The amount of time that a customer spends waiting at an airp

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean = 8.2 minutes and standard deviation = 1.5 minutes. Suppose that a random sample of n = 49 customers is taken. Compute the approximate probability that the average waiting time for these customers is:

(a) Less than 10 minutes.

(b) Between 5 and 10 minutes.

(c) Less than 6 minutes.

Solution

Normal Distribution
Mean ( u ) =8.2
Standard Deviation ( sd )=1.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 10) = (10-8.2)/1.5
= 1.8/1.5= 1.2
= P ( Z <1.2) From Standard Normal Table
= 0.8849                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 5) = (5-8.2)/1.5
= -3.2/1.5 = -2.1333
= P ( Z <-2.1333) From Standard Normal Table
= 0.01645
P(X < 10) = (10-8.2)/1.5
= 1.8/1.5 = 1.2
= P ( Z <1.2) From Standard Normal Table
= 0.88493
P(5 < X < 10) = 0.88493-0.01645 = 0.8685                  

c)
P(X < 6) = (6-8.2)/1.5
= -2.2/1.5= -1.4667
= P ( Z <-1.4667) From Standard Normal Table
= 0.0712