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Unformatted text preview: IE 300 / GE 331 Discussion Problem Set 9 72 ~(100, 102 ), = 1, ... ,25. ~( , 2 ), = 100, = (  1.8 + ) = 1.8 723(a,b,d,e)  1 = (1)  (1.8) = 0.8054 25 10 = 2. a) A point estimate for the mean is obtained by the sample mean: 425 + 431 + 416 + + 413 + 416 = = 423.33 24 b) A point estimate for the standard deviation is obtained by the sample standard deviation: 24 (  423.33)2 = =1 = 9.08 24  1 d) A point estimate for the median is obtained by the sample median: 425 + 423 = = 424 2 e) A point estimate for the required proportion is obtained by the sample proportion, which is 7/24 since 7 of the measurements exceed 430 Angstroms 752 5 37  40 . ( 37) = = (5.36) = 4e  8 16 5/16 ~(40,5), i = 1, ... ,16. ~ 40, 753 (use the central limit theorem) Denote the defective rate by . = 1 with probability ; = 0 with probability 1  . = ( ) = , 2 = ( ) = (1  ). From the central limit theorem, is approximately normal: ~ , (1  ) 100 ...
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This note was uploaded on 05/04/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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