Chapter 5 - z-Scores: Location of Scores and Standardized Distributions - Problems - Page 146: 4

A. When μ = 80, σ = 10 , X = 75 is 5 points below the mean, and it is slightly below average, therefore z = -0.5 X = 100 is 20 points above the mean, and it is in the extreme right-hand tail, therefore z = +2 X = 60 is 20 points below the mean, and it is in the extreme left-hand tail, therefore z = -2 X = 95 is 15 points above the mean, and it is in the extreme right-hand tail, therefore z = +1.5 X = 50 is 30 points below the mean, and it is in the extreme left-hand tail, therefore z = -3 X = 85 is 5 points above the mean, and it is slightly above average, therefore z = +0.5 B. When μ = 80, σ = 10 Z = 1.00 denoted that the score is located above the mean. . It corresponds to a position exactly 1 standard deviation above the mean. Therefore , X = 80+10 = 90 Z = 0.20 denoted that the score is located above the mean. It corresponds to a position exactly $\frac{1}{5}$ standard deviation above the mean. Therefore , X = 80+2 = 82 Z = 1.50 denoted that the score is located above the mean. It corresponds to a position exactly 1.5 standard deviations above the mean. Therefore , X = 80+15 = 95 Z = -0.50 denoted that the score is located below the mean. It corresponds to a position exactly 1.5 standard deviations below the mean. Therefore , X = 80-5 = 75 Z = -2.00 denoted that the score is located below the mean. It corresponds to a position exactly 2 standard deviations below the mean. Therefore , X = 80-20 = 60 Z = -1.50 denoted that the score is located below the mean. It corresponds to a position exactly -1.5 standard deviations below the mean. Therefore, X = 80-15= 65

--