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tutorial3sol

# tutorial3sol - Stat 211 Tutorial 3 Solution/Answers 1(a x-5...

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Stat 211 - Tutorial 3 Solution/Answers 1. (a) x -5 5 20 P ( X = x ) 25/36 10/36 1/36 (b) E ( X ) = - \$1 . 53 and V ar ( X ) = 33 . 08256 . 2. (a) x 100 200 P ( X = x ) 0.6 0.4 (b) mean, μ x = \$140 ; variance, σ 2 x = 2400 ; standard deviation, σ x = \$48 . 99 . (c) mean, μ y = \$147 ; variance, σ 2 y = 2646 ; standard deviation, σ y = \$51 . 44 . (d) mean, μ p = \$150 ; variance, σ 2 p = 2400 ; standard deviation, σ p = \$48 . 99 . (e) mean, μ m = \$140 ; variance, σ 2 m = 1200 ; standard deviation, σ m = \$34 . 64 . (f) i. \$144 ii. E ( X ) = 200 - 100 p iii. ˆ p = 0 . 56 . 3. 95% of parts produced by a machine in a day are within the required specification. In a given day, ten parts are randomly selected. Find the probability that (a) exactly 1 part is not within specification. Solution : X = number of parts in a set of 10 to be within specification. X Binomial (10 , 0 . 95) . If exactly 1 part is not within specification, then exactly 9 parts must be within specification. Therefore, the required probability is P ( X = 9) = 0 . 3151 (b) at least one part is within specification. Solution : P ( X 1) = 1 - P ( X < 1) = 1 - P ( X = 0) 1 . (c) Let X denote the number of parts in the sample of 10 that are within specifica- tion. Find E ( X ) , E (2 X

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