Problem 17-2Complete the following tableInstructions Enter y.pdf
Problem The Times of London reported on 4 January 2002 that Polish .pdf
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Problem 4S-5 The guidance system of a ship is controlled by a computer that has three major modules. In order for the computer to function properly, all three modules must function. Two of the modules have reliabilities of .95, and the other has a reliability of .99. What is the reliability of the computer? (Round your answer to 4 decimal places.) A backup computer identical to the one being used will be installed to improve overall reliability. Assuming the new computer automatically functions if the main one fails, determine the resulting reliability. (Round your intermediate calculations and final answers to 4 decimal places.) If the backup computer must be activated by a switch in the event that the first computer fails, and the switch has a reliability of .95, what is the overall reliability of the system? (Both the switch and the backup computer must function in order for the backup to take over.) (Round your intermediate calculations and final answers to 4 decimal places.) The guidance system of a ship is controlled by a computer that has three major modules. In order for the computer to function properly, all three modules must function. Two of the modules have reliabilities of .95, and the other has a reliability of .99. Solution What is the reliability of the computer? (Round your answer to 4 decimal places.) a. 0.95*0.95*0.99 = 0.893475 b. Failure rate of 1st is 0.106525, 2nd will function for 0.106525* 0.893475 = , so resulting gap is 0.095177424375, reliability is 0.904823 c. switch and 2nd comp work in 0.106525* 0.893475*0.95= 0.0904194019575, so gap is 0.009, reliability is : 0.991. What is the reliability of the computer? (Round your answer to 4 decimal places.).
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Problem The Times of London reported on 4 January 2002 that Polish .pdf
- Problem: The Times of London reported on 4 January 2002 that Polish mathematicians Gliszczynski and Zawadowski recorded 140 heads in 250 spins (56%) of the Belgian one euro coin, suggesting that it was unbalanced; this touched off an international media frenzy. The story is reported in the 11.02 issue of Chance News, available at www.dartmouth.edu/~chance Test the null hypothesis that the true probability of heads is equal to 0.5. My Question: Could someone please show me how this problem is done with explained work or steps? First person to respond with correct work will receive life saver points. Solution The test hypothesis is Ho:p=0.5 Ha:p not equal to 0.5 The test statistic is Z=(phat-p)/[p*(1-p)/n] =(0.56-0.5)/sqrt(0.5*0.5/250) =1.9 If a=0.05, the critical value is |Z(0.025)|=1.96 (check standard normal table) Since Z=1.9 is less than 1.96, we do not reject Ho.