Question

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 more than 196 oz. Carl decides to answer the following questions about the population of vegetables from these sample statistics: Carl calculates the z-score corresponding to the weight 196 oz. Using the table (column .00), Carl sees the area associated with this z-score is 0. Carl rounds this value to the nearest thousandth or 0. Now, Carl subtracts 0.50 - 0. = %.

Answer 1

To find the z-score for a weight of 196 oz., use

$z=\frac{x-\mu}{\sigma}=\frac{196-180}{8}=\frac{16}{8}=2$

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2.  Carl is wondering about the percentage of boxes with weights ABOVE z = 2.  The total area under the normal curve is 1, so subtract .9772 from 1.0000.

1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.