QUESTION 5 [4]
A courier service company has found that their delivery time of parcels to clients is approximately normally distributed with a mean delivery time of 30 minutes and a variance of 25 minutes (squared).
a) What is the probability that a randomly selected parcel will take between 26 and 42 minutes to deliver? (2)
b) What is the maximum delivery time (minutes) for the 2.5% of parcels with the shortest time to deliver? (2)
Answers
Given : A courier service company has found that their delivery time of parcels to clients is approximately normally distributed with a mean delivery time of 30 minutes and a variance of 25 minutes (squared).
To find : a) What is the probability that a randomly selected parcel will take between 26 and 42 minutes to deliver
b) What is the maximum delivery time (minutes) for the 2.5% of parcels with the shortest time to deliver
Solution:
mean delivery time = 30 minutes
variance of 25 minutes
=> Standard deviation = √25 = 5 minutes
Z score = ( Value - Mean)/SD
Z score for 26 = (26 - 30)/5 = -0.8 => 0.2119
Z score for 42 = (42 - 30)/5 = 2.4 => 0.9918
Probability = 0.9918 - 0.2119 = 0.7799
0.7799 is the probability that a randomly selected parcel will take between 26 and 42 minutes to deliver
2.5% of parcels with the shortest time to deliver
Z score = -1.96
-1.96 = (value - 30)/5
=> -9.8 = Value - 30
=> Value = 20.2
20.2 minutes is maximum delivery time (minutes) for the 2.5% of parcels with the shortest time to deliver.
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Answer ... nice answer sir