Exercise 6.3

Exercise 6.3

Miscellaneous Practice Problems

1. Draw a pie chart for the given table.

Solution:
Converting the area in percentage into components parts of 360°, we have.

Continental Area.

2. The data on modes of transport used by the students to come to school are given below. Draw a pie chart for the data.

Solution:

Converting the percentage into components parts of 360°, we have

Mode of Transport by students.

3. Draw a histogram for the given frequency distribution.

Solution:

The given distribution is discontinuous.

Lower boundary = lower limit – 1/2 (gap between the adjacent class interval)

= 41− 1/2 (l) = 40.5

Upper boundary = Upper limit + 1/2 (gap between the adjacent class interval)

= 45 + 1/2(1) = 45.5

Now continuous frequency table is as below

 4. Draw a histogram and the frequency polygon in the same diagram to represent the following data.

Solution:

The given distribution is discontinuous.

Lower boundary = lower limit – 1/2 (gap between the adjacent class interval)

= 50 – 1/2(1) = 49.5  

Upper boundary = Upper limit + 1/2 (gap between the adjacent class interval)

= 55 + 1/2(1) = 55.5

∴  The continuous frequency table is as below.

Challenging problems

5. Form a continuous frequency distribution table and draw histogram from the following data.

Solution:

Converting into continuous distribution we have

6. A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.

Solution:

1 Rupee = 100 paise.

Expenditure of a cloth manufacturing company.

7. Draw a histogram for the following data.

Solution:

Since mid values are given, the given distributors is discontinuous.

Lower boundary = lower limit – 1/2 (gap between the adjacent class interval)

= 15 – 1/2(10) = 10

Upper boundary = Upper limit + 1/2(gap between the adjacent class interval)

= 15 + 1/2(10) = 20

The continuous distribution will be as follows.