Weights of the vegetables in a field are normally distributed. from a sample carl cornfield determines the mean weight of a box of vegetables to be 172 oz. with a standard deviation of 8 oz. he wonders what percent of the vegetable boxes he has grouped for sale will have a weight less than 164 oz. carl calculates the z-score corresponding to the weight 164 oz. using the table (column .00), carl sees the area associated with this z-score is 0. carl rounds this value to the nearest hundredth or 0. now, carl decides to find the percent associated with boxes less than 164 oz. since 0.50 represents the part of the curve to the left of the mean (a z-score of 0) and the z-score carl just calculated represents the part of the curve from the mean to the point representing 164 oz., then carl can find the part of the curve to the left of 164 oz. by subtracting. so, carl subtracts 0.50 - 0.34 = 0. a3 or a4 %. this is the percent of vegetable boxes that weigh less than 164 oz.