Chapter 6 - Section 6.4 - Determining the z and x Values When an Area Under the Normal Distribution Curve Is Known - Exercises - Page 256: 6.39

a. mean = 550, standard deviation = 75 Area to the left of x = 0.025, the z value for .025 is -1.96. x=μ+zσ= 550+(-1.96)(75) = 403 b. Area to the right of x value = 1.0- 0.9345=0.0655, the z value for 0.0655 is -1.51. x=μ+zσ= 550+(-1.51)(75) = 436.75 c. Area to the right of x value = 1.0- 0.0275= 0.9725, the z value for this is 1.92. x=μ+zσ= 550+(1.92)(75) = 694 d. Area to the left of x = 0.9600, the z value for .96 is 1.75. x=μ+zσ= 550+(1.75)(75) = 681.25 e. Area from μ to x = 0.47 , the area to the left of x = 0.5-0.47=0.03, the z-value for 0.97 is -1.88. x =μ+zσ= 550+(-1.88)(75) = -41 f. Area from μ to x = 0.41 , the area to the left of x = 0.41 +0.5 = 0.91, the z-value for 0.91 is 1.34 x =μ+zσ= 550+(1.34)(75) = 650.5

a. mean = 550, standard deviation = 75 Area to the left of x = 0.025, the z value for .025 is -1.96. x=μ+zσ= 550+(-1.96)(75) = 403 b. Area to the right of x value = 1.0- 0.9345=0.0655, the z value for 0.0655 is -1.51. x=μ+zσ= 550+(-1.51)(75) = 436.75 c. Area to the right of x value = 1.0- 0.0275= 0.9725, the z value for this is 1.92. x=μ+zσ= 550+(1.92)(75) = 694 d. Area to the left of x = 0.9600, the z value for .96 is 1.75. x=μ+zσ= 550+(1.75)(75) = 681.25 e. Area from μ to x = 0.47 , the area to the left of x = 0.5-0.47=0.03, the z-value for 0.97 is -1.88. x =μ+zσ= 550+(-1.88)(75) = -41 f. Area from μ to x = 0.41 , the area to the left of x = 0.41 +0.5 = 0.91, the z-value for 0.91 is 1.34 x =μ+zσ= 550+(1.34)(75) = 650.5