2.2 Assignment

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Feb 20, 2024

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1 2.2 Assignment : Module 2 Assignment Hyunjo Kwak Embry-Riddle Aeronautical University RSCH 665: Statistical Anyls Professor Samuel Benavides, PhD January 21, 2024

2 Measures of Central tendency and Variability Emmett Mean: (94 + 74 + 63 + 59 + 74 + 92) / 6 = 76 Mode: 74 (occurs twice) Median: Arrange the scores in ascending order: 59, 63, 74, 74, 92, 94 Median = (74 + 74) / 2 = 74 Kiernan Mean: (84 + 83 + 73 + 80 + 88 + 80) / 6 = 81.33 Mode: No mode (all scores are unique) Median: Arrange the scores in ascending order: 73, 80, 80, 83, 84, 88 Median = (80 + 83) / 2 = 81.5 Brennan Mean: (92 + 68 + 69 + 82 + 92 + 78) / 6 = 80.17 Mode: 92 (occurs twice) Median: Arrange the scores in ascending order: 68, 69, 78, 82, 92, 92 Median = (78 + 82) / 2 = 80 Suggestions for each student Emmett: Given that the scores are relatively close, any of the measures (mean, mode, or median) would be reasonable. Emmett might choose the mean as it provides an average value. Kiernan: Since there is no mode, and the scores are somewhat evenly distributed, the median might be a good representation of the central tendency. Brennan: With repeated occurrences of 92, the mode might be a suitable choice for Brennan to highlight the most frequent score.

3 Measures of Variance and Standard Deviation Class 1: Mean 78.67 , Variance 197.75, Std. Dev 14.06 Class 2: Mean 80.56, Variance 62.53, Std. Dev 7.91 Class 3: Mean 85.11, Variance 48.36, Std. Dev 6.95 Class 4: Mean 87.11, Variance 90.11, Std. Dev 9.49 Class 5: Mean 84.44, Variance 138.02, Std. Dev 11.75 No.3 Question Mean : 15.04, Variance : 82.3, Std. Dev : 9.07 Difference between Sample and Population Standard Deviation Sample standard deviation: This is an estimate of the population standard deviation based on a limited set of data (in this case, the 24 airport security guards). It provides an approximation of how spread out the values are in the sample. Population standard deviation: This is the true measure of how spread out the values are in the entire population of airport security guards. It is calculated using the same formula as the sample standard deviation, but instead of dividing by n-1, we divide by n (the total number of individuals in the population). Therefore, the sample standard deviation we calculated tells us about the variability of the security risks identified by these specific 24 guards, while the population standard deviation would tell us about the variability of risk identification for all airport security guards.

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4 Z table Questions 17 years old candidate Calculate the z-score: z-score = (x - μ) / σ x = 72.1 (graduation rate of the specific flight academy) μ = 80 (mean graduation rate for 17-year-olds) σ = 2.1 (standard deviation for 17-year-olds) z-score = (72.1 - 80) / 2.1 = -3.76 Use the Z-table: Look up the area to the left of z = -3.76 on the Z-table. The area is approximately 0.0001, meaning only 0.01% of flight academies have a graduation rate lower than 72.1 for 17-year-olds. Calculate the percentage with a higher rate: Subtract the area to the left from 100% to find the percentage with a higher rate. 100% - 0.01% = 99.99% Therefore, approximately 99.99% of flight academies have a higher graduation rate for 17-year- old candidates than 72.1. 34 years old candidate Calculate the z-score: z = (83.5 - 83.1) / 1.3 = 0.31 Look up the area to the left of z = 0.31 on the Z-table. This area is approximately 0.6217.

5 Calculate the percentage with a lower rate: 100% - 62.17% = 37.83% Therefore, approximately 37.83% of flight academies have a lower graduation rate for 34-year- old candidates than 83.5%. 29 years old candidate Calculate the z-score: z = (93.1 - 91.1) / 1.0 = 2 Look up the area to the left of z = 2 on the Z-table. This area is approximately 0.9772. Calculate the percentage below the academy's rate: 100% - 97.72% = 2.28% Since 2.28% of academies have a higher rate than 93.1, the academy is not in the top 10%. Their claim is false. 18 year old candidate This decision depends on the academy's definition of "success" and the cost of the new methods. The academy's graduation rate of 82.6 is above the average for 18-year-olds (78.5). However, if they want to significantly increase their graduation rate or compete with the top academies (whose rates are in the 85-90% range), the new methods might be worthwhile. Ultimately, the decision should be based on a cost-benefit analysis considering the potential improvement in graduation rate, the cost of the new methods, and the academy's overall goals.

6 27 years old candidate Find the z-scores for both periods: 2000-2009: z = (85.1 - 87.5) / 1.8 = -1.33 2010-2018: z = (88.3 - 87.5) / 1.8 = 0.44 Look up the areas to the left of these z-scores on the Z-table: 2000-2009: Approximately 0.0917 2010-2018: Approximately 0.67 Convert these areas to percentiles: 2000-2009: 9.17th percentile 2010-2018: 67th percentile This means that: In 2000-2009, the academy's graduation rate was below 9.17% of all academies for 27-year-olds. In 2010-2018, the academy's graduation rate was above 67% of all academies for 27-year-olds. Therefore, the new instructors and clearer instruction significantly improved the academy's performance. They moved from the bottom 10% to the top 35% in terms of graduation rate for 27-year-old candidates.

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