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

Answer 1

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 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.

now you can solve it....

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