Class 6 - Mathematics
Chapter - Data Handling : Exercise 3.1
Top Block 1
Question: 1.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 | Akhi | 5.1 |
| 5 | Riya | 5.2 |
| 6 | Akhsat | 5.3 |
| 7 | Abhishek | 5.1 |
| 8 | Mayank | 4.7 |
| 9 | Rahul | 4.9 |
| 10 | Ayush | 4.5 |
Answer :
Range = Highest height – Lowest height
= 5.3 – 4.2
= 1.1 feet.
Question: 2.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 lowest?
(iv) Find the arithmetic mean
Answer :
| S.No. | Marks | Tally Marks | Frequency (No. of Students) |
|---|---|---|---|
| 1 | 1 | Ι | 1 |
| 2 | 2 | ΙΙ | 2 |
| 3 | 3 | Ι | 1 |
| 4 | 4 | ΙΙΙ | 3 |
| 5 | 5 | 5 | |
| 6 | 6 | ΙΙΙΙ | 4 |
| 7 | 7 | ΙΙ | 2 |
| 8 | 8 | Ι | 1 |
| 9 | 9 | Ι | 1 |
(ii)The lowest number is 1.
(iii)The range of the data is 9 – 1 = 8
(iv)
Sum of numbers
Arithmetic mean = ──────────────
Total number
4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7
= ─────────────────────────
20
100
= ────── = 5
20
Question: 3.Find the mean of the first five whole numbers.
Answer :
The first five whole numbers are 0, 1, 2, 3 and 4.
Sum of numbers
Mean of first five whole numbers = ──────────────
Total number
0+1+2+3+4
= ─────────
5
10
= ────── = 2
5
Thus, the mean of first five whole numbers is 5.
Question: 4.A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100 Find the mean score.
Answer :
Number of innings = 8
Sum of scores
Mean of score = ──────────────
Number of innings
58+76+40+35+46+45+0+100
= ──────────────
8
400
= ────── = 50
8
Thus, the mean score is 50.
Question: 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 |
(i) Find the mean to determine A’s average number of points scored per game.Now answer the following Questions:
(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?
Answer :
Sum of scores by A
Mean of player A = ──────────────
No. of games played by A
14+16+10+10
= ─────────
4
50
= ────── = 12.5
4
(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.
Sum of scores by B
∴ Mean of player B = ──────────────
No. of games played by B
0+8+6+4
= ─────────
4
18
= ────── = 4.5
4
(iv) To find the best performer, we should know the mean of all players.
Mena of player A = 12.5
Mean of player B = 4.5
Sum of scores by C
∴ Mean of player C = ──────────────
No. of games played by C
8+11+13
= ─────────
3
32
= ────── = 10.67
3
Therefore, on comparing means of all players, player A is the best performer.
Question: 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:
(i)The highest and the lowest marks obtained by the students.
(ii)Range of the marks obtained
(iii) Mean marks obtained by the group.
Answer :
(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)
Sum of marks
Mean of obtained marks = ──────────────
Total number of marks
85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75
= ──────────────────────────
10
730
= ────── = 73
10
Thus, mean marks obtained by the group of students is 73.
Question: 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.
Answer :
Sum of numbers of enrolment
Mean enrolment = ─────────────────
Total number of enrolment
1555 + 1670 + 1750 + 2013 + 2540 + 2820
= ────────────────────────
6
12348
= ────── = 2058
6
Thus, the mean enrolment of the school is 2,058.
Mddle block 1
Question: 8.The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Rainfall(in mm) | 0 | 12.2 | 2.1 | 0 | 20.5 | 5.5 | 1 |
(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?
Answer :
(i) The range of the rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0 = 20.5 mm
Sum of rainfall recorded
Mean Rainfall = ──────────────
Total number of days
0.0 + 12.2 + 2.1 + 2.2 + 20.5 + 5.5 + 1.0
= ────────────────────────
7
41.3
= ────── = 5.9
7
(iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls were less than the mean rainfall.
Question: 9.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?
Answer :
(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
Sum of heights of the girsl
The mean height = ──────────────
Total numebr of girls
135+150+139+128+151+132+146+149+143+141
= ─────────────────────────
10
1414
= ────── = 141.4
10
(v) Five girls, i.e., 150, 151, 146, 149, 143 have heights (in cm) more than the mean height.