Samacheer Kalvi Class 8 Maths Solutions Chapter 6 Statistics Exercise 6.3

Miscellaneous Practice Problems

Question 1. Draw a pie chart for the given table.
Answer: We need to convert the area percentages into degrees for a 360° circle to draw the pie chart. Here is the table with the calculated central angles:

🎯 Exam Tip: Always ensure the sum of all central angles equals 360° to confirm your calculations are correct before drawing the pie chart.

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: We need to convert the percentage of students into degrees for a 360° circle to draw the pie chart. Below is the table showing the calculated central angles for each mode of transport:

🎯 Exam Tip: When drawing a pie chart, always start measuring angles from the same point (usually the 12 o'clock position) and move clockwise to ensure accuracy.

Question 3. Draw a histogram for the given frequency distribution.
Answer: The given frequency distribution (Age 41-45, 46-50, etc.) is discontinuous because there is a gap between the end of one class interval and the start of the next (e.g., 45 to 46). To draw a histogram, we need a continuous frequency distribution.

To make it continuous, we find the "correction factor." The gap between intervals is \( 46 - 45 = 1 \). We divide this gap by 2: \( \frac{1}{2} = 0.5 \). We then subtract 0.5 from the lower limit of each class and add 0.5 to the upper limit of each class to get the continuous class boundaries.

For example, for the first class (41-45):

Lower boundary \( = 41 - 0.5 = 40.5 \)
Upper boundary \( = 45 + 0.5 = 45.5 \)

Applying this, the continuous frequency table is:

🎯 Exam Tip: Remember that in a histogram, there should be no gaps between the bars, as it represents continuous data. The height of each bar shows the frequency of that class.

Question 4. Draw a histogram and the frequency polygon in the same diagram to represent the following data.
Answer: The given distribution (Weight in kg 50-55, 56-61, etc.) is discontinuous because there is a gap between the end of one class interval and the start of the next (e.g., 55 to 56). To draw a histogram and frequency polygon, we need a continuous frequency distribution.

To make it continuous, we find the "correction factor." The gap between intervals is \( 56 - 55 = 1 \). We divide this gap by 2: \( \frac{1}{2} = 0.5 \). We then subtract 0.5 from the lower limit of each class and add 0.5 to the upper limit of each class to get the continuous class boundaries.

For example, for the first class (50-55):

Lower boundary \( = 50 - 0.5 = 49.5 \)
Upper boundary \( = 55 + 0.5 = 55.5 \)

Applying this, the continuous frequency table is:

🎯 Exam Tip: When drawing a frequency polygon with a histogram, connect the midpoints of the top of each bar. Ensure the polygon starts and ends on the X-axis by extending it to the midpoints of the hypothetical classes before and after the actual data.

Challenging Problems

Question 5. Form a continuous frequency distribution table and draw histogram from the following data.
Answer: The given data is in a cumulative "less than" format. First, we need to convert it into regular class intervals with their simple frequencies. Then, we will ensure the intervals are continuous to draw the histogram. The class width for each interval is 5 years.

Here's how we find the class intervals and frequencies:

• Under 5: 1 person. This is the class 0-5.
• Under 10: 12 persons. So, for the class 5-10, we have \( 12 - 1 = 11 \) persons.
• Under 15: 19 persons. So, for the class 10-15, we have \( 19 - 12 = 7 \) persons.
• Under 20: 26 persons. So, for the class 15-20, we have \( 26 - 19 = 7 \) persons.
• Under 25: 27 persons. So, for the class 20-25, we have \( 27 - 26 = 1 \) person.
• Under 30: 35 persons. So, for the class 25-30, we have \( 35 - 27 = 8 \) persons.
• Under 35: 38 persons. So, for the class 30-35, we have \( 38 - 35 = 3 \) persons.
• Under 40: 45 persons. So, for the class 35-40, we have \( 45 - 38 = 7 \) persons.
• Under 45: 48 persons. So, for the class 40-45, we have \( 48 - 45 = 3 \) persons.
• Under 50: 53 persons. So, for the class 45-50, we have \( 53 - 48 = 5 \) persons.

This forms our continuous frequency distribution table:

🎯 Exam Tip: When converting 'less than' cumulative frequency to a simple frequency distribution, remember that each class frequency is found by subtracting the cumulative frequency of the previous class from the current one.

Question 6. A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.
Answer: We need to represent the distribution of a rupee (100 paise) in a cloth manufacturing company using a pie chart. This means we convert each amount of paise into a central angle out of 360°. Based on the calculations used in the solution, here's the table:

🎯 Exam Tip: Always double-check that the sum of all paise equals 100 and the sum of all central angles equals 360° to catch any calculation errors before drawing.

Question 7. Draw a histogram for the following data.
Answer: The data provides mid-values for different classes. To draw a histogram, we first need to find the continuous class intervals and their frequencies. The difference between consecutive mid-values (e.g., \( 25 - 15 = 10 \)) gives us the class width, which is 10.

For each mid-value (x), the class interval is calculated as \( (x - \frac{\text{class width}}{2}) \) to \( (x + \frac{\text{class width}}{2}) \). For example, for the mid-value 15 with a class width of 10:

Lower limit \( = 15 - \frac{10}{2} = 15 - 5 = 10 \)
Upper limit \( = 15 + \frac{10}{2} = 15 + 5 = 20 \)

Thus, the first class interval is 10-20. We apply this to all mid-values to get our continuous frequency table:

🎯 Exam Tip: When given mid-values, remember to first calculate the class width and then use it to find the class boundaries for the histogram.