posted by Anonymous .
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 180 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 between 169 oz. and 191 oz. Carl decides to answer the following questions about the population of vegetables from these sample statistics. You will use some answers more than once. Carl calculates the z-score corresponding to the weight 169 oz. (to the nearest tenth). Using the table, Carl sees the percentage associated with this z-score is
z-score for the 169 = (169-180)/8 = -1.375
x-score for the 191 = (191-180)/8 = +1.375
look up value for 1.375 and the value for -1.375
subtract the two values
You should get .8309
I recommend this webpage for these type of questions.
The beauty of this page is that it lets you enter the mean and SD directly without even finding the z-score.
Personally, I don't see any difference in going to some table that "somebody" created ages ago, and using this page. In either case you are not actually doing the arithmetic.
notice you can just enter 180 for the mean, 8 for the sd, click on the "between" and enter 169 and 191 to get 0.8309
Bookmark this webpage for future use.