SOLUTION: The average time for a courier to travel from pittsburgh to harrisburg is 200 minutes, and the stnadard deviationis 10 minutes. If one of thes trips is selected at random, find th

Question 1191748: The average time for a courier to travel from pittsburgh to harrisburg is 200 minutes, and the stnadard deviationis 10 minutes. If one of thes trips is selected at random, find the probability that the courier will have the following travel time. Assume that variable is normally distributed. a. at least 180 minutes b. at most 205 minutes

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
the mean is 200 minutes and the standard deviation is 10 minutes.

z-score for 180 minutes is (180 - 200) / 10 = -20/10 = -2.
if it's going to be at least 180 minutes, then you want the area to the right of that z-score.
the area to the left is equal to .02275, rounded to 5 decimal digits.
the area to the right is equal to 1 minus that = .97725.
that means the probability that the travel time will be greater than 180 minutes is 97.725%.

z-score for 205 minutes is (205 - 200) / 10 = .5
if it's going to be at most 205 minutes, then you want the area to the left of that z-score.
the area to the left is equal to .69146, rounded to 5 decimal places,
that means 69.146% probability that the travel time will be less than 205 minutes.

here's a table that i normally use.
https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf

the tables are designed to show you the area to the left.
the area to the right is equal to 1 minus the area to the left.

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