A frequency distribution table is one way you can organize data so that it make mare sense. By counting frequencies, we can make a frequency distribution table. Here are some test scores from a math class.
80 88 91 65 74 93 82 87 70 89
99 89 70 88 78 83 59 69 87 54
94 84 96 98 46 70 90 96 88 72
89 83 74 94 80 67 77 82 92 70
The above data can be organized into frequency distribution in several ways. One method is to use intervals as a basis. The smallest value in the above data is 46 and largest is 99. The interval from 46 to 99 is broken up into smaller subinterval (called class interval) for each class interval. For each class interval, the amount of data item falling in this interval is counted. The number is called the frequency of the class interval. The results are tabulated as frequency table as follows:
Class interval | Tally Marks | Frequency |
40-50 | 1 | |
50-60 | 3 | |
60-70 | 5 | |
70-80 | 8 | |
80-90 | 13 | |
90-100 | 9 |
Now, we can see the biggest number of the score are between 80 to 90 and the least number of scores between 40 and 50.
In pie-chart, the circle is divided into different sectors.It is a circular chart. Each sector represents an item in a data.
Reading pie-chart
The pie chart shows the result of a survey that was carried out to find out how student travels to school.
Constructing pie-charts
To construct a pie chart you need to work out the fraction of the total that the sector represents. You can then convert this to an angle and draw the sector on the chart. Look at the following example.
The table below shows the grades achieved by 30 pupils in their final exams.
Grade | A | B | C | D | E |
Frequency | 7 | 11 | 6 | 4 |
2 |
To show the information in a pie chart, take the following steps:
A line graph is most useful for displaying data or information.It is plotted as a series of points, which are then joined with the straight line.The end of line graph does not have to join to the axis.
How to make the line graph?
a) Upper-class limit of 30 - 35 is 35.
b) Lower- class limit of 10 - 15 is 10.
c) If the upper and lower limits of a class interval are 26 and 20 respectively, the class interval is 6.
d) If the lower and upper limits of a class interval are 5 and 10 respectively, the class interval is 5.
Renu's math test grades are given in the following table. Draw a line graph to represent the data.
Test | Grade |
1 | 75 |
2 | 80 |
3 | 100 |
4 | 95 |
5 | 60 |
6 | 95 |
Solution:
The data from the Table is used to create the following line graph.
Draw a line graph to represent this information:
Class | 5 | 6 | 7 | 8 | 9 | 10 |
Number of students | 30 | 40 | 35 | 44 | 50 | 45 |
Solution:
Here, the given information, classes are presented in x- axis and Number of students are presented in y- axis.
Draw a line graph to represent this information.
Time | 6 am | 8 am | 10 am | 12 pm | 2 pm | 6 pm |
Temperature | 10^{o}c | 12^{o}c | 18^{o}c | 25^{o}c | 25^{o}c | 17^{o}c |
Solution:
Here, the given information, Time is presented in x- axis and Temperature is presented in y- axis.
The table below shows the Admission rate of Buddhi Memorial School of Class-10.
Year | 2064 | 2065 | 2066 | 2067 | 2068 | 2069 |
Admission Rate | 22 | 24 | 21 | 18 | 15 | 12 |
Solution:
The given information of Buddha Memorial School, Year is presented in x- axis and Admission rate is presented in y- axis.
A Circus sold the following number of tickets for the four shows during a day. Draw in pie chart.
Show | First | Second | Third | Fourth |
Number of tickets sold | 200 | 500 | 300 | 1000 |
Solution:
Total Number of Tickets sold = 2000
Then,
Tickets sold on first show (in degree)
= \(\frac{200}{2000}\)\(\times\) 360^{o}
= 36^{o}
Tickets sold on second show (in degree)
= \(\frac{500}{2000}\)\(\times\)360^{o}
= 90^{o}
Tickets sold on Third show (in degree)
= \(\frac{300}{2000}\)\(\times\)360^{o}
= 54^{o}
Tickets sold on fourth show (in degree)
= \(\frac{1000}{2000}\)\(\times\)360^{o}
= 180^{o}
^{}Pie chart:-
Saroja spends her income as shown in the following table.
Source of Expenditure | Wages | Raw Material | Extras | Fuel |
Expenditure | 1500 | 1200 | 500 | 400 |
Solution:
Given information,
Wages= 1500
Raw Material = 1200
Extras = 500
Fuel= 400
Total Expenditure = 3600
Now, Expenditure in Wages (in degree)
= \(\frac{1500}{3600}\)\(\times\)360^{o}
= 150^{o}
Expenditure on Raw Material (in degree)
= \(\frac{1200}{3600}\)\(\times\)360^{o}
= 120^{o}
Expenditure on Extras (in degree)
= \(\frac{500}{3600}\)\(\times\)360^{o}
= 50^{o}
Expenditure on Fuel (in degree)
= \(\frac{400}{3600}\)\(\times\)360^{o}
= 40^{o}
Showing in the pie- chart
Solution:
According to the question, Here's Cumulative frequency distribution table for the above set of data:
Class | Frequency | Scores less than | Cumulative Frequency |
0-10 | 3 | 10 | 3 |
10-20 | 8 | 20 | 3+8=11 |
20-30 | 8 | 30 | 11+8=19 |
30-40 | 6 | 40 | 19+6=25 |
40-50 | 5 | 50 | 25+5=30 |
This is less than Cumulative Frequency distribution.
Solution:
According to the question, Here's cumulative frequency distribution table for the above set of data.
Class | Frequency | Scores less than | Cumulative Frequency |
40-50 | 1 | 50 | 1 |
50-60 | 2 | 60 | 1+2=3 |
60-70 | 3 | 70 | 3+3=6 |
70-80 | 10 | 80 | 6+10=16 |
80-90 | 14 | 90 | 16+14=30 |
90-100 | 10 | 100 | 30+10=40 |
This is less than Cumulative frequency distribution.
Draw a line graph to represent the data.
Time | 12pm | 2 pm | 4 pm | 6 pm | 8 pm | 10 pm | 12 pm |
Temperature | 26^{o}C | 28^{O}C | 25^{O}C | 21^{O}C | 15^{O}C | 12^{O}C | 10^{O}C |
Solution:
Here, Time is presented in x- axis and Temperature is presented in Y-axis in the following graph.
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