chapter 1

Chapter 1: Overview and Descriptive Statistics1= ∑ in50. First, we need x x = ( 20,179) = 747. 37deviation127( 20,179)24,657,511−s =27= 606.8926x + 2 s = 747.37 + 2(606.89) = 1961.16bit less than the $3.5 million that was awarded originally.2. Then we need the sample standard. The maximum award should be, or in dollar units, $1,961,160. This is quite a51.a. Σx = 2563 and Σx2 = 368, 501, so22 [368,501−(2563) /19]s == 1264.766 and s = 35. 56418c =b. If y = time in minutes, then y = cx where , so602 2 2 1264.76635.564sy= c sx= = .351 and sy= csx= = . 593360060152. Let d denote the fifth deviation. Then . 3 .9 + 1.0 + 1.3+= 0+ d or 3 .5 + d = 0 , sod = −3.5 . One sample for which these are the deviations is x1= 3.8,x2= 4.4,x = 4.5, ,3x4= 4.8 = 0 5.x (obtained by adding 3.5 to each deviation; adding any othernumber will produce a different sample with the desired property)53.a. lower half: 2.34 2.43 2.62 2.74 2.74 2.75 2.78 3.01 3.46upper half: 3.46 3.56 3.65 3.85 3.88 3.93 4.21 4.33 4.52Thus the lower fourth is 2.74 and the upper fourth is 3.88.b. f = 3 .88 − 2.74 = 1. 14sc. fswouldn’t change, since increasing the two largest values does not affect the upperfourth.d. By at most .40 (that is, to anything not exceeding 2.74), since then it will not change thelower fourth.e. Since n is now even, the lower half consists of the smallest 9 observations and the upperhalf consists of the largest 9. With the lower fourth = 2.74 and the upper fourth = 3.93,f = 1.19 .s25