02_14ans - Hypergeometric distribution. without replacement...

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STAT 400 Examples for 02/14/2011 Spring 2011 4. a) No . Without replacement Trials are not independent. b) 15 005 5 70 21 6 15 4 8 2 7 , C C C = 0.2937 . b B 7 8 nut plain ± ± 2 4 OR 10 5 11 6 12 7 13 8 14 6 15 7 2 6 C 0.2937 . c) Find the probability that at most 2 nut bars were selected. P( X 2 ) = 6 15 4 8 2 7 6 15 5 8 1 7 6 15 6 8 0 7 C C C C C C C C C + + = 005 5 70 21 005 5 56 7 005 5 28 1 , , , + + 0.3776 . 5. With replacement Binomial distribution. Without replacement
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Unformatted text preview: Hypergeometric distribution. without replacement with replacement a) 3 10 1 6 2 4 C C C = 0.30 . ( ) ( ) 1 2 60 . 40 . 2 3 C = 0.288 . b) 3 100 1 60 2 40 C C C 0.289425 . ( ) ( ) 1 2 60 . 40 . 2 3 C = 0.288 . c) 3 1000 1 600 2 400 C C C 0.288144 . ( ) ( ) 1 2 60 . 40 . 2 3 C = 0.288 ....
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