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

Answer: i got this correct on test

2

0.4773

0.477

.023 or 2.3%

Step-by-step explanation:

(196-180)/8=2

look at 2.0 on the chart to get the 0.4773

then subtract .50 from .477 to get 0.023

move decimal over 2 to get 2.3%

:)

Answer 2

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


`z=(x-\mu)/(\sigma)=(196-180)/(8)=(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.