Suppose a soft drink bottler delivery time for the service r

Suppose a soft drink bottler delivery time for the service routes in a distribution system follows a normal distribution. This service activity includes stocking the machine with beverage products and minor maintenance or housekeeping. The expected delivery time is 22.38 minutes with a standard deviation of 15.52 minutes. Find:

a. The probability that the delivery time takes more than 18.11 minutes.

b. The probability that the delivery time takes less than 9.5 minutes.

c. The probability that the delivery time takes more than 52.32 minutes.

Solution

Normal Distribution
Mean ( u ) =22.38
Standard Deviation ( sd )=15.52
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)                  
P(X > 18.11) = (18.11-22.38)/15.52
= -4.27/15.52 = -0.2751
= P ( Z >-0.275) From Standard Normal Table
= 0.6084                  

b)
P(X < 9.5) = (9.5-22.38)/15.52
= -12.88/15.52= -0.8299
= P ( Z <-0.8299) From Standard Normal Table
= 0.2033

c)
P(X > 52.32) = (52.32-22.38)/15.52
= 29.94/15.52 = 1.9291
= P ( Z >1.929) From Standard Normal Table
= 0.0269