Homework 2 solutions

# Homework 2 solutions - ENME 392 Fall 2009 Statistical...

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ENME 392 Fall 2009 - Statistical Methods for Product and Process Development Tues – Thurs: Section 0101 11:00-12:15 room EGR 2107; Section 0201 9:30-10:45 room ANS 0408 Prof. D. B. Barker, room 2112 ENG, (301) 405-5264, [email protected] 8/31/09 page 1 Homework #2 – due Tuesday September 15, at beginning of class Note: answers to odd numbered problems are shown in textbook. Problem numbering, please follow for ease of grading: 1(1.3 c-d) - this is means homework problem 1 which is Walpole problem 1.3 parts c thru d inclusive. 1 (2.53, p 55 Walpole 500 envelopes total 75 contain \$100 150 contain \$25 275 contain \$10 500 – 75 – 150 – 275 = 0 no empty envelopes Cost of envelope = \$25. S = {\$10, \$25, \$100} P(\$100) = 75/500 = 0.15 P(\$25) = 150/500 = 0.30 P(\$10) = 275/500 = 0.55 P(<\$100) = P(\$25 U \$10) = 0.30 + 0.55 = 0.85 2 (2.54, p 55 Walpole This problem is most readily done using a Venn diagram. The one on the left contains the numbers given in the problem, the one on the right contains the numbers of each subsection. 500 210 258 216 S D E 52 97 83 122 66 43 105 88 S D E 52 45 31 70

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ENME 392 Fall 2009 - Statistical Methods for Product and Process Development Tues – Thurs: Section 0101 11:00-12:15 room EGR 2107; Section 0201 9:30-10:45 room ANS 0408 Prof. D. B. Barker, room 2112 ENG, (301) 405-5264, [email protected] 8/31/09 page 2 using left diagram using right diagram a) P(S ∩D’) (210-122)/500 = 88/500 43+45/500 b) P(E ∩D∩S’) (83-52)/500 = 31/500 31/500 c) P(S’
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Homework 2 solutions - ENME 392 Fall 2009 Statistical...

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