Solutions to Ex. 3.1:
Q2. Organise the following marks in a class assessment, in a tabular form.
Ans:
| Marks |
Tally Bars |
Number of students |
| 1 |
I |
1 |
| 2 |
II |
2 |
| 3 |
I |
1 |
| 4 |
III |
3 |
| 5 |
IIIII |
5 |
| 6 |
IIII |
4 |
| 7 |
II |
2 |
| 8 |
I |
1 |
| 9 |
I |
1 |
(i) Highest marks is 9.
(ii) Lowest marks is 1.
(iii) Range = Highest marks – lowest marks
= 9 – 1 = 8
(iv) Total marks = 4 + 6 + 7 + 5 + 3 + 5 + 4 + 5 + 2 + 6 +2 + 5 + 1 + 9 + 6 + 5 + 8 + 4 + 6 +7 = 100
Total students = 20
Arithmetic mean = Total marks / Total students
= 100 / 20 = 5
Q3. Find the mean of the first five whole numbers.
Ans:
First five whole numbers are = 0, 1, 2, 3, 4
Sum of the digits is = 0 + 1 + 2 + 3 + 4 = 10
Total digits = 5
Mean = Total sum / total digits = 10 /2 = 5
Q4. A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Ans:
Total runs scored = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Total innings = 8
Mean score = Total runs scored / Total innings
= 400 / 8 = 50
Q5. Following table shows the points of each player scored in four games. Now answer the following questions:
| Player |
Game 1 |
Game 2 |
Game 3 |
Game 4 |
| A |
14 |
16 |
10 |
10 |
| B |
0 |
8 |
6 |
4 |
| C |
8 |
11 |
Not played |
13 |
Ans:
(i) Total points of A = 14 + 16 + 10 + 10 = 50
Total games = 4
Mean = total points / total games
= 50 / 4 = 12.5
(ii) For C, divide the total points by 3 as C did not play 1 game. Hence total games played by C = 3.
(iii) Total points of B = 0 + 8 + 6 + 4 = 18
Total games = 4
Mean = total points / total games
= 18 / 4 = 4.5
(iv) 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:
Ans:
(i) Highest marks obtained is 95
Lowest marks obtained is 39
(ii) Range = highest marks – lowest marks
= 95 – 39 = 56
(iii) Total marks = 85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75 = 730
Total students = 10
Mean = total marks / total students
= 730 / 10 = 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:
Total enrollment = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348
Total years = 6
Mean = total enrollment / total years
= 12348 / 6 = 2058
Q8. 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 (mm) |
0 |
12.2 |
2.1 |
0 |
20.5 |
5.5 |
1 |
Ans:
(i) Highest rain = 20.5mm
Lowest rain = 0
Range = highest rain – lowest rain
= 20.5 – 0 = 20.5mm
(ii) Total rainfall = 0 + 12.2 + 2.1 + 0 + 20.5 + 5.5 + 1 = 41.3
Total days = 7
Mean = total rainfall / total days
= 41.3 / 7 = 5.9
(iii) 5 days the rainfall is less than the mean rainfall.
Q9. 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.
Ans:
(i) Height of the tallest girl = 151cm
(ii) Height of the shortest girl = 128cm
(iii) Range = height of tallest girl – height of the shortest girl
= 151 -128 = 23cm
(iv) Total height = 1414 cm
Total girls = 10
Mean = Total height / total girls
= 1414 / 10 = 141.4cm
(v) 5 girls have height more than the mean height.