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 data. 2. **Given Data:** | IQ Score | Frequency | |----------|-----------| | 80-94 | 10 | | 95-109 | 55 | | 110-124 | 93 | | 125-139 | 33 | | 140-154 | 9 | 3. **Class Boundaries:** Class boundaries are found by averaging the upper limit of one class and the lower limit of the next. - Between 94 and 95: $\frac{94 + 95}{2} = 94.5$ - Between 109 and 110: $\frac{109 + 110}{2} = 109.5$ - Between 124 and 125: $\frac{124 + 125}{2} = 124.5$ - Between 139 and 140: $\frac{139 + 140}{2} = 139.5$ So, class boundaries are: - 79.5 - 94.5 - 94.5 - 109.5 - 109.5 - 124.5 - 124.5 - 139.5 - 139.5 - 154.5 4. **Class Width:** Class width = upper boundary - lower boundary of any class For first class: $94.5 - 79.5 = 15$ 5. **Class Marks (Midpoints):** Class mark = $\frac{\text{Lower limit} + \text{Upper limit}}{2}$ - 80-94: $\frac{80 + 94}{2} = 87$ - 95-109: $\frac{95 + 109}{2} = 102$ - 110-124: $\frac{110 + 124}{2} = 117$ - 125-139: $\frac{125 + 139}{2} = 132$ - 140-154: $\frac{140 + 154}{2} = 147$ 6. **Frequency Table:** | Class Boundaries | Frequency | Class Mark | |------------------|-----------|------------| | 79.5 - 94.5 | 10 | 87 | | 94.5 - 109.5 | 55 | 102 | | 109.5 - 124.5 | 93 | 117 | | 124.5 - 139.5 | 33 | 132 | | 139.5 - 154.5 | 9 | 147 | 7. **Mode:** Mode class is the class with highest frequency: 110-124 (frequency 93). 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: $$\text{Mode} = 109.5 + \frac{(93 - 55)}{(2 \times 93 - 55 - 33)} \times 15 = 109.5 + \frac{38}{(186 - 88)} \times 15 = 109.5 + \frac{38}{98} \times 15$$ $$= 109.5 + 5.816 = 115.316$$ 8. **Median:** Total frequency $N = 200$ Median class is where cumulative frequency $\\geq N/2 = 100$ Cumulative frequencies: - 10 - 65 - 158 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) - $f = 93$ (frequency of median class) - $h = 15$ Calculate: $$\text{Median} = 109.5 + \frac{100 - 65}{93} \times 15 = 109.5 + \frac{35}{93} \times 15 = 109.5 + 5.645 = 115.145$$ 9. **Quartiles:** - $Q_1$ position: $\frac{N}{4} = 50$ - $Q_3$ position: $\frac{3N}{4} = 150$ $Q_1$ class: cumulative frequency $\\geq 50$ is 95-109 (cumulative 65) $$Q_1 = 94.5 + \frac{50 - 10}{55} \times 15 = 94.5 + \frac{40}{55} \times 15 = 94.5 + 10.909 = 105.409$$ $Q_3$ class: cumulative frequency $\\geq 150$ is 125-139 (cumulative 158) $$Q_3 = 124.5 + \frac{150 - 158}{33} \times 15 = 124.5 + \frac{-8}{33} \times 15 = 124.5 - 3.636 = 120.864$$ 10. **Standard Deviation:** Calculate mean $\bar{x}$: $$\bar{x} = \frac{\sum f x}{N} = \frac{10 \times 87 + 55 \times 102 + 93 \times 117 + 33 \times 132 + 9 \times 147}{200}$$ $$= \frac{870 + 5610 + 10881 + 4356 + 1323}{200} = \frac{23040}{200} = 115.2$$ Calculate variance: $$\sigma^2 = \frac{\sum f (x - \bar{x})^2}{N}$$ Calculate each term: - $(87 - 115.2)^2 = 795.24$, sum: $10 \times 795.24 = 7952.4$ - $(102 - 115.2)^2 = 174.24$, sum: $55 \times 174.24 = 9583.2$ - $(117 - 115.2)^2 = 3.24$, sum: $93 \times 3.24 = 301.32$ - $(132 - 115.2)^2 = 282.24$, sum: $33 \times 282.24 = 9313.92$ - $(147 - 115.2)^2 = 1008.64$, sum: $9 \times 1008.64 = 9077.76$ Sum all: $7952.4 + 9583.2 + 301.32 + 9313.92 + 9077.76 = 36128.6$ Variance: $$\sigma^2 = \frac{36128.6}{200} = 180.643$$ Standard deviation: $$\sigma = \sqrt{180.643} = 13.44$$ **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: 120.864 - Standard deviation: 13.44