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
1. $120.66+15.55=$136.21
2. $700.00-$136.21=?
3. The answer is in the 1st and 2nd question
The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed. However, it all depends on what you are trying to show from your data.
