document.write( "Question 352363: The amount of fill in a half liter soft drink bottle is normally distributed. The process has a std. deviation of 5 ml. The mean is adjustable. a) Where should the mean be set to ensure a 95 percent probability that a half liter bottle will not be underfilled? b) A 99 percent probability? c) A 99.9 percent probability? Explain.
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Algebra.Com's Answer #251955 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The amount of fill in a half liter soft drink bottle is normally distributed. The process has a std. deviation of 5 ml. The mean is adjustable. \n" ); document.write( "--------------------- \n" ); document.write( "a) Where should the mean be set to ensure a 95 percent probability that a half liter bottle will not be underfilled? \n" ); document.write( "---- \n" ); document.write( "A value that is 1.645 std below the mean will have 95% of the data above it. \n" ); document.write( "1.645*5ml = 8.2243ml \n" ); document.write( "----- \n" ); document.write( "1/2 liter = 500 ml \n" ); document.write( "500-8.2243 = 491.78ml \n" ); document.write( "--- \n" ); document.write( "Set the mean at 491.78ml\r \n" ); document.write( "\n" ); document.write( "b) A 99 percent probability? \n" ); document.write( "Same as above but use z value of 2.326. That z will have 99% above it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "c) A 99.9 percent probability? Explain. \n" ); document.write( "Same as above but use z value of 3.09. That will have 99.9% above it. \n" ); document.write( "==================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |