Mathematics
A random survey of the number of children of various age groups playing in a park was found as follows:
| Age (in years) | Number of children |
|---|---|
| 1 - 2 | 5 |
| 2 - 3 | 3 |
| 3 - 5 | 6 |
| 5 - 7 | 12 |
| 7 - 10 | 9 |
| 10 - 15 | 10 |
| 15 - 17 | 4 |
Draw a histogram to represent the data above.
Answer
From the given data, we can observe that the class intervals have varying widths. The areas of the rectangles should be proportional to the frequencies in a histogram.
For example, when the class size is 5, the length of the rectangle is 10. So when the class size is 1, the length of the rectangle will be 10/5 × 1 = 2
| Age (in years) | Number of children | Width of class | Length of rectangle |
|---|---|---|---|
| 1 - 2 | 5 | 1 | (5 x 1)/1 = 5 |
| 2 - 3 | 3 | 1 | (3 x 1)/1 = 3 |
| 3 - 5 | 6 | 2 | (6 x 1)/2 = 3 |
| 5 - 7 | 12 | 2 | (12 x 1)/2 = 6 |
| 7 - 10 | 9 | 3 | (9 x 1)/3 = 3 |
| 10 - 15 | 10 | 5 | (10 x 1)/5 = 2 |
| 15 - 17 | 4 | 2 | (4 x 1)/2 = 2 |
Steps of construction :
Take the age of children on x-axis, using scale 1 unit = 1 year.
Take proportion of children per 1 year interval on y-axis, using scale 1 unit = 1 child.
To represent our first Head, i.e., (1 - 2) we draw a rectangular bar with width 1 unit and length 5 units.
Similarly, other Heads are represented without leaving a gap between two consecutive bars, according to the table.
Related Questions
The following table gives the life times of 400 neon lamps:
Life time (in hours) Number of lamps 300 - 400 14 400 - 500 56 500 - 600 60 600 - 700 86 700 - 800 74 800 - 900 62 900 - 1000 48 (i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
The following table gives the distribution of students of two sections according to the marks obtained by them:
Section A
Marks Frequency 0 - 10 3 10 - 20 9 20 - 30 17 30 - 40 12 40 - 50 9 Section B
Marks Frequency 0 - 10 5 10 - 20 19 20 - 30 15 30 - 40 10 40 - 50 1 Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Number of ball Team A Team B 1 - 6 2 5 7 - 12 1 6 13 - 18 8 2 19 - 24 9 10 25 - 30 4 5 31 - 36 5 6 37 - 42 6 3 43 - 48 10 4 49 - 54 6 8 55 - 60 2 10 Represent the data of both the teams on the same graph by frequency polygons. [Hint : First make the class intervals continuous.]
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
Number of letters Number of surnames 1 - 4 6 4 - 6 30 6 - 8 44 8 - 12 16 12 - 20 4 (i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.