document.write( "Question 981477: A company believes that the annual demand for a product will follow a normal random variable with a mean of 900 pieces and a standard deviation of 60 pieces. If the company produces 1000 pieces, what is the chance that it will run out of the product? Assume that the only way to meet the demand for the product is to use this year's production number.\r
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document.write( "I don't understand - if they make 1000 pieces, they should always have enough, right? Thanks! \n" );
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Algebra.Com's Answer #602451 by jim_thompson5910(35256) ![]() You can put this solution on YOUR website! No there is the chance that the demand exceeds 1000.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x = 1000 \n" ); document.write( "mu = 900 \n" ); document.write( "sigma = 60 \n" ); document.write( "----------------- \n" ); document.write( "z = (x-mu)/simga \n" ); document.write( "z = (1000 - 900)/60 \n" ); document.write( "z = 100/60 \n" ); document.write( "z = 1.67\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Use a calculator or table to find that P(Z < 1.67) = 0.95254 \r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So,\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(Z > 1.67) = 1 - P(Z < 1.67) \n" ); document.write( "P(Z > 1.67) = 1 - 0.95254 \n" ); document.write( "P(Z > 1.67) = 0.04746 \n" ); document.write( "P(X > 1000) = 0.04746\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There is a 4.746% chance that they won't have enough to meet demand. \n" ); document.write( " |