Assume that a specific emergency department has a length of stay with behavior n
ID: 3171904 • Letter: A
Question
Assume that a specific emergency department has a length of stay with behavior normally distributed; mean of 5 hours and a standard deviation of 3 hours. Use standard normal random variables to resolve these probabilities. (a) What is the probability of a length of stay greater than 12 hours? (b) What length of stay is exceeded by 20% of the visits? (c) From the normally distributed model, what is the probability of a length of stay less than 0 hours? Comment on the normally distributed assumption in this example.Explanation / Answer
from normal distribution
zscore =(X-mean)/std deviation
a) P(X>12) =1-P(X<12)=1-P(Z<(12-5)/3)=1-P(Z<2.3333)=1-0.9902=0.0098
b)for 20 percentile ,z=-0.8416
hence corresponding score =mean +z*std deviation =2.475
c)P(X<0)=P(Z<(0-5)/3)=P(Z<-1.667)=0.0478
as time can not be negative, for which as per nornal distribution, there is still some probabilty of that happening, hence it can not be normally distributed but right skewed,