Statistics Homework Solutions 20

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

Ask a homework question - tutors are online