Chapters 1-3:In order to control costs, a company wishes to study the amount of money its sales force spends entertaining clients. The following is a random sample of six entertainment expenses (dinner costs for four people) from expense reports submitted by members of the sales force. $157, $132, $109, $145, $125, $139. Calculate the mean and sample variance(s^2) and standard deviation. Mean = 807/6 = 134.5. Sample Variance = (109925 – (807^2/6)/6-1 = (109925 – 108541)/5 = 1384/5 = 276.8. Standard Deviation = √276.8 = 16.6373. ***the 109925 is all values of x individually squared and then summed together. ***the 6-1 is because it is a sample, if this were a population it would just be 6. ***the 807 is the sum off all x. Coefficient of Variation = (16.63/134.5)*100 = 12.3643. Calculate estimates of tolerance intervals containing 68.26, 95.44, and 99.73 percents. Mean ± 1 SD (68.26%) = 134.5 ± 16.63 = [117.87, 151.13]. Mean ± 2 SD (95.44%) = 134.5 ± 33.26 = [101.24, 167.76]. Mean ± 3 SD (99.73%) = 134.5 ± 49.89 = [84.61, 184.39]. Compute and interpret some of the Z-scores. (157-134.5)/16.63 = 1.35 standard deviations above the mean. (109-134.5)/16.63 = -1.53 standard deviations below the mean. Mean is the average of all the data. Mode is the number that occurs most frequently in the data set. Median is the middle value or average of the two middle values when the data is arranged in order from smallest to larges. Chapter 4: Basic Probability Concepts:In an organization of 30 people, we wish to elect 3 officers. How many different groups of officers are possible? 30*29*28=24,360 (if only 1 person per office). Or 30*30*30=27,000 (if 1 person can hold more than one office). Combinations: number of different combinations of N items taken n at a time: N!/n!(N-n)!. How many ways can we elect an executive committee of three from an organization of 30 members? 30!/3!(30-3)! = 24360/6 = 4060. We have four different flags and four slots on...

Continue Reading
Please join StudyMode to read the full document