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Unformatted text preview: Stat 211  Tutorial 3 Solution/Answers 1. (a) x5 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 , . 95) . If exactly 1 part is not within specification, then exactly 9 parts must be within specification. Therefore, the required probability isspecification....
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This note was uploaded on 02/15/2012 for the course MATH 211 taught by Professor Javid during the Winter '12 term at Waterloo.
 Winter '12
 Javid
 Standard Deviation, Variance

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