1. **Problem Statement:** Find the class boundaries, class width, class marks, construct the frequency table, histogram, and determine mode, median, quartiles Q1 and Q3, and standard deviation for the IQ scores. 2. **Class Boundaries:** Class boundaries are the actual limits of the classes, found by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. For example, for 80-94: Lower boundary = 80 - 0.5 = 79.5 Upper boundary = 94 + 0.5 = 94.5 Similarly for others: 95-109: 94.5 - 109.5 110-124: 109.5 - 124.5 125-139: 124.5 - 139.5 140-154: 139.5 - 154.5 3. **Class Width:** Class width = Upper boundary - Lower boundary for any class = 94.5 - 79.5 = 15 4. **Class Marks (Midpoints):** Class mark = (Lower limit + Upper limit) / 2 For 80-94: (80 + 94)/2 = 87 95-109: 102 110-124: 117 125-139: 132 140-154: 147 5. **Frequency Table:** | Class Interval | Frequency | Class Boundaries | Class Mark | | 80-94 | 10 | 79.5 - 94.5 | 87 | | 95-109 | 55 | 94.5 - 109.5 | 102 | | 110-124 | 93 | 109.5 - 124.5 | 117 | | 125-139 | 33 | 124.5 - 139.5 | 132 | | 140-154 | 9 | 139.5 - 154.5 | 147 | 6. **Mode:** Mode class is the class with highest frequency = 110-124 Mode formula for grouped data: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ Where: $L=109.5$ (lower boundary of modal class), $f_1=93$ (frequency of modal class), $f_0=55$ (frequency before modal class), $f_2=33$ (frequency after modal class), $h=15$ (class width) Calculate: $$\text{Mode} = 109.5 + \frac{(93 - 55)}{(2 \times 93 - 55 - 33)} \times 15 = 109.5 + \frac{38}{(186 - 88)} \times 15 = 109.5 + \frac{38}{98} \times 15$$ $$= 109.5 + 5.816 = 115.316$$ 7. **Median:** Total frequency $N=200$ Median class is where cumulative frequency $rac{N}{2} = 100$ lies. Cumulative frequencies: 10, 65, 158, 191, 200 Median class = 110-124 Median formula: $$\text{Median} = L + \frac{\frac{N}{2} - F}{f} \times h$$ Where: $L=109.5$, $F=65$ (cumulative frequency before median class), $f=93$, $h=15$ Calculate: $$\text{Median} = 109.5 + \frac{100 - 65}{93} \times 15 = 109.5 + \frac{35}{93} \times 15 = 109.5 + 5.645 = 115.145$$ 8. **Quartiles:** $Q_1$ position = $\frac{N}{4} = 50$ $Q_3$ position = $\frac{3N}{4} = 150$ $Q_1$ class: cumulative frequency just greater than 50 is 65 (95-109) $$Q_1 = 94.5 + \frac{50 - 10}{55} \times 15 = 94.5 + \frac{40}{55} \times 15 = 94.5 + 10.909 = 105.409$$ $Q_3$ class: cumulative frequency just greater than 150 is 158 (110-124) $$Q_3 = 109.5 + \frac{150 - 65}{93} \times 15 = 109.5 + \frac{85}{93} \times 15 = 109.5 + 13.710 = 123.210$$ 9. **Standard Deviation:** Use formula for grouped data: $$\bar{x} = \frac{\sum f x}{N}$$ Calculate mean: $$\sum f x = 10 \times 87 + 55 \times 102 + 93 \times 117 + 33 \times 132 + 9 \times 147 = 870 + 5610 + 10881 + 4356 + 1323 = 23040$$ $$\bar{x} = \frac{23040}{200} = 115.2$$ Calculate variance: $$\sigma^2 = \frac{\sum f x^2}{N} - \bar{x}^2$$ Calculate $\sum f x^2$: $$10 \times 87^2 + 55 \times 102^2 + 93 \times 117^2 + 33 \times 132^2 + 9 \times 147^2$$ $$= 10 \times 7569 + 55 \times 10404 + 93 \times 13689 + 33 \times 17424 + 9 \times 21609$$ $$= 75690 + 572220 + 1272877 + 574992 + 194481 = 2596260$$ Variance: $$\sigma^2 = \frac{2596260}{200} - 115.2^2 = 12981.3 - 13271.04 = -289.74$$ Since variance cannot be negative, re-check calculations or use corrected formula: Use: $$\sigma^2 = \frac{\sum f (x - \bar{x})^2}{N}$$ Calculate $\sum f (x - \bar{x})^2$: $$(87 - 115.2)^2 \times 10 = 7958.4$$ $$(102 - 115.2)^2 \times 55 = 9497.4$$ $$(117 - 115.2)^2 \times 93 = 318.2$$ $$(132 - 115.2)^2 \times 33 = 9246.7$$ $$(147 - 115.2)^2 \times 9 = 8995.2$$ Sum = 7958.4 + 9497.4 + 318.2 + 9246.7 + 8995.2 = 36015.9 Variance: $$\sigma^2 = \frac{36015.9}{200} = 180.08$$ Standard deviation: $$\sigma = \sqrt{180.08} = 13.42$$ --- **Final answers:** - Class boundaries: 79.5-94.5, 94.5-109.5, 109.5-124.5, 124.5-139.5, 139.5-154.5 - Class width: 15 - Class marks: 87, 102, 117, 132, 147 - Mode: 115.316 - Median: 115.145 - Quartile Q1: 105.409 - Quartile Q3: 123.210 - Standard deviation: 13.42