Frequency 04:3 59:5 10-14:9 1516:6 2024:2
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Answer
Step-by-step solution
4 stepsHere'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) | Cumulative Frequency (cf) | |||
|---|---|---|---|---|---|---|
| 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 |
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) | Cumulative Frequency (cf) | |||
|---|---|---|---|---|---|---|
| 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 |
Self-correction: , not 22.
Let's correct the and for the last row.
.
.
The is correct: .
Corrected Table:
| Class Interval | Midpoint (x) | Frequency (f) | Cumulative Frequency (cf) | |||
|---|---|---|---|---|---|---|
| 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 |
Part a· 2 steps
Mean
Calculate the sum of (frequency midpoint) and the total frequency.
From the table, and .
Apply the formula for the mean ().
Part b· 2 steps
Median
Determine the median position and median class.
The total frequency is . The median position is the -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.Identify the values for the median formula for grouped data.
Lower boundary of median class () = 9.5
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