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?

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.

First, find the decimal of 3/4 so that 0,75 multiplied by 500 is 375 seats are filled.