Statistics Homework Solutions 20

Statistics Homework Solutions 20 - 20 CHAPTER 2 3 10 £...

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Unformatted text preview: 20 CHAPTER 2 3 10 £ ¡¥ 15. (a) P A ¤ (b) Given that A occurs, there are 9 components remaining, of which 2 are defective. 3 10 2 9 1 15 ¡¥ £ ¦¥ ¤ £¤ ¡¥ ¤ P A P BA £ ¡¥ ¤ ¥ ¤ ¡¥ (c) P A B 2 9. £ Therefore P B A ¤ ¡ (d) Given that Ac occurs, there are 9 components remaining, of which 3 are defective. ¥ ¤ 7 30. 3 10 £ ¡ ¡¥ ¤ ¡ £ ¢ £ £ ¡¥ ¤ ¡ ¡¥ ¢¥ ¤ ¤ ¡ ¥ ¤ P A P B ]. ¤ ¤ ¡¥ P B A [or equivalently, P A B ¥ ¡¥ ¤ ¡ ¥ ¤ ¡¥ ¤ n 10 000. The two components are a simple random sample from the population. When the population is large, the items in a simple random sample are nearly independent.  ¡ 17. ¤ (f) No. P B 7 30 ¡¥ 1 15 £¤ P Ac B 7 10 3 9 £ ¦¥ ¤ PA B P Ac P B Ac ¡¥ (e) P B 3 9. Now P Ac B £ Therefore P B A 19. (a) On each of the 24 inspections, the probability that the process will not be shut down is 1 0 05 0 95. Therefore P not shut down for 24 hours 0 95 24 0 2920. It follows that P shut down at least once 1 0 2920 0 7080. ¡ 0 1, ¡¥ ¡ ¤ ¨ ¡ ¥ ¨ ¤ ¡¥ ¤ ¨ 0 009255. ¡ ¥ ¡ ¤ ¡¥ ¡ ¨ ¡ ¤ ¤ ¥¨ ¥ ¡ ¡¥ P C R c P Rc 09 1 01 ¥ ¨ ¤ ¤ ¦¥ ¨ ¢¥ ¤ ¤ ¨ ¤ ¢¥ ¨ ¥ ¡¥ ¤ ¤¥ ¨ ¤ 0 89 ¤ PCRPR 08 01 ¨ 0 80. Solving for p yields p ¨ 24 ¡ p ¨ 1 Let R denote the event of a rainy day, and let C denote the event that the forecast is correct. Then P R PCR 0 8, and P C Rc 0 9. ¨ ¡ ¡ ¥ ¤ ¡ (a) P C ¡¥ ¨ ¤ 21. ¨ (b) P not shut down for 24 hours ...
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