document.write( "Question 960750: A Wendy’s fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is normally distributed with a mean of 450 and standard deviation of 80 and the demand for chicken sandwiches is normally distributed with a mean of 120 and standard deviation of 30. \r
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document.write( "How many hamburgers must the restaurant stock to be 99% sure of not running out on a given day? \n" );
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Algebra.Com's Answer #587126 by mathmate(429)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Given: \n" ); document.write( "Hamburgers sold daily are normally distributed and have a mean of 450 and standard deviation of 80, i.e. N(450,80). \n" ); document.write( " \n" ); document.write( "Require: \n" ); document.write( "Daily stock of hamburgers necessary in order not to run out 99 times out of 100. \n" ); document.write( " \n" ); document.write( "Let X=number required, then we need \n" ); document.write( "P(x>X)=0.99 which means \n" ); document.write( "P(Z>(X-450)/80)>0.99 \n" ); document.write( "From normal distribution tables, Z=2.327, \n" ); document.write( "solve for X \n" ); document.write( "2.3263=(X-450)/80 \n" ); document.write( "X-450=80*2.3263=186.1 \n" ); document.write( "X=450+186.2=636.1 \n" ); document.write( " \n" ); document.write( "Answer: \n" ); document.write( "637 hamburgers should be stocked. \n" ); document.write( " |