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?
Answers
Given :
Mean delivery time = 30 minutes
Variance = 25 minutes
To find:
Probability that a randomly selected parcel will take between 26 and 42 minutes to deliver
Solution:
Variance = 25 minutes
Therefore, Standard deviation = √25
= 5 minutes
Z score = ( Given value - Mean)/Standard deviation
Z score for 26 = (26 - 30)/5
= -4/5
= -0.8
= 0.2119
Similarly,
Z score for 42 = (42 - 30)/5
= -8/5
= 2.4
= 0.9918
Probability of the random selected particle -
= 0.9918 - 0.2119
= 0.7799
Answer: The probability of a randomly selected parcel taking between 26 and 42 minutes to deliver is 0.7799.
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
Solution:
mean delivery time = 30 minutes
variance of 25 minutes
=> Standard deviation = √variance = √25 = 5 minutes
Z score = ( Value - Mean)/SD
Z score for 26 = (26 - 30)/5 = -0.8 => 0.2119 Refer attached z score table
Z score for 42 = (42 - 30)/5 = 2.4 => 0.9918 Refer attached z score table
=> Between 26 & 42 mins = 0.9918 - 0.2119
Probability = 0.7799
0.7799 is the probability that a randomly selected parcel will take between 26 and 42 minutes to deliver
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