Question

Records show that in a certain hospital the distribution of the “length of stay” of its patients is normal with a mean of 10.5 days and a standard deviation of 2 days. a. What percentage of the patients stayed 8 days? b. What is the probability that a patient stays in the hospital between 9 and 11 days? 

Answer 1

$\mu=10.5 \\ \sigma = 2$

a. The normal distribution is contibuous, so P(X≤x) = P(X<x) for all x. In other words, P(X=x) = 0 for all x.

P(X=8) = 0

b.

$P(9<X<11) = P(X<11) -P(X<9) \\ =P(Z< \frac{11-10.5}{2}) -P(Z< \frac{9-10.5}{2}) \\ = P(Z< 0.25) -P(Z< -0.75) \\ = 0.5987 -0.2266 \\ = 0.3721$