Draw a histogram of the data.

The data for Question 1 (English marks) is not provided in the image. Therefore, I cannot answer questions 1.6, 1.7, 1.8, and 1.9. Let's proceed with Question 2 based on the provided table. Question 2 The given data is: | Distance (x) | Frequency (f) | |---|---| | 0 < x 5 | 3 | | 5 < x 10 | 6 | | 10 < x 15 | 8 | | 15 < x 20 | 7 | | 20 < x 25 | 1 | 2.1 Draw a histogram of the data. To draw a histogram, plot the distance intervals on the x-axis and the frequency on the y-axis. Each bar should correspond to a class interval, with its height representing the frequency. The bars should be adjacent. X-axis (Distance in km): Intervals 0-5, 5-10, 10-15, 15-20, 20-25. Y-axis (Frequency): Values from 0 to 8. Bars: For 0 < x 5: Height = 3 For 5 < x 10: Height = 6 For 10 < x 15: Height = 8 For 15 < x 20: Height = 7 For 20 < x 25: Height = 1 2.2 Draw the line graph of the data. To draw a line graph (frequency polygon), first find the midpoint of each class interval. Then plot these midpoints against their corresponding frequencies and connect the points with straight lines. To close the polygon, add midpoints for the classes before and after with a frequency of 0. Midpoints (m): 0 < x 5: m = (0+5)/(2) = 2.5 5 < x 10: m = (5+10)/(2) = 7.5 10 < x 15: m = (10+15)/(2) = 12.5 15 < x 20: m = (15+20)/(2) = 17.5 20 < x 25: m = (20+25)/(2) = 22.5 Points to plot (midpoint, frequency): (2.5, 3) (7.5, 6) (12.5, 8) (17.5, 7) (22.5, 1) To close the polygon: Add ( -2.5, 0) for the class before ( -5 < x 0). Add (27.5, 0) for the class after (25 < x 30). 2.3 Determine the approximate average distance of the girls' trips. Step 1: Calculate the midpoint (m) for each class interval and the product of midpoint and frequency (fm). | Distance (x) | Midpoint (m) | Frequency (f) | fm | |---|---|---|---| | 0 < x 5 | 2.5 | 3 | 2.5 × 3 = 7.5 | | 5 < x 10 | 7.5 | 6 | 7.5 × 6 = 45.0 | | 10 < x 15 | 12.5 | 8 | 12.5 × 8 = 100.0 | | 15 < x 20 | 17.5 | 7 | 17.5 × 7 = 122.5 | | 20 < x 25 | 22.5 | 1 | 22.5 × 1 = 22.5 | Step 2: Calculate the sum of frequencies ( f) and the sum of fm ( fm). f = 3 + 6 + 8 + 7 + 1 = 25 fm = 7.5 + 45.0 + 100.0 + 122.5 + 22.5 = 297.5 Step 3: Calculate the approximate average distance (mean). Mean = ( fm)/( f) = (297.5)/(25) = 11.9 The approximate average distance is 11.9 km. 2.4 What is the modal group of the data? The modal group is the class interval with the highest frequency. From the table, the highest frequency is 8, which corresponds to the distance group 10 < x 15. The modal group is 10 < x 15. 2.5 Comment on the spread of the data. The data ranges from 0 km to 25 km. The frequencies are concentrated in the middle intervals, particularly between 10 km and 15 km, which has the highest frequency. The frequencies decrease towards the extreme ends (0-5 km and 20-25 km), indicating that very short and very long trips are less common. The data shows a moderate spread, with most trips falling within a reasonable range around the average. Send me the next one 📸