The lodge at a ski resort has 50 suites In the winter months

The lodge at a ski resort has 50 suites. In the winter months, the occupancy rate for the suites is approximately 90%. What is the probability that 45 or fewer rooms are occupied on a given day?

Solution

Normal Distribution
Proportion ( P ) =0.9
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.9*0.1/50)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P(X >= 0.8) = (0.8-0.9)/0.0424
= -0.1/0.0424 = -2.3585
= P ( Z >-2.358) From Standard Normal Table
= 0.9908