set4solutions - IE 300 / GE 331 Discussion Problem Set 4...

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IE 300 / GE 331 Discussion Problem Set 4 4-5 (a), (e), (f) a) The distribution of X is symmetric around 0. Therefore, ( 0 ) 0.5. PX <= e) ( ± < 0 ²³ ´ > 0.5) = µ ( 1< <1)= 1. f) 1 1 2 33 ( ) 1. 5 0. 5 0. 5 0. 5 0.05, 0.9655. x x P x X x d x x xx < = = = - == 4-6 (a), (b), (d) Let X denote the time to failure a) /100 0 /100 03 3000 3000 ( 3000 ) /100 0 0.05. P X e d x ee - -- = = - b) 2000 2000 /100 0 /100 0 12 1000 1000 (100 0 2000 ) 0 0.233. P X e d x e - - < < = = - =-= d) /100 0 /100 0 /1000 0 0 ( ) 0 1 0.10, 105.36. x x x P X x e d x e ex - < = = - = -== 4-14 (c), (d) c) Let X denote the waiting time until a customer arrives. · = ¸ ( ¹ < 40) = º »¼ / ½¾ /10 ¿À ÁÂ Ã = −Ä ÅÆ / ÇÈ | É ÊË =1 −Ì ÍÎ = 0.9817. Out of five, the number of customers that arrive in 40 minutes is binomial: 1 − Ï 5 0 ÐÑ Ò (1 −Ó ) Ô −Õ 5 1 Ö× Ø (1 −Ù ) Ú = 0.9999994. d) Û ( Ü ) = ∫ Ý Þß / àá /10 âã ä å = −æ çè / éê | ë ì −í î ï ðñ . ò (15 ≤ó≤ 30) = ô (30) −õ (15)= 0.1733. PDF created with pdfFactory Pro trial version www.pdffactory.com
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IE 300 / GE 331 Discussion Problem Set 4 4-30 7 0 70 70 1 11 70 a) ( ) ( ) 70/(6 9 ) l n 4.3101. 69 E X x f x d x x d xx = = == ∫∫ 7 0 70 2 2 22 ( ) ( ) 70/6
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This note was uploaded on 03/16/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.

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set4solutions - IE 300 / GE 331 Discussion Problem Set 4...

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