document.write( "Question 1139689: A company installs 5000 light bulbs, each with an average life of 500 hours, standard deviation of 100 hours, and distribution approximated by a normal curve. Find the percentage of bulbs that can be expected to last the period of time. Round to the nearest hundredth, if necessary.\r
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document.write( "Less than 520 hours
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Algebra.Com's Answer #760193 by Theo(13342) You can put this solution on YOUR website! mean is 500 hours. \n" ); document.write( "standard deviation is 100 hours. \n" ); document.write( "number of light bulbs installed is 5000.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z is the z-score \n" ); document.write( "x is the raw score \n" ); document.write( "m is the raw mean \n" ); document.write( "s is the standard error, which, in this case, is the standard deviation, since you are dealing with the population of 5000 light bulbs rather than a sample of a much larger population.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in this problem, the formula becomes z = (520 - 500) / 100.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for z to get z = 20 / 100 = .2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "with a z-score of .2, the percentage of bulbs that can be expected to last less than 520 hours would be equal to .5792596878 * 100 = 57.92596878%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "round to 2 decimal places and the solution is 57.93%\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |