If the mean time to clean a motel room is 160 min and the st

If the mean time to clean a motel room is 16.0 min and the standard deviation is 1.5 min, what percentage of the rooms will take less than 13.0 min, what percentage of the rooms will take less than 13.0 min to complete? What percentage of the rooms will take more than 20.0 min to complete? What percentage of the rooms will take between 13.0 and 20.5 min to complete? The data are normally distributed.

Solution

Normal Distribution
Mean ( u ) =16
Standard Deviation ( sd )=1.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 13) = (13-16)/1.5
= -3/1.5= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228                  
b)
P(X > 20) = (20-16)/1.5
= 4/1.5 = 2.6667
= P ( Z >2.667) From Standard Normal Table
= 0.0038                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 13) = (13-16)/1.5
= -3/1.5 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 20.5) = (20.5-16)/1.5
= 4.5/1.5 = 3
= P ( Z <3) From Standard Normal Table
= 0.99865
P(13 < X < 20.5) = 0.99865-0.02275 = 0.9759