1. Statement: When the mean is computed for individual data, all values in the data set are used. Explanation: True. The mean (average) is found by summing all values and dividing by the count of values, so every data point affects the mean. 2. Statement: The mean cannot be found for grouped data when there is an open class. Explanation: True. An open class means the class interval has no definite upper or lower bound, so it's impossible to know exact values for midpoint calculations, preventing accurate mean calculation. 3. Statement: A single, extremely large value can affect the median more than the mean. Explanation: False. The mean is sensitive to extremely large values (outliers) and will shift more, while the median is resistant since it depends on the middle position, not the actual value. 4. Inches of Rain Frequency Distribution: Given classes and frequencies: | Class Interval | Frequency | | 5.5–20.5 | 2 | | 20.5–35.5 | 3 | | 35.5–50.5 | 8 | | 50.5–65.5 | 6 | | 65.5–80.5 | 3 | | 80.5–95.5 | 3 | Total frequencies sum: $2+3+8+6+3+3=25$ 4a. Mean: Calculate midpoints for each class: $\frac{5.5+20.5}{2}=13$, $\frac{20.5+35.5}{2}=28$, $\frac{35.5+50.5}{2}=43$, $\frac{50.5+65.5}{2}=58$, $\frac{65.5+80.5}{2}=73$, $\frac{80.5+95.5}{2}=88$ Multiply midpoints by frequencies and sum: $13*2+28*3+43*8+58*6+73*3+88*3=26+84+344+348+219+264=1285$ Mean = $\frac{1285}{25}=51.4$ 4b. Modal Class: The modal class is the class with the highest frequency, which is 8 for $35.5–50.5$ inches. 4c. Variance: Calculate variance using midpoints. Calculate mean=51.4 (from above). Compute $\sum f(x-\bar{x})^2$: For each class: $(13-51.4)^2*2=1480.96*2=2961.92$ $(28-51.4)^2*3=544.36*3=1633.08$ $(43-51.4)^2*8=70.56*8=564.48$ $(58-51.4)^2*6=43.56*6=261.36$ $(73-51.4)^2*3=462.76*3=1388.28$ $(88-51.4)^2*3=1334.56*3=4003.68$ Sum = $2961.92+1633.08+564.48+261.36+1388.28+4003.68 = 10812.8$ Variance = $\frac{10812.8}{25} = 432.512$ 4d. Standard Deviation: Standard deviation = $\sqrt{432.512} \approx 20.8$ 5. Reaction Times Data: Count data and values: 2.4 appears 9 times 3.1 appears 12 times 3.8 appears 7 times 4.5 appears 5 times 5.2 appears 2 times 5.9 appears 1 time Total n = 9+12+7+5+2+1=36 5a. Variance: Calculate mean first: Mean = $\frac{2.4*9 + 3.1*12 + 3.8*7 + 4.5*5 + 5.2*2 + 5.9*1}{36} = \frac{21.6 + 37.2 + 26.6 + 22.5 + 10.4 + 5.9}{36} = \frac{124.2}{36} \approx 3.45$ Calculate $\sum f(x-\bar{x})^2$: $(2.4-3.45)^2*9 = ( -1.05)^2*9 = 1.1025*9=9.9225$ $(3.1-3.45)^2*12 = (-0.35)^2*12=0.1225*12=1.47$ $(3.8-3.45)^2*7 = 0.35^2*7=0.1225*7=0.8575$ $(4.5-3.45)^2*5 = 1.05^2*5=1.1025*5=5.5125$ $(5.2-3.45)^2*2 = 1.75^2*2=3.0625*2=6.125$ $(5.9-3.45)^2*1 = 2.45^2*1=6.0025$ Sum = $9.9225 + 1.47 + 0.8575 + 5.5125 + 6.125 + 6.0025 = 29.89$ Variance = $\frac{29.89}{36} \approx 0.83$ 5b. Standard Deviation: $\sqrt{0.83} \approx 0.91$ Final answers summary: 1. True 2. True 3. False 4a. Mean = 51.4 4b. Modal Class = 35.5–50.5 4c. Variance = 432.512 4d. Std Dev = 20.8 5a. Variance = 0.83 5b. Std Dev = 0.91