= 166 – 100
= 66 cm
2. 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
Note: First we have to arrange the given marks in ascending order:
i.e. 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9
Let us form a frequency table of the given data.
Marks (xi) | Tally Marks | Frequency (fi) | fi xi |
1 | I | 1 | 1 |
2 | II | 2 | 4 |
3 | I | 1 | 3 |
4 | III | 3 | 12 |
5 | 5 | 25 | |
6 | IIII | 4 | 24 |
7 | II | 2 | 14 |
8 | I | 1 | 8 |
9 | I | 1 | 9 |
Total – | 20 | 100 |
(i) Which number is the highest?
Solution:
By observing the given data, the highest number is = 9
(ii) Which number is the lowest?
Solution:
By observing the given data, the lowest number is = 1
(iii) What is the range of the data?
Solution:
Range = Highest value – Lowest value
= 9 – 1
= 8
Hence, the range of the data is 8.
(iv) Find the arithmetic mean.
Solution:

= 100/20
= 5
3. Find the mean of the first five whole numbers.
Solutions:

4. A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Solution:

5. Following table shows the points of each player scored in four games:
Player | Game 1 | Game 2 | Game 3 | Game 4 |
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.
Solution:
A’s average number of points scored per day = total points scored by a in 4 games/Total number of games
= (14+16+10+10)/4
= 50/4 = 12.5 points
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
Solution:
we divide by 3 to find the mean number of points per game for C because C played only 3 games.
(iii) B played in all the four games. How would you find the mean?
Solution:
Total number of games = 4
Given, B played in all the four games
So, To find out the mean we will divide the total points by 4
Now,
Mean of B’s score = Total point scored by B in 4 games/Total number of games
= (0+8+6+4)/4
= 18/4 = 4.5 points
(iv) Who is the best performer?
Solution:
To find the best performer among 3 player
First, we have to find the average point of C = (8+11+13)/3
= 32/3 = 10.66
Now, we have average points scored by A is 12.5 which is more than B and C.
Hence, it is clear that A is the best performer among the 3 player.
6. 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:
First of all, we have to arrange the marks in ascending order:
= 39,48,56,75,76,81,85,85,90,95
(i) Highest and the lowest marks obtained by the students.
Solution:
By observing the given marks:
The highest marks obtained by the student = 95
The lowest marks obtained by the student=39
(ii) Range of the marks obtained.
Solution:
Range of marks = Highest marks – Lowest marks
= 95-39 = 56
(iii) Mean marks obtained by the group.
Solution:
Mean of marks obtained by group = (sum of all marks obtained by the group of students)/(Total number of marks)
= (39+48+56+75+76+81+85+85+90+95)/10
= 730/10 = 73
7. 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.
Solution:
Mean enrolment = (Sum of all observation )/ (Number of Observation)
= ( 1555 + 1670 + 1750 + 2013 + 2540 + 2820)/6
= 12348/6 = 2058
Hence, the required mean enrolment of a school for the given period is 2058
8. 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.
Solution:
Range of rainfall for the given data = highest rainfall – lowest rainfall
= 20.5 – 0.0 = 20.5 mm
(ii) Find the mean rainfall for the week.
Solution:
Mean of rainfall for the given data = (Sum of all observation) / (Total number of observation)
= (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0) / 7
= 41.3 / 7 = 5.9 mm
(iii) On how many days was the rainfall less than the mean rainfall.
Solution:
After observing the average rainfall,
We may find the rainfall of 5 days was less than the average rainfall i.e Monday, Wednesday, Thursday, Saturday, and Sunday.
9. 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.
We have to arrange the given data in ascending order:
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) What is the height of the tallest girl?
Solution:
By observing the given data the height of the tallest girl = 151 cm
(ii) What is the height of the shortest girl?
Solution:
By observing the given data the height of the shortest girl = 128 cm
(iii) What is the range of the data?
Solution:
Range of the given data = Tallest height – Shortest height
= 151 – 128 = 23 cm
(iv) What is the mean height of the girls?
Solution:
Mean height of the girl = (Sum of the height of all-girl) / Total number of girls)
= (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151) / 10
= 1414 / 10 = 141.4 cm
(v) How many girls have heights more than the mean height.
Solution:
The mean height is 141.4 cm
Hence, 5 girls have a height more than the mean height.
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