1. **Problem Statement:** Construct a frequency distribution for the given 50 Math 111 scores using a class interval of 5. 2. **Step 1: Determine the range and class intervals.** - Minimum score = 25 - Maximum score = 81 - Class interval = 5 - Classes: 25-29, 30-34, 35-39, 40-44, 45-49, 50-54, 55-59, 60-64, 65-69, 70-74, 75-79, 80-84 3. **Step 2: Tally the scores into these classes:** - 25-29: 25, 28 - 30-34: None - 35-39: 39, 39 - 40-44: 40, 43 - 45-49: 46, 46, 45, 48, 47 - 50-54: 52, 52, 53, 54 - 55-59: 55, 55, 55, 57, 57, 58, 58, 59 - 60-64: 60, 61, 62, 62, 63, 63, 63 - 65-69: 67, 68, 68, 69, 69, 69 - 70-74: 70, 70, 70, 71 - 75-79: 79 - 80-84: 80, 80, 81, 81 4. **Step 3: Frequency distribution table:** | Class Interval | Frequency | |----------------|-----------| | 25 - 29 | 2 | | 30 - 34 | 0 | | 35 - 39 | 2 | | 40 - 44 | 2 | | 45 - 49 | 5 | | 50 - 54 | 4 | | 55 - 59 | 8 | | 60 - 64 | 7 | | 65 - 69 | 6 | | 70 - 74 | 4 | | 75 - 79 | 1 | | 80 - 84 | 4 | 5. **Step 4: Calculate percentage for each class:** Percentage = $\frac{\text{Frequency}}{50} \times 100$% 6. **Step 5: Calculate midpoints for each class:** Midpoint = $\frac{\text{Lower limit} + \text{Upper limit}}{2}$ 7. **Step 6: Calculate mean using short method:** - Choose assumed mean $A = 54.5$ (midpoint of 50-54 class) - Calculate $d = \text{midpoint} - A$ - Calculate $fd = \text{frequency} \times d$ - Mean formula: $$\bar{x} = A + \frac{\sum fd}{\sum f}$$ | Class | Midpoint (x) | d = x - 54.5 | Frequency (f) | fd | |-------|--------------|--------------|---------------|----| |25-29 | 27 | -27.5 | 2 | -55| |30-34 | 32 | -22.5 | 0 | 0 | |35-39 | 37 | -17.5 | 2 | -35| |40-44 | 42 | -12.5 | 2 | -25| |45-49 | 47 | -7.5 | 5 | -37.5| |50-54 | 52 | -2.5 | 4 | -10| |55-59 | 57 | 2.5 | 8 | 20 | |60-64 | 62 | 7.5 | 7 | 52.5| |65-69 | 67 | 12.5 | 6 | 75 | |70-74 | 72 | 17.5 | 4 | 70 | |75-79 | 77 | 22.5 | 1 | 22.5| |80-84 | 82 | 27.5 | 4 | 110| Sum $f = 50$, Sum $fd = 107.5$ Mean: $$\bar{x} = 54.5 + \frac{107.5}{50} = 54.5 + 2.15 = 56.65$$ 8. **Step 7: Calculate median:** - Median class is where cumulative frequency reaches $\frac{50}{2} = 25$ - Cumulative frequencies: 2, 2, 4, 6, 11, 15, 23, 30, 36, 40, 41, 45 - Median class: 60-64 (cumulative frequency 30) - Median formula: $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times c$$ Where: - $L=59.5$ (lower boundary of median class) - $N=50$ - $F=23$ (cumulative frequency before median class) - $f_m=7$ (frequency of median class) - $c=5$ (class width) Calculate: $$\text{Median} = 59.5 + \left(\frac{25 - 23}{7}\right) \times 5 = 59.5 + \frac{2}{7} \times 5 = 59.5 + 1.43 = 60.93$$ 9. **Step 8: Calculate mode:** - Modal class is class with highest frequency = 55-59 (frequency 8) - Mode formula: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times c$$ Where: - $L=54.5$ (lower boundary of modal class) - $f_1=8$ (frequency of modal class) - $f_0=4$ (frequency before modal class) - $f_2=7$ (frequency after modal class) - $c=5$ Calculate: $$\text{Mode} = 54.5 + \frac{8 - 4}{2 \times 8 - 4 - 7} \times 5 = 54.5 + \frac{4}{16 - 11} \times 5 = 54.5 + \frac{4}{5} \times 5 = 54.5 + 4 = 58.5$$ 10. **Step 9: Calculate quartiles:** - $Q_1$ position = $\frac{N}{4} = 12.5$ - $Q_2$ is median = 60.93 (already calculated) - $Q_3$ position = $\frac{3N}{4} = 37.5$ Cumulative frequencies: 2, 2, 4, 6, 11, 15, 23, 30, 36, 40, 41, 45 - $Q_1$ class: 50-54 (cumulative frequency 15) - $Q_3$ class: 70-74 (cumulative frequency 40) Calculate $Q_1$: $$Q_1 = L + \left(\frac{12.5 - F}{f}\right) \times c = 49.5 + \left(\frac{12.5 - 11}{4}\right) \times 5 = 49.5 + \frac{1.5}{4} \times 5 = 49.5 + 1.875 = 51.375$$ Calculate $Q_3$: $$Q_3 = L + \left(\frac{37.5 - F}{f}\right) \times c = 69.5 + \left(\frac{37.5 - 36}{4}\right) \times 5 = 69.5 + \frac{1.5}{4} \times 5 = 69.5 + 1.875 = 71.375$$ **Final answers:** - Mean = 56.65 - Median = 60.93 - Mode = 58.5 - $Q_1 = 51.38$ - $Q_2 = 60.93$ - $Q_3 = 71.38$