Refer to the given set of data below to answer Questions 6 and 7. Then refer to the given table below to answer Questions 8 - 10. Finally, refer to the donations summary and the company director's question for question 11.

Here are the solutions to the questions: For Questions 6 and 7: Step 1: Arrange the scores in ascending order. The given scores are: 15, 7, 8, 12, 9, 13, 13. Arranging them from smallest to largest: 7, 8, 9, 12, 13, 13, 15 6. Arrange the scores in ascending order. Answer: 7, 8, 9, 12, 13, 13, 15 Step 2: Find the median of the set of scores. The median is the middle value in an ordered set of numbers. There are 7 scores (an odd number). The median is the ((n+1)/(2))-th term. Median position = (7+1)/(2) = (8)/(2) = 4-th term From the ordered list (7, 8, 9, 12, 13, 13, 15), the 4th term is 12. 7. What is the median of the set of scores? Answer: 12 For Questions 8 - 10: Step 3: Determine the value of 'a'. The value 'a' is the lower class limit for the class interval 16-20. Observing the pattern of class limits: For 1-5, class limits are 0.5 - 5.5 For 6-10, class limits are 5.5 - 10.5 For 11-15, class limits are 10.5 - 15.5 Following this pattern, the lower class limit for 16-20 should be the upper class limit of the preceding class, which is 15.5. 8. What must be the value of a in the table? Answer: 15.5 Step 4: Determine the value of 'b'. The value 'b' is the "Greater than cumulative frequency" (>cf) for the class 6-10. This is the sum of the frequencies of the current class and all classes with higher values. Frequencies (f) are: 16-20: 5 11-15: 7 6-10: 9 1-5: 4 b = Frequency(6-10) + Frequency(11-15) + Frequency(16-20) b = 9 + 7 + 5 = 21 9. What must be the value of b in the table? Answer: 21 Step 5: Find the median of the set of grouped data. The total frequency N = 25. The median position is (N)/(2) = (25)/(2) = 12.5. We look at the "Less than cumulative frequency" (<cf) column to find the median class: For 1-5, <cf = 4 For 6-10, <cf = 13 (since 4+9=13). This class contains the 12.5th observation. So, the median class is 6-10. Now, apply the median formula for grouped data: Median = L + ((N)/(2) - CF_bf_m) × w Where: L = Lower boundary of the median class = 5.5 (from class limits for 6-10) N = Total frequency = 25 CF_b = Cumulative frequency of the class before* the median class = 4 (for 1-5) f_m = Frequency of the median class = 9 (for 6-10) w = Class width = Upper boundary - Lower boundary = 10.5 - 5.5 = 5 Median = 5.5 + ((25)/(2) - 49) × 5 Median = 5.5 + ((12.5 - 4)/(9)) × 5 Median = 5.5 + ((8.5)/(9)) × 5 Median = 5.5 + (0.9444...) × 5 Median = 5.5 + 4.7222... Median ≈ 10.22 10. What is the median of the set of data? Answer: 10.22 For Question 11: Step 6: Determine the best measure of central tendency. The donations are: K10: 8 people K20: 24 people K100: 2 people Total number of people = 8 + 24 + 2 = 34. Let's find the mean, median, and mode: Mode: The most frequent donation is K20, contributed by 24 people. Median: The middle value. Since there are 34 people, the median is the average of the 17th and 18th values. The first 8 donations are K10. The next 24 donations (from 9th to 32nd) are K20. Both the 17th and 18th donations are K20. So, the median is K20. Mean: Total amount = (8 × 10) + (24 × 20) + (2 × 100) = 80 + 480 + 200 = 760 Mean = (760)/(34) ≈ 22.35 The mean (K22.35) is higher than what most people contributed due to the two K100 donations. The median (K20) and mode (K20) are identical and represent the most common contribution. When data is skewed or has outliers, the median or mode are generally better measures of central tendency to describe a "typical" value than the mean. Since the question asks "how much did each employee contributed" and there's a clear most frequent amount, the mode best describes the most common contribution. 11. The company's director asked one of the missionaries how much did each employee contributed. What measure of central tendency best describes the answer? Answer: Mode 3 done, 2 left today. You're making progress.