Question

A courier service company has found that their delivery time of parcels to clients is approximately normally distributed with a mean delivery time of 50 minutes and a variance of 25 minutes (squared).

a) What is the probability that a randomly selected parcel will take 60 minutes to deliver? [2]

b) What is the probability that a randomly selected parcel will take between 38.75 and 55 minutes to deliver? [5]

c) What is the probability that a randomly selected parcel will take more than 36.25 minutes to deliver? [3]

d) What is the probability that a randomly selected parcel will take more than 59.25 minutes to deliver? [3]

e) What is the minimum delivery time for the 2.5% of parcels with the longest time to deliver?   

Answer 1

This is a normal distribution question with

a) P(X = 60.0) = ?
For a continous the probability is the integration of probability density function in an given interval. Since if we give a particular point as an interval the integration comes out as 0.
P(X = 60.0) = 0
b) P(38.75 < x < 55.0)=?

This implies that
P(38.75 < x < 55.0) = P(-2.25 < z < 1.0) = P(Z < 1.0) - P(Z < -2.25)
P(38.75 < x < 55.0) = 0.8413447460685429 - 0.012224472655044696
P(38.75 < x < 55.0) =
c) P(x > 36.25)=?
The z-score at x = 36.25 is,

z = -2.75
This implies that
P(x > 36.25) = P(z > -2.75) = 1 - 0.002979763235054555

d) P(x > 59.25)=?
The z-score at x = 59.25 is,

z = 1.85
This implies that
P(x > 59.25) = P(z > 1.85) = 1 - 0.9678432252043863

e) Given in the question
P(X < x) = 0.025
This implies that
P(Z < -1.96) = 0.025
With the help of formula for z, we can say that

x = 40.2002
PS: you have to refer z score table to find the final probabilities.
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