document.write( "Question 993926: In a bottle-filling process, the amount of drink injected into 16 oz bottles is normally distributed with a mean of 16 oz and a standard deviation of .32 oz. Bottles containing less than 15.86 oz do not meet the bottler’s quality standard. What percentage of filled bottles do not meet the standard? (Round the z value to 2 decimal places. Round your answer to 2 decimal places.)\r
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document.write( " P(x < 15.86) % \n" );
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Algebra.Com's Answer #613097 by Theo(13342)![]() ![]() You can put this solution on YOUR website! the z factor is equal to (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x is the raw score. \n" ); document.write( "m is the mean. \n" ); document.write( "s is the standard deviation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the critical z factor is the z-score of the cutoff value.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if the bottle contains less than 15.86 ounces of drink, then it is rejected.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that makes the cutoff z factor equal to (15.86 - 16) / .32 = -.44\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "look up a z-score of -.44 in the z-score table and you will get .33\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that means that 33% of the z-scores in the normal distribution table are below a z-score of .44.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that means that 33% of the bottles would be rejected if the amount of drink in the bottles was normally distributed with a mean of 16 ounces and the cutoff was any bottle that had less than 15.86 ounces of drink in it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you can see the results in the following z-score calculator outputs.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the first output is based on the raw score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the second output is based on the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the different in percentage is due to rounding of the z-score.\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( " |