The length of a stay at a specific emergency department in Phoenix, Arizona in 2009 had a mean of 4.6 hours with a standard deviation of 2.9. Assume that the length of stay is normally distributed, what is the probability of a length of a stay greater than 4.6 hours Solution P(X>4.6) = P((X-mean)/s >(4.6-4.6)/2.9) =P(Z>0) =0.5 (from standard normal table).

1. The length of a stay at a specific emergency department in Phoenix, Arizona in 2009 had a mean
of 4.6 hours with a standard deviation of 2.9. Assume that the length of stay is normally
distributed, what is the probability of a length of a stay greater than 4.6 hours
Solution
P(X>4.6) = P((X-mean)/s >(4.6-4.6)/2.9)
=P(Z>0)
=0.5 (from standard normal table)