Question

We wish to investigate the level of service provided at these three airports. The level of service depends on departure and arrival delays, among other things. Assume that the length of departure delays for flights from the NYC airports in 2013 follows a normal distribution with mean 12.64 minutes and standard deviation 40.21 minutes.

(a) What is the maximum length of departure delays for 20% of flights that departed from the NYC airports in 2013? Give your answer correct to one decimal place. Remember to define the variable of interest and state the distribution. Show all working.

(b) Given the number of flights that depart from all three major airports in NYC, departure delays are inevitable. We categorize departures that are delayed by more than an hour as ‘long delays’

i) What is the likelihood that a flight chosen at random experiences a “long delay”?

ii) If there were 336,800 flights that departed from the NYC airports in 2013, how many flights are expected to have “long delays”? Show all working.

(c)  A random sample of 72 flights was taken from flights that departed from the NYC airports in 2013. Calculate the probability that the average delay of this sample is not more than 20 minutes. Show all working.

(d) If the distribution of delay times is right skewed, (instead of normally distributed) with the same mean and standard deviation, will your answers to

i) change or remain unchanged? Briefly explain

ii) change or remain unchanged? Briefly explain

Answer 1

a)

µ =    12.64                              
σ =    40.21                              
n=   1                              
proportion=   0.8000                              
                                  
Z value at    0.8   =   0.842   (excel formula =NORMSINV(   0.80   ) )          
z=(x-µ)/(σ/√n)                                  
so, X=z * σ/√n +µ=   0.842   *   40.21   / √    1   +   12.64   =   46.48 minutes

b)

i)

Z =   (X - µ )/(σ/√n) = (   60   -   12.64   ) / (    40.21   / √   1   ) =   1.178  
                                          
P(X ≥   60   ) = P(Z ≥   1.18   ) =   P ( Z <   -1.178   ) =    0.11943489           (answer)

ii) flights are expected to have “long delays” = 0.1194*336800=40225.67158 ≈ 40226

c)

µ =    12.64                                      
σ =    40.21                                      
n=   72                                      
                                          
X =   20                                      
                                          
Z =   (X - µ )/(σ/√n) = (   20   -   12.64   ) / (   40.210   / √   72   ) =   1.553  
                                          
P(X ≤   20   ) = P(Z ≤   1.553   ) =   0.9398                       (answer)

d)

remain unchanged

because of central limit theorem (sample size is large enough to assume that population is approx normally distributed)