1. A supermarket manager has determined that the amount of time customers spend in the supermarket is approximately normally distributed with a mean of 45 minutes and a standard deviation of 6 minutes. Find the probability that: a. A customer spends less than 48 minutes in the supermarket(3pts) b. Customer spends between 39 and 43 minutes in the supermarket(3pts)

a)
P(x < 48)=
P(z < (48-45)/6)=
P(z < 0.5)=0.6914 

b)
P(39 < x < 43)=
P((39-45)/6 < z < (43-45)/6)=
P(-1 < z < -1/3)=0.2108