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...
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.
1 answer