Answer:
d) 0.135% of total arrivals were much more delayed than others.
e) 0.135% of total flights arrived much earlier than scheduled compared to others.
Explanation:
With the question not totally complete, we will solve the available parts as well as possible.
d) How many arrivals were much more delayed than others (i.e., positive outliers with z-score of time difference > 3)?
P(z > 3)
Using the normal distribution table
P(z > 3) = 1 - P(z ≤ 3) = 1 - 0.99865 = 0.00135 = 0.135% of total arrivals were much more delayed than others.
e) How many flights arrived much earlier than scheduled compared to others (i.e., negative outliers with z-score of time difference < -3)?
P(z < -3)
Using the normal distribution table
P(z < -3) = 0.00135 = 0.135% of total flights arrived much earlier than scheduled compared to others.
Hope this Helps!!!