1. Problem 54: Find the median score of the seven students with scores 48, 62, 62, 98, 76, 88, 70. 2. Sort the scores in ascending order: 48, 62, 62, 70, 76, 88, 98. 3. Since there are 7 scores (odd number), median is the middle value, which is the 4th score. 4. The 4th score is 70. Answer: B. 70 --- 5. Problem 55: Find mean, median, and mode for hourly pay rates with frequencies. 6. List pay rates with frequencies: 9.00 (4), 12.50 (29), 15.50 (7), 20.00 (4). 7. Total employees = 4 + 29 + 7 + 4 = 44. 8. Calculate mean: $$\text{Mean} = \frac{(9*4) + (12.5*29) + (15.5*7) + (20*4)}{44} = \frac{36 + 362.5 + 108.5 + 80}{44} = \frac{587}{44} \approx 13.34$$ 9. Find median: - The median position is the $\frac{44 + 1}{2} = 22.5$th value. - Count cumulative frequencies: - 9.00: first 4 employees - 12.50: next 29 employees (positions 5 to 33), so median lies in 12.50 class. - Thus, median is 12.50. 10. Find mode: The pay rate with highest frequency is 12.50 (29 times). Answer: Mean: 13.34, Median: 12.50, Mode: 12.50 Corresponds to C. --- 11. Problem 56: Find the midrange of temperatures 54, 58, 65, 69, 71, 67, 65, 65, 60. 12. Midrange is average of minimum and maximum values. 13. Minimum = 54, Maximum = 71. 14. Midrange = $\frac{54 + 71}{2} = \frac{125}{2} = 62.5$. Answer: D. 62.5 --- 15. Problem 57: Find needed fifth quiz score for mean 87 given four quiz scores: 89, 90, 82, 77. 16. Use formula: $\text{Mean} = \frac{\text{Sum of scores}}{5}.$ 17. Sum of first 4 scores: $89 + 90 + 82 + 77 = 338.$ 18. Let fifth score be $x.$ Then: $$87 = \frac{338 + x}{5} \Rightarrow 338 + x = 435 \Rightarrow x = 435 - 338 = 97.$$ Answer: E. 97% --- 19. Problem 58: Find weighted average grade for Maegan. 20. Weights: each test = 3 quizzes, final exam = 2 tests = 6 quizzes. 21. Maegan's scores: - Tests: 87, 96 - Quizzes: 86, 72, 83 - Final: 93 22. Convert all to quiz-equivalent points: - Tests: $(87 + 96) * 3 = (183)*3 = 549$ quiz-points - Quizzes: $86 + 72 + 83 = 241$ quiz-points - Final exam: $93 * 6 = 558$ quiz-points 23. Total quiz-points: $549 + 241 + 558 = 1348$ 24. Total weight: $2*3 + 3 + 6 = 6 +3 +6 = 15$ 25. Weighted mean = $\frac{1348}{15} \approx 89.87$ rounded to 90. Answer: B. 90 --- 26. Problem 59: Science of collecting, organizing, analyzing, interpreting data is Statistics. Answer: B. Statistics --- 27. Problem 60: Collection of all outcomes, responses, counts of interest is Population. Answer: E. Population --- 28. Problem 61: Data entry with greatest frequency is called Mode. Answer: E. Mode --- 29. Problem 62: Mean with varying weights is Weighted Mean. Answer: C. Weighted Mean --- 30. Problem 63: Range of starting salaries 41, 38, 39, 45, 47, 41, 44, 41, 37, 42. 31. Max = 47, Min = 37 32. Range = 47 - 37 = 10. Answer: D. 10 --- 33. Problem 64: If $x - 6 = 9$, find $3x + 1$. 34. Solve $x = 9 + 6 = 15$ 35. Compute $3x + 1 = 3*15 +1 = 45 +1 = 46$. Note: No matching option; re-check original problem. Oops, options probably typo or question misread. Answer none of above. But with given options, closest reading mistake? If $x-6=9$ then $x=15$. $3x+1=46$, no option matches. Assuming typo in options or question. --- 36. Problem 65: Which expression equals $b + c - d$? Options involve expressions with products which don't simplify to the sum $b + c - d$. Answer is none matches. --- 37. Problem 66: Area under standard normal curve to right of $z=1$ is approximately 0.158. Answer: A. 0.158 --- 38. Problem 67: Area to right of $z=0$ is 0.5. Answer: A. 0.50 --- 39. Problem 68: For mean 60, SD 10, probability of value greater than 70. Calculate z-score: $$z=\frac{70-60}{10} = 1$$ Area to right of $z=1$ is 0.158 or 15.8%. Answer: A. 15.8% --- 40. Problem 69: Mean 120, SD 15, probability variable > 150. Calculate $z=\frac{150-120}{15} = 2$ Area to right of $z=2$ is approx 0.023. Answer: A. 0.023 --- 41. Problem 70: Heights mean 160, SD 10, probability taller than 170. $z=\frac{170-160}{10} = 1$ Area to right $z=1$ approx 0.158. Answer: A. 0.158 --- 42. Problem 36: Longest length among 5.5cm, 5.5m, 0.0055km, 5555mm, 0.0005dm. Convert all to cm: - 5.5 cm = 5.5 cm - 5.5 m = 550 cm (1 m = 100 cm) - 0.0055 km = 550 cm (1 km = 100000 cm) - 5555 mm = 555.5 cm (1 mm = 0.1 cm) - 0.0005 dm = 0.005 cm (1 dm = 10 cm) Largest is 550 cm (5.5 m or 0.0055 km). Answer: B. 5.5 m --- 43. Problem 37: Radius of circle with area $25 \pi$ cm². Area formula: $A = \pi r^2$ Given $25 \pi = \pi r^2$, divide both sides by $\pi$: $r^2 = 25$ $r = \sqrt{25} = 5$ cm. Answer: D. 5 cm --- 44. Problem 38: Pancake cut in 8 pieces, 3 eaten, fraction left. Pieces left = 8 -3 = 5 pieces. Fraction left = $\frac{5}{8}$. Answer: C. 5/8 --- 45. Problem 39: First day Sept is Wednesday, find last day. September has 30 days. 30 mod 7 = 2 remainder days. Add 29 days to Wednesday gives same weekday. So day 1 Wednesday, day 29 Wednesday + 28 days= Wednesday. Day 30 = day after Wednesday (Thursday). Answer: C. Thursday --- 46. Problem 40: Angle is 25 more than supplement. Let angle be $x$, supplement is $180 - x$. Given $x = (180 - x) + 25$ $2x = 205$ $x = 102.5$ (larger angle). Answer: C. 102.5 --- 47. Problem 41: Ruben’s grades 88,89,84,90,91,86; find Math grade for average 88. Let Math grade be $x$. Sum existing grades = 88 + 89 + 84 + 90 + 91 + 86 = 528. Total grades = 7. Mean = 88 = (528 + x)/7 Multiply: $528 + x = 616$ $x=88$ Answer: B. 88 --- 48. Problem 42: Aquarium volume with length 9cm, width 5cm, height 7cm. Volume = $l \times w \times h = 9*5*7 = 315$ cm³. Answer: C. 315 cm³ --- 49. Problem 43: Median score of 8 students: 12,10,13,11,15,20,19,17. Sort: 10,11,12,13,15,17,19,20. Median for even n=8 is average of 4th and 5th values. Median = $\frac{13+15}{2} = 14$. Answer: B. 14 --- 50. Problem 44: Area=64, length=16, find perimeter. Width $w = \frac{64}{16} = 4$. Perimeter $= 2(l + w) = 2(16 +4) = 2*20 = 40$. Answer: D. 40 --- 51. Problem 45: Number ways to seat 5 students at round table. Number of arrangements for n people around a round table = $(n-1)! = 4! = 24$. Answer: D. 24 --- 52. Problem 46: Midpoint joining (6,4) and (3,-4). Midpoint $= \left(\frac{6+3}{2}, \frac{4+(-4)}{2}\right) = \left(\frac{9}{2}, 0\right)$. Answer: D. (9/2, 0) --- 53. Problem 47: Digit in thousandths place of 5672.134987 is 9. Answer: C. 9 --- 54. Problem 48: 20 men build in 60 days, how many men needed for 15 days? Using inverse proportionality: $20 * 60 = x * 15$ $x = \frac{20 * 60}{15} = 80$ men. Answer: C. 80 --- 55. Problem 49: Median score interval from frequency table. Sum frequencies: 2+4+6+12+10+8+6+4= 52 Median position: 26.5 Cumulative frequencies: 16-23: 2 24-31: 6 32-39: 12 40-47: 24 48-55: 34 Median lies in 48-55 (since 26.5 between 24 and 34) Answer: C. 48-55 --- 56. Problem 50: How much larger is $2^5$ than $5^{2n}$ (without value of n given, insufficient data). Cannot solve. --- 57. Problem 51: Mode of data 12,6,9,8,10,5,12,7 is 12 (occurs twice). Answer: E. 12 --- 58. Problem 52: Same as 41 above: grade must be 88. Answer: B. 88 --- 59. Problem 53: Median of math test scores same as 43 above is 14. Answer: B. 14