Business and Application Solutions - Statistics Project Example

Business Statistics and Applications Solutions to Midterm Examination a. i. Total number of managers who completed four or more trainings = 30 Total number of Managers 100 ii. For A. Total number of Managers with an “A” rating = 40 Total number of Managers 100 For B. Total number of Managers with a “B” rating = 10 Total number of Managers 100 For C. Total number of Managers with a “C” rating = 20 Total number of Managers 100 For D. Total number of Managers with a “D” rating = 20 Total number of Managers 100 iii. Total number of Manager who has completed 4 or more training w/ “A” rating Total number of Managers with an “A” rating = 10/40 = ¼ =.25 iv. = (15 x 1) + (30 x 2) + (15 x 3) + (15 x 4) + (10 x 5) + (6 x 1) 100 = 2.36 v. (1 x 8) + (2 x 17) + (5 x 3) + (7 x 4) + (5 x 2) + (6 x 1) 40 = 2.525 vi. There is no relationship between performance ratings and number of trainings attended. The performance rating of a manager is not dependent on the trainings he went through. It can be noted that the number of managers who performed better does not increase with the number of trainings. From the preceding number, it was obtained that managers given a rating of one only has a mean attendance in 2.525 (approximately 3) trainings. b. Given: P= 0.30 Q= 0.70 p = 0.90 q = 0.10 n= 10 i. not defective = (.90) (.70) = .63 = 63% therefore, defective = 1 – (.63) = 0.37 we now have = p = 0.37 & q = 0.63 ii. P (9 out of 10) = (10!/ (1!) (10!)) * (0.10).37 (0.90) .63 iii. mean = np = (100) (0.37) = 37 standard deviation = (npq)1/2 = 4.82 2. Given: Population Sample Observation = 500 Observation = 100 Mean = $1050 Mean = - Standard deviation = $1300 Standard deviation = $200 i. The sample mean balance is allowed only to stay within $200 within the population mean w/c implies that the sample mean should be $1050 +/- $200 or $850 Read More