document.write( "Question 1203889: The lengths of pregnancies in a small rural village are normally distributed with a mean of 270 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 270 and a standard deviation of 15.\r
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document.write( "What percentage of pregnancies last fewer than 257 days?
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document.write( "P(X < 257 days) = %\r
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document.write( "Enter your answer as a percent accurate to 1 decimal place (do not enter the \"%\" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. \n" );
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Algebra.Com's Answer #839827 by Theo(13342)![]() ![]() You can put this solution on YOUR website! population mean is 270 days. \n" ); document.write( "population standard deviation is 15 days. \n" ); document.write( "test value = 257 days.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "use z-score formula. \n" ); document.write( "z = (x - m) / 15 becomes z = (257 - 270) / 15 = -.86667 \n" ); document.write( "z is the z-score \n" ); document.write( "x is the test value \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( "area to the left of z-score of -.86667 = .1931. \n" ); document.write( "this means there is a 19.3% probability of a pregnancy lasting less than 257 days. \n" ); document.write( "you would enter 19.3 as your solution is what i think based on the instructions.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |