sec_5_4_sol

sec_5_4_sol - 250-240 1440 P ( Z . 26) = 0 . 6026 b. P ( S...

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APPM 4/5570 Solutions to Problems from Section 5.4 46. a. P (11 . 99 X 12 . 01) = P ± 11 . 99 - 12 0 . 04 / 16 X - μ σ/ n 12 . 01 - 12 0 . 04 / 16 ² = P ( - 1 Z 1) = Φ(1) - Φ( - 1) = 0 . 8413 - 0 . 1587 = 0 . 6826 b. P ( X > 12 . 01) = P ± X - μ σ/ n > 12 . 01 - 12 0 . 04 / 25 ² = P ( Z > 1 . 25) = 1 - Φ(1 . 25) = 0 . 1056 49. We have the time to grade each paper: X 1 , X 2 , . . . , X 40 , where, for each i = 1 , 2 , . . . , 40, E [ X i ] = 6 and V ar [ X i ] = 36. In this problem, we are not given a distribution name. Fortunately, due to the large sample size, we know that X = 1 40 40 X i =1 X i is approximately normally distributed because of the wonderful central limit theorem! However, this problem does not ask us to compute probabilities involving X . We want probabilities involving 40 i =1 X i . Fortunately, this is just 40 X ; a constant times a normal which we have seen is still normal. The mean is E [ 40 X i =1 X i ] = 40 X i =1 E [ X i ] = 40 X i =1 6 = (40)(6) = 240 . Similarly, V ar [ 40 X i =1 X i ] indep = 40 X i =1 V ar [ X i ] = 40 X i =1 6 = (40)(36) = 1440 . We are now ready to answer the questions. Let’s use S to denote the sum 40 i =1 X i . S approx N (240 , 1440) .
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a. P ( S 250) = P ± S - 240 1440
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Unformatted text preview: 250-240 1440 P ( Z . 26) = 0 . 6026 b. P ( S > 260) = P S-240 1440 > 260-240 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 11-10 2 / 5 = P ( Z 1 . 12) = 0 . 8686 Similarly, P ( X 2 11) = P X 1- / n 2 11-10 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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sec_5_4_sol - 250-240 1440 P ( Z . 26) = 0 . 6026 b. P ( S...

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