How many arrivals were much more delayed than others (i. e., positive outliers with z-score of time difference > 3)? (e) How many flights arrived much earlier than scheduled compared to others (i. e., negative outliers with z-score of time difference < -3)?

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.

Step-by-step 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!!!