document.write( "Question 1182192: A global financial institution transfers a large data file every evening from offices around the world to its London headquarters. Once the file is recieved, it must be cleaned and partitioned before being stored in a company's data warehouse. Each file is the same size and the time required to transfer, clean, and partition a file is normally distributed, with a mean of 1.5 hours and standard deviation of 15 minutes. If a manager must be present until 85% of the files are transferred, cleaned, and partitioned, how long will the manager need to be there? \n" ); document.write( "
Algebra.Com's Answer #812560 by Theo(13342)![]() ![]() You can put this solution on YOUR website! mean is 1.5 hours and standard deviation is 15 minutes = .25 hours. \n" ); document.write( "z-score for 85% area under the normal distribution curve to the left of it is equal to 1.03643338. \n" ); document.write( "z-score formula is: \n" ); document.write( "z = (x - m) / s \n" ); document.write( "z is the z-score \n" ); document.write( "x is the raw score \n" ); document.write( "m is the mean \n" ); document.write( "s is the standard deviation. \n" ); document.write( "solve for raw score to get: \n" ); document.write( "x = z * s + m = 1.03643338 * .25 + 1.5 = 1.759108345 hours. \n" ); document.write( "that's how long the manager must be present. \n" ); document.write( "this is what it looks like on the statistical calculator found at https://davidmlane.com/hyperstat/z_table.html\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |