Exercise 6.3 - Chapter 6 - Statistics - 8th Maths Guide Samacheer Kalvi Solutions

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Miscellaneous Practice Problems
Question $1 .$
Draw a pie chart for the given table.

Answer:

Question 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.

Answer:

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

Question $3 .$
Draw a histogram for the given frequency distribution.

Answer:
The given distribution is discontinuous.
Lower boundary $=$ lower limit $-\frac{1}{2}$ (gap between the adjacent class interval) $=41-\frac{1}{3}(1)=40.5$
Upper boundary $=$ Upper limit $+\frac{1}{2}$ (gap between the adjacent class interval) $=45+\frac{1}{2}(1)=45.5$
Now continuous frequency table is as below

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

Answer:
The given distribution is discontinuous.
Lower boundary $=$ lower limit $-\frac{1}{2}$ (gap between the adjacent class interval) $=50-\frac{1}{2}(1)=49.5$
Upper boundary $=$ Upper limit $+\frac{1}{2}$ (gap between the adjacent class interval) $=55+\frac{1}{2}(1)=55.5$
$\therefore$ The continuous frequency table is as below.

Challenging problems
Question $5 .$
Form a continuous frequency distribution table and draw histogram from the following data.

Answer:

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

Answer:

Question $7 .$
Draw a histogram for the following data.

Answer:
Since mid values are given, the given distributors is discontinuous.
Lower boundary lower limit $-\frac{1}{2}$ (gap between the adjacent class interval) $=15-\frac{1}{2}(10)=10$
Upper boundary $=$ Upper limit $+$ (gap between the adjacent class interval) $=15+\frac{1}{2}(10)=20$
The continuous distribution will he as follows.