8th Maths : Chapter 6 : Statistics : Exercise 6.3 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Numerical problems, Exercises Questions with Answers, Solution

**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.

Tags : Questions with Answers, Solution | Statistics | Chapter 6 | 8th Maths , 8th Maths : Chapter 6 : Statistics

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8th Maths : Chapter 6 : Statistics : Exercise 6.3 | Questions with Answers, Solution | Statistics | Chapter 6 | 8th Maths

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