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Unformatted text preview: 250240 1440 P ( Z . 26) = 0 . 6026 b. P ( S > 260) = P S240 1440 > 260240 1440 P ( Z > . 53) = 0 . 2981 51. Let X 1 be the sample mean time for the Frst day. Since this is a sample mean for 5 individuals, we know that X 1 N (10 , 4 / 5) . Let X 2 be the sample mean time for the Frst day. Since this is a sample mean for 6 individuals, we know that X 1 N (10 , 4 / 6) . We want to compute P ( X 1 ) 11 , X 2 11). Since X 1 and X 2 are independent, this is simply P ( X 1 11) P ( X 2 11) P ( X 1 11) = P X 1 / n 1 1110 2 / 5 = P ( Z 1 . 12) = 0 . 8686 Similarly, P ( X 2 11) = P X 1 / n 2 1110 2 / 6 = P ( Z 1 . 22) = 0 . 8888 So, the answer is (0 . 8686)(0 . 8888) . 7720...
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
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