document.write( "Question 1075099: Please help me solve the problem.I will appreciate your solution.
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document.write( "Question 1.\r
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document.write( " The weight of a certain type of chicken, when ready for the market, follows a normal distribution with mean 1.0 kg and a standard deviation of 0.20 kg. Find the probability that a chicken weighs less than 1.50 kilogram. \n" );
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Algebra.Com's Answer #689812 by Theo(13342) You can put this solution on YOUR website! mean is 1 kg. \n" ); document.write( "standard deviation is .2 kg.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you want probability that the chicken weighs less than 1.5 kg.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z-score = (x-m)/s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x = 1.5 kg \n" ); document.write( "m = 1 kg \n" ); document.write( "s = .2 kg.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "formula becomes z = (1.5 - 1) / .2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "result is z = 2.5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "probability of a z-score less than 2.5 is equal to .9938\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in other words, the vast majority of the chickens will weight less than 1.5 kg if the mean is 1.0 kg and the standard deviation is .2 kg.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "here's the results using z-scores and using raw scores.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the results are the same, as they should be.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" );
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