Question
Bob has entered his giant pumpkin into the tops field fair in New England. To qualify for the finals, pumpkins must meet a minimum weight requirement, which is based on the weights of all of the entries. This year, 55.17% of all entries will make it to the finals. With a standard deviation of 674.29lbs and an average weight of 1029.51lbs will bobs 938.23lb pumpkin make it to the finals? Provide justification and show all work.
Answers
Scott
the distance below the mean of Bob's pumpkin (z-score) is
... (938.23 - 1029.51) / 674.29
use a z-score table to find if the value lies in the upper 55.17% of entries
looks like he just missed...
... (938.23 - 1029.51) / 674.29
use a z-score table to find if the value lies in the upper 55.17% of entries
looks like he just missed...