document.write( "Question 1121297: A commercial jet aircraft has four engines. For an aircraft in flight to land safely, at least two engines should be in working condition. Each engine has an independent reliability of p=92%.
\n" );
document.write( "a) if the probability of landing safely must be at least 99.5% what is the minimum value for p ?
\n" );
document.write( "b)...
\n" );
document.write( "c)...
\n" );
document.write( "The question is quoted from the book \"Complete Business Statistics\" 7th Edition by Sounderpandian (Asian edition)
\n" );
document.write( "Chapter 3, problem 3-41 (page no.134) \n" );
document.write( "
Algebra.Com's Answer #737135 by Theo(13342)![]() ![]() You can put this solution on YOUR website! it looks like 89% reliability is the magic number.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that will give you an overall probability of 2 or more engines working at the same time of .995115.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "anything less than that will result in an overall probability of less than .995.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i used the binomial theorem and kept dropping the individual reliability percent down by 1 until the overall probability became less than 99.5%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in other words, i started at p = .92, then .91, then .90, then .89, then .88.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( ".89 was above 99.5% overall. \n" ); document.write( ".88 was below 99.5% overall.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the calculations are shown in the following spreadsheet printouts.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i'm not exactly sure if this is the right formula to use, but if it is, there's a high probability that the answer will be correct.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the binomial formula tells you the probability of x occurrences out of n, given the probability that the event will occur or not occur.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula is p(x) = p^x * q^(n-x) * c(n,x).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n is 4 in this problem. \n" ); document.write( "x ranges from 0 to 4. \n" ); document.write( "p is .92 or .91 or .90 or .89 or .88 \n" ); document.write( "q is 1 - p \n" ); document.write( "c(n,x) is the numbe of ways you can get sets of x elements out of a set of n elements when order is not important. \n" ); document.write( "that formula is c(n,x) = n! / (x! * (n-x)!)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |