1. **State the problem:** We have IQ scores of 60 children and need to form a frequency distribution with class intervals of width 15, then locate the mode graphically. 2. **Determine class intervals:** The minimum IQ is 43 and the maximum is 153. Start from 40 (a convenient lower bound) and add intervals of width 15: $$40-54, 55-69, 70-84, 85-99, 100-114, 115-129, 130-144, 145-159$$ 3. **Count frequencies for each class:** - 40-54: 43, 54 → 2 - 55-69: 67 → 1 - 70-84: 72, 73, 76, 76, 79, 79, 80, 81, 84, 84 → 10 - 85-99: 85, 87, 88, 89, 89, 90, 93, 94, 94, 94, 95, 96, 96, 97, 97, 98, 98, 99 → 18 - 100-114: 100, 101, 101, 102, 102, 102, 103, 103, 105, 107, 109, 109, 110, 110, 111, 112, 114, 115, 115 → 19 - 115-129: 118, 119, 120, 122, 123, 127, 128 → 7 - 130-144: 132 → 1 - 145-159: 146, 153 → 2 4. **Frequency distribution table:** | Class Interval | Frequency | |----------------|-----------| | 40 - 54 | 2 | | 55 - 69 | 1 | | 70 - 84 | 10 | | 85 - 99 | 18 | | 100 - 114 | 19 | | 115 - 129 | 7 | | 130 - 144 | 1 | | 145 - 159 | 2 | 5. **Locate the mode graphically:** The mode corresponds to the class with the highest frequency, which is 100-114 with frequency 19. To locate the mode more precisely, we use the formula for modal class: $$\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ where: - $L$ = lower boundary of modal class = 99.5 - $f_1$ = frequency of modal class = 19 - $f_0$ = frequency of class before modal = 18 - $f_2$ = frequency of class after modal = 7 - $h$ = class width = 15 Calculate: $$\text{Mode} = 99.5 + \left(\frac{19 - 18}{2 \times 19 - 18 - 7}\right) \times 15 = 99.5 + \left(\frac{1}{38 - 18 - 7}\right) \times 15 = 99.5 + \left(\frac{1}{13}\right) \times 15 = 99.5 + 1.15 = 100.65$$ 6. **Interpretation:** The mode IQ is approximately 100.65, indicating the most frequent IQ range is just above 100.