Question 1172257: Suppose the average length of stay in a chronic disease hospital of a certain type of patient is 80 days with a standard deviation of 20. If it is reasonable to assume an approximately normal distribution of lengths of stay, find the probability that a randomly selected patient from this group will have a length of stay :
(a). Greater than 60 days
(b) Less than 40 days
(c) Between 35 and 55 days
(d) Greater than 100 days
(e) And draw their both normal distribution curve and standard normal distribution
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
a. z=(60-80)/20=-1 The probability is that of z > -1 or 0.8413
b. less than 40 days is z <-2 or 0.0228
c. between 35 and 55 days is -2.25 < z < -1.25 From the probability table or VARS2ENTER. Prob. is 0.0934.
d. Greater than 100 days is a z > 1 or 0.1587.
Draw the problem's distribution with a normal curve with mean 80 and 100, 120,140 for sd +1,2,3
and 60,40,10 for -1,-2,-3.
Then use the standard normal curve with mean 0 and sd 1 to compare.
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