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Unformatted text preview: left of z is 0.90. This z is 1.28. Then we solve the equation below for x . x-3 . 5 . 8 = 1 . 28 So x = (1 . 28)(0 . 8) + 3 . 5 = 4 . 52 inches. Problem 6.21 First, we need to nd z such that the area to the right of z is 0.02 and the area to the left of z is 0.98. This z is 2.05. Then we solve the equation below for x . x-100 15 = 2 . 05 So x = (2 . 05)(15) + 100 = 130 . 75. Problem 6.25 Here we have a normal distribution with = 6 . 8 hours and = 0 . 6 hours. a. P ( x > 8) = P ( z > 8-6 . 8 . 6 ) = P ( z > 2) = 1-. 9772 = 0 . 0228 b. P ( x 6) = P ( z 6-6 . 8 . 6 ) = P ( z -1 . 33) = 0 . 0918 c. P (7 < x < 9) = P ( 7-6 . 8 . 6 < z < 9-6 . 8 . 6 ) = P (0 . 33 < z < 3 . 67) = 0 . 9998-. 6293 = . 3705 2...
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- Spring '10
- Jager,AbigailL
- Left-wing politics, 3.5 inches, 0.8 inches, 3.01%, 0.6 hours, 4.52 inches
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