Question 1: A histogram shows the number of students achieving different score ranges. The class intervals are 50-59, 60-74, and 75-89 with frequencies 12, 18, and 15 respectively. The class widths are unequal. Calculate the frequency density for each class.

Answer: 50-59: 12/10=1.2; 60-74: 18/15=1.2; 75-89: 15/14=1.07

Question 2: Construct a histogram using the following data: Class intervals and frequency densities—10-19 (2.0), 20-39 (1.5), 40-59 (1.0), 60-79 (2.5). Indicate the unequal class widths.

Answer: Students draw bars with heights proportional to frequency densities, with widths matching class intervals.

Question 3: A histogram shows the number of hours students spend on a task. The class intervals are 0-2, 3-5, 6-9, with frequencies 8, 15, 12 respectively. The class widths are 3, 3, and 4 hours. Calculate the total area (frequency) represented by the histogram.

Answer: Calculate each area: (frequency density) * (class width); then sum for total.

Question 4: Explain why calculating the frequency density is important when classes have unequal widths. Provide a brief example to support your answer.

Answer: Frequency density allows comparison across classes with different widths; otherwise, tall bars might be misleading.

Question 5: Given the class intervals 0-4, 5-9, 10-14 with frequencies 24, 30, and 18 respectively, find the frequency density for each class and construct the corresponding histogram on the grid.

Answer: Calculate density: 24/4=6, 30/5=6, 18/5=3. Then sketch bars accordingly.

Question 6: A histogram with unequal class widths shows the following data: Class intervals and frequency densities—20-27 (2.0), 28-35 (1.2), 36-43 (1.8). Which class has the highest total frequency? Show your calculations.

Answer: Calculate total frequency: density * width for each; highest total corresponds to highest product.

Question 7: A real-world scenario involves measuring the daily sales (in units) in a store over a week. The class intervals are 0-50, 51-100, 101-150 with frequencies 5, 7, 3. The class widths are unequal. Calculate the frequency densities and interpret the data.

Answer: Density = frequency / class width; interpret the relative sales in each interval.

Question 8: Identify and correct the mistake in the following calculation: The class interval 10-20 with frequency 15 and class width 10 has a frequency density of 15/10=1.5. The mistake is that the class width should be 20-10=10. Is this correct? Explain.

Answer: Yes, calculation is correct; the mistake would be if they used wrong interval difference.

Question 9: For a histogram with class intervals 0-5, 6-10, 11-15, and respective frequencies 10, 20, 15, the class widths are 6, 5, and 5. Calculate the frequency densities and draw the histogram on the grid.

Answer: Density = frequency / width; then draw bars based on these densities.

Question 10: A histogram with unequal class widths shows that the total area of all bars corresponds to 120 students. If one class with width 8 has a frequency density of 2.5, what is the frequency for that class? Show your working.

Answer: Frequency = frequency density * class width = 2.5 * 8 = 20 students.