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# sol12 - Chapter 12 Exercises 12.1 1 4 4 1 E X = 0 1 = 5 5 5...

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381 Chapter 12 Exercises 12.1 1. E ( X ) = 0 1 5 ⎟ + 1 4 5 ⎟ = 4 5 V ( X ) = 0 4 5 2 1 5 ⎟ + 1 4 5 2 4 5 = 20 125 = .16 Standard deviation .16 .4 = = 2. E ( X ) = 1 4 9 + 2 4 9 + 3 1 9 = 15 9 = 5 3 V ( X ) = 1 5 3 2 4 9 + 2 5 3 2 4 9 + 3 5 3 2 1 9 = 36 81 = 4 9 Standard deviation 36 6 2 81 9 3 = = = 3. a. E ( X ) = 4(.5) + 6(.5) = 5 V ( X ) = (4 5) 2 (.5) + (6 5) 2 (.5) = 1 b. E ( X ) = 3(.5) + 7(.5) = 5 V ( X ) = (3 5) 2 (.5) + (7 5) 2 (.5) = 4 c. E ( X ) = 1(.5) + 9(.5) = 5 V ( X ) = (1 5) 2 (.5) + (9 5) 2 (.5) = 16 As the difference between maximum and minimum values increases, so does the variance. 4. a. E ( X ) = 2(.1) + 4(.4) + 6(.4) + 8(.1) = 5 V ( X ) = (2 5) 2 (.1) + (4 5) 2 (.4) + (6 5) 2 (.4) + (8 5) 2 (.1) = 2.6 b. E ( X ) = 2(.3) + 4(.2) + 6(.2) + 8(.3) = 5 V ( X ) = (2 5) 2 (.3) + (4 5) 2 (.2) + (6 5) 2 (.2) + (8 5) 2 (.3) = 5.8 Values of (b) farther from E ( X ) have higher probabilities. Thus the variance is larger.

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