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 what? carl calculates the z-score corresponding to the weight 191 oz. (to the nearest tenth). using the table below, carl sees the percentage associated with this z-score is what? adding these together, carl sees the percentage between 169 oz. and 191 oz. is what?

The z-score corresponding to the weight 169 oz is -1.4

Using the table, the percentage associated with this z-score is 41.92%

The z-score corresponding to the weight 191 oz is 1.4

Using the table, the percentage associated with this z-score is 41.92%

The percentage between 169 oz. and 191 oz. is 83.84%

Step-by-step explanation:

We are informed that the weights of vegetables in a field are normally distributed with the following characteristics;

Mean weight = 180

Standard deviation = 8

We are to determine the percent of the vegetable boxes grouped for sale with a weight between 169 oz. and 191 oz. To do this, we shall perform a number of calculations involving z-cores;

The first step is to compute the z-score corresponding to the weight of 169 oz;

Using the table, the percentage associated with this z-score is 41.92% since the standard normal curve is symmetric.

The next step is to compute the z-score corresponding to the weight of 191 oz;

Using the table, the percentage associated with this z-score is 41.92%

Finally, the percentage between 169 oz. and 191 oz. is;

41.92% + 41.92% = 83.84%.