document.write( "Question 1059036: 5. Assume the average length of stay in a hospital for chronic disease is 60 days with a standard deviation of 15. If the stay has a normal distribution, find out the probability that a randomly selected patient from this group will have a length of stay that is:
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document.write( "a. Less than 30 days
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document.write( "b. Between 30 and 60 days
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document.write( "c. Greater than 90 days
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Algebra.Com's Answer #674111 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "for the first \n" ); document.write( "z=(30-60)/15 \n" ); document.write( "want p(z<-2) and that is 0.0228 \n" ); document.write( "---------------------- \n" ); document.write( "want the probability of z being between -2 and +2. That is the complement of double the first or 0.9544 \n" ); document.write( "--------------------- \n" ); document.write( "Greater than 90 days is the same as fewer than 30 days or 0.0228. The normal distribution function is symmetrical. \n" ); document.write( " |