Question

1. Flight 202's arrival time is normally distributed with a mean arrival time of 4:30

Answer 1

Answer:

Step-by-step explanation:

1) Let the random time variable, X = 45min; mean, ∪ = 30min; standard deviation, α = 15min

By comparing P(0 ≤ Z ≤ 30)

P(Z ≤ X - ∪/α) = P(Z ≤ 45 - 30/15) = P( Z ≤ 1)

Using Table

P(0 ≤ Z ≤ 1) = 0.3413

P(Z > 1) = (0.5 - 0.3413) = 0.1537

∴ P(Z > 45) = 0.1537

2)  By compering (0 ≤ Z ≤ 15) ( that is 4:15pm)

P(Z ≤ 15 - 30/15) = P(Z ≤ -1)

Using Table

P(-1 ≤ Z ≤ 0) = 0.3413

P(Z < 1) = (0.5 - 0.3413) = 0.1587

∴ P(Z < 15) = 0.1587

3) By comparing P(0 ≤ Z ≤ 60) (that is for 5:00pm)

P(Z ≤ 60 - 30/15) = P(Z ≤ 2)

Using Table

P(0 ≤ Z ≤ 1) = 0.4772

P(Z > 1) = (0.5 - 0.4772) = 0.0228

∴ P(Z > 60) = 0.0228