1. **State the problem:** We have weights of 42 students and need to a) Prepare a frequency table with class intervals of width 5. b) Draw an ogive and locate the median. c) Calculate median and mode from the frequency table. 2. **Organize data into classes:** Find minimum and maximum weights: Minimum weight = 106.3 Maximum weight = 139.6 Classes of width 5: $105-110$, $110-115$, $115-120$, $120-125$, $125-130$, $130-135$, $135-140$ 3. **Tally data into frequency table:** Class Interval | Frequency 105-110 | Count weights between 105 and <110 110-115 | Count weights between 110 and <115 115-120 | Count weights between 115 and <120 120-125 | Count weights between 120 and <125 125-130 | Count weights between 125 and <130 130-135 | Count weights between 130 and <135 135-140 | Count weights between 135 and <140 Counting each: 105-110: 106.3,106.3,107.8,106.9,109.7,109.8,107.9 → 7 110-115: 115.6 → 1 115-120: 118.9,117.3,119.0,118.3 → 4 120-125: 120.5,120.8,120.8,120.4,121.9,121.8,122.5,123.7,123.8,124.2,124.5,124.5,124.5,124.6,124.7,124.7,125.8,125.9 → 18 125-130: 126.2,127.5,128.9,129.0,129.5 → 5 130-135: 131.8,134.7,134.8 → 3 135-140: 136.8,138.9,139.5,139.6 → 4 4. **Frequency table:** | Class | Frequency | |------------|-----------| | 105 - 110 | 7 | | 110 - 115 | 1 | | 115 - 120 | 4 | | 120 - 125 | 18 | | 125 - 130 | 5 | | 130 - 135 | 3 | | 135 - 140 | 4 | Total observations = 42 5. **Calculate cumulative frequency (CF) for median:** | Class | Freq | CF | |------------|-------|-----| | 105 - 110 | 7 | 7 | | 110 - 115 | 1 | 8 | | 115 - 120 | 4 | 12 | | 120 - 125 | 18 | 30 | | 125 - 130 | 5 | 35 | | 130 - 135 | 3 | 38 | | 135 - 140 | 4 | 42 | 6. **Locate Median Class:** Median position = $\frac{42}{2} = 21$ The CF just greater or equal to 21 is 30 at class 120-125, so median class is $120-125$ 7. **Calculate Median using formula:** $$\text{Median} = L + \left(\frac{\frac{n}{2} - F}{f}\right) \times h$$ Where: $L$ = lower boundary of median class = 120 $n$ = total frequency = 42 $F$ = CF before median class = 12 $f$ = frequency of median class = 18 $h$ = class width = 5 Calculate: $$\text{Median} = 120 + \left(\frac{21 - 12}{18}\right) \times 5 = 120 + \frac{9}{18} \times 5 = 120 + 0.5 \times 5 = 120 + 2.5 = 122.5$$ 8. **Calculate Mode approximately by formula:** Mode class is the class with highest frequency = $120-125$ Let: $d_1$ = difference between frequency of modal class and previous class = $18 - 4 = 14$ $d_2$ = difference between frequency of modal class and next class = $18 - 5 = 13$ Mode formula: $$\text{Mode} = L + \left(\frac{d_1}{d_1 + d_2}\right) \times h = 120 + \left(\frac{14}{14 + 13}\right) \times 5 = 120 + \frac{14}{27} \times 5$$ Calculate: $$= 120 + 0.5185 \times 5 = 120 + 2.59 = 122.59$$ 9. **Ogive:** Plot cumulative frequencies against upper class boundaries: 110 (7), 115 (8), 120 (12), 125 (30), 130 (35), 135 (38), 140 (42) Median corresponds to CF = 21, can be located on the curve between 120 and 125. **Final answers:** Median = 122.5 Mode = 122.59