Question 1197417: Sun Courier - a parcel delivery company - has found that the delivery time of parcels
to clients in Durban metropolitan area after airport collection is normally distributed
with a mean delivery time equal to 45 minutes and a standard deviation of 8 minutes.
For a newly arrived consignment at Durban airport, what is the probability that a
randomly selected parcel will take:
a. Between 45 and 51 minutes to deliver to the client?
b. Less than 48 minutes to deliver?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
you have:
m = 45
s = 8
when x = 45, z = (45 - 45) / s = 0
when x = 51, z = (51 - 45) / s = 6 / 8 = 3/4 = .75
area to the left of z = 0 = .5
area to the left of z = .75 = .7734 rounded to 4 decimal places.
area in between z = 0 and z = .75 = .7734 - .5 = .2734.
that's the probability that the package will take between 45 and 51 minutes to deliver to the client.
when x = 48, z = (x - m) / s becomes z = (48 - 45) / 8 which becomes z = 3 / 8 = .375.
area to the left of z = .375 = .6462 rounded to 4 decimal places.
that's the probability that the package will take less than 48 minutes to deliver to the client.
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