SOLUTION: 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 t
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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:
a. Less than 30 days
b. Between 30 and 60 days
c. Greater than 90 days
You can put this solution on YOUR website! z=(x-mean)/sd
for the first
z=(30-60)/15
want p(z<-2) and that is 0.0228
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want the probability of z being between -2 and +2. That is the complement of double the first or 0.9544
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Greater than 90 days is the same as fewer than 30 days or 0.0228. The normal distribution function is symmetrical.
