Question

please check my workkkk:)!!!!!!! PLEASE:)

Answer 1

an rtf file eh, can you save that as a word doc?

Answer 2

yes:) hold on

Answer 3

what about now??

Answer 4

odt? maybe

Answer 5

what is odt?

Answer 6

dunno, thats the file extension on it

Answer 7

hmmm

Answer 8

at any rate, tell me how you did the zscore of 196

Answer 9

of 196?

Answer 10

ohh...

Answer 11

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. = 2.0

Using the table (column .00), Carl sees the area associated with this z-score is 0.4772

Carl rounds this value to the nearest thousandth or 0.477

Now, Carl subtracts 0.50 - 0.477 = 2.3 %.

Answer 12

are my answers correct?

Answer 13

their correctness is what we are determining, but im not sure how you got the answers, the process you took is more informative. Otherwise, we simply cannot tell if you are trying to verify someone elses work

Answer 14

how do we find the zscore of 196 if we know the mean and sd ?

Answer 15

do you know the chart for zscore

Answer 16

it looks like the ztable they are using is measured from the mean, as opposed to tails

Answer 17

i guess....

Answer 18

Im asuming its correct:)

Answer 19

since we want to know whats in the tail above our z, the symmetric nature of the curve tells us that that will be the same probability as if it were to the left of -z