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.
Y=1x+6 right? idk i think that’s right