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BSTAT 3321 Assignment 2 20 points each question 1. Problem 3.62 a, b, and d Excel output: Time Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count First Quartile Third Quartile interquartile range c.v (a) mean = 43.89 (b) range = 76 43.88889 4.865816 45 17 25.28352 639.2564 -0.90407 0.517268 76 16 92 1185 27 18 63 45 57.61% median = 45 interquartile range = 45 standard deviation = 25.28 (d) 1st quartile = 18 3rd quartile = 63 variance = 639.2564 coefficient of variation = 57.61% The mean approval process takes 43.89 days with 50% of the policies being approved in less than 45 days. 50% of the applications are approved between 18 and 63 days. About 67% of the applications are approved between 18.6 to 69.2 days. 2. Problem 3.64 a, b, and d Excel output: Width Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count First Quartile Third Quartile Interquartile Range CV 8.420898 0.006588 8.42 8.42 0.046115 0.002127 0.035814 -0.48568 0.186 8.312 8.498 412.624 49 8.404 8.459 0.055 0.55% (a) mean = 8.421, median = 8.42, range = 0.186 and standard deviation = 0.0461. On average, the width is 8.421 inches. The width of the middle ranked observation is 8.42. The difference between the largest and smallest width is 0.186 and majority of the widths fall between 0.0461 inches around the mean of 8.421 inches. (b) Minimum = 8.312, 1st quartile = 8.404, median = 8.42, 3rd quartile = 8.459 and maximum = 8.498 (d) All the troughs fall within the limit of 8.31 and 8.61 inches. 3. Problem 3.67 a, b, c, and e Excel output: Teabags Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count First Quartile Third Quartile Interquartile Range CV 5.5014 0.014967 5.515 5.53 0.10583 0.0112 0.127022 -0.15249 0.52 5.25 5.77 275.07 50 5.44 5.57 0.13 1.9237% (a) mean = 5.5014, median = 5.515, first quartile = 5.44, third quartile = 5.57 (b) range = 0.52, interquartile range = 0.13, variance = 0.0112, standard deviation = 0.10583, coefficient of variation = 1.924% (c) The mean weight of the tea bags in the sample is 5.5014 grams while the middle ranked weight is 5.515. The company should be concerned about the central tendency because that is where the majority of the weight will cluster around. The average of the squared differences between the weights in the sample and the sample mean is 0.0112 whereas the square-root of it is 0.106 gram. The difference between the lightest and the heaviest tea bags in the sample is 0.52. 50% of the tea bags in the sample weigh between 5.44 and 5.57 grams. According to the empirical rule, about 68% of the tea bags produced will have weight that falls within 0.106 grams around 5.5014 grams. The company producing the tea bags should be concerned about the variation because tea bags will not weigh exactly the same due to various factors in the production process, e.g. temperature and humidity inside the factory, differences in the density of the tea, etc. Having some idea about the amount of variation will enable the company to adjust the production process accordingly. (e) On average, the weight of the teabags is quite close to the target of 5.5 grams. Even though the mean weight is close to the target weight of 5.5 grams, the standard deviation of 0.106 indicates that about 75% of the teabags will fall within 0.212 grams around the target weight of 5.5 grams. The interquartile range of 0.13 also indicates that half of the teabags in the sample fall in an interval 0.13 grams around the median weight of 5.515 grams. The process can be adjusted to reduce the variation of the weight around the target mean. 4. Problem 3.42 Excel output: Kilowatt Hours Mean 12999.22 Standard Error 546.9863 Median 13255 Mode #N/A Standard Deviation 3906.264 Sample Variance 15258895 Kurtosis 0.232784 Skewness 0.468115 Range 18171 Minimum 6396 Maximum 24567 Sum 662960 Count 51 (a) mean = 12999.2158, variance = 14959700.52, std. dev. = 3867.7772 (b) 64.71%, 98.04% and 100% of these states have average per capita energy consumption within 1, 2 and 3 standard deviations of the mean, respectively. (c) This is consistent with the 68%, 95% and 99.7% according to the empirical rule. (d) Excel output: Kilowatt Hours Mean 12857.74 Standard Error 539.0489 Median 12999 Mode #N/A Standard Deviation 3811.651 Sample Variance 14528684 Kurtosis 0.464522 Skewness 0.494714 Range 18171 Minimum 6396 Maximum 24567 Sum 642887 Count 50 (d) (a) mean = 12857.7402, variance = 14238110.67, std. dev. = 3773.3421 (b) 66%, 98% and 100% of these states have average per capita energy consumption within 1, 2 and 3 standard deviation of the mean, respectively. (c) This is consistent with the 68%, 95% and 99.7% according to the empirical rule. 5. Problem 3.43 (a) Excel output (in $ millions): Pop Mean Pop Std Dev Market Cap 96822.61 70547.85603 (b) On average, the market capitalization for this population of 30 companies is $96.8226 billions. The average distances between the market capitalization and the mean market capitalization for this population of 30 companies $70.5479 billions.