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Unformatted text preview: ≤ 250240 √ 1440 ² ≈ P ( Z ≤ . 26) = 0 . 6026 b. P ( S > 260) = P ± S240 √ 1440 > 260240 √ 1440 ² ≈ 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 1μ σ/ √ n 1 ≤ 1110 2 / √ 5 ² = P ( Z ≤ 1 . 12) = 0 . 8686 Similarly, P ( X 2 ≤ 11) = P ± X 1μ σ/ √ n 2 ≤ 1110 2 / √ 6 ² = P ( Z ≤ 1 . 22) = 0 . 8888 So, the answer is (0 . 8686)(0 . 8888) ≈ . 7720...
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 '08
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 Central Limit Theorem, Normal Distribution, Probability theory

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