Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. assume also that this time is normally distributed with a standard deviation of 2 minutes. what percentage of the students will take between 2 and 6 minutes to find a parking spot in the main parking lot?

Question

Martha Fox

Mathematics

1 May, 01:40

Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. assume also that this time is normally distributed with a standard deviation of 2 minutes. what percentage of the students will take between 2 and 6 minutes to find a parking spot in the main parking lot?

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Jaxon · Answer 1

Mean=5

sd=2

Using Z (x) = (x-mean) / sd

Z (2) = (2-5) / 2=-1.5

Z (6) = (6-5) / 2=0.5

Therefore

probability of finding a spot between 2 and 6 minutes is

P (x<6) - P (x<2)

=P (Z<0.5) - P (Z<-1.5)

=0.6915-0.0668

=0.6247

=>

More than half of the students (62.5%) finds a space between 2 and 6 minutes.