This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here's the solution for the given frequency distribution: First, let's organize the data into a frequency table with midpoints (x), frequencies (f), and necessary calculations for grouped data. | Class Interval | Midpoint (x) | Frequency (f) | fx | Cumulative Frequency (cf) | x^2 | fx^2 | | :------------- | :----------- | :------------ | :--- | :------------------------ | :---- | :----- | | 0-4 | 2 | 3 | 6 | 3 | 4 | 12 | | 5-9 | 7 | 5 | 35 | 8 | 49 | 245 | | 10-14 | 12 | 9 | 108 | 17 | 144 | 1296 | | 15-19 | 17 | 6 | 102 | 23 | 289 | 1734 | | 20-24 | 22 | 2 | 44 | 25 | 484 | 968 | | Total | | f = 25 | fx = 295 | | | fx^2 = 4255 | Correction: The interval "15_16" is unusual. Assuming it means 15-19 to maintain consistent class widths (or 15-16 as a distinct class). Given the previous intervals (0-4, 5-9, 10-14), a class width of 5 seems intended. If 15-16 is a class, its width is 2. Let's re-evaluate the class intervals to ensure consistency. 0-4 (width 5) 5-9 (width 5) 10-14 (width 5) 15-16 (width 2) - This breaks the pattern. 20-24 (width 5) Given the inconsistency, I will assume the user intended standard class intervals with a width of 5, and "15_16" was a typo for "15_19". If "15_16" is strictly adhered to, the calculations for median and mode become more complex due to varying class widths. For simplicity and common practice in statistics, I will assume the class 15-19 with midpoint 17. Let's re-calculate the table with the assumption of 15-19 for the fourth class: | Class Interval | Midpoint (x) | Frequency (f) | fx | Cumulative Frequency (cf) | x^2 | fx^2 | | :------------- | :----------- | :------------ | :--- | :------------------------ | :---- | :----- | | 0-4 | 2 | 3 | 6 | 3 | 4 | 12 | | 5-9 | 7 | 5 | 35 | 8 | 49 | 245 | | 10-14 | 12 | 9 | 108 | 17 | 144 | 1296 | | 15-19 | 17 | 6 | 102 | 23 | 289 | 1734 | | 20-24 | 22 | 2 | 22 | 25 | 484 | 968 | | Total | | f = 25 | fx = 273 | | | fx^2 = 4255 | Self-correction: 22 × 2 = 44, not 22. Let's correct the fx and fx for the last row. 22 × 2 = 44. fx = 6 + 35 + 108 + 102 + 44 = 295. The fx^2 is correct: 12 + 245 + 1296 + 1734 + 968 = 4255. Corrected Table: | Class Interval | Midpoint (x) | Frequency (f) | fx | Cumulative Frequency (cf) | x^2 | fx^2 | | :------------- | :----------- | :------------ | :--- | :------------------------ | :---- | :----- | | 0-4 | 2 | 3 | 6 | 3 | 4 | 12 | | 5-9 | 7 | 5 | 35 | 8 | 49 | 245 | | 10-14 | 12 | 9 | 108 | 17 | 144 | 1296 | | 15-19 | 17 | 6 | 102 | 23 | 289 | 1734 | | 20-24 | 22 | 2 | 44 | 25 | 484 | 968 | | Total | | f = 25 | fx = 295 | | | fx^2 = 4255 | a) Mean: Step 1: Calculate the sum of (frequency × midpoint) and the total frequency. From the table, fx = 295 and f = 25. Step 2: Apply the formula for the mean (x). x = ( fx)/( f) x = (295)/(25) x = 11.8 b) Median: Step 1: Determine the median position and median class. The total frequency is N = f = 25. The median position is the (N)/(2) = (25)/(2) = 12.5-th value. From the cumulative frequency column, the 12.5th value falls in the 10-14 class (cf=17). So, the median class is 10-14. Step 2: Identify the values for the median formula for grouped data. Lower boundary of median class (L) = 9.5