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.
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...
To determine if Bob's pumpkin will make it to the finals, we need to compare its weight to the minimum weight requirement.
First, let's find the minimum weight requirement by calculating the average weight of all entries:
1. Calculate the total weight of all entries:
Average weight = 1029.51 lbs
2. To find the minimum weight requirement, we multiply the average weight by the percentage of entries that will make it to the finals:
Minimum weight requirement = Average weight * (55.17/100)
= 1029.51 lbs * 0.5517
≈ 567.95 lbs
The minimum weight requirement for the finals is approximately 567.95 pounds.
Next, we can compare Bob's pumpkin weight of 938.23 pounds to the minimum weight requirement. If his pumpkin weighs more than the requirement, it will make it to the finals; otherwise, it will not.
To determine if his pumpkin meets the requirement, we can use the concept of Z-scores. A Z-score measures how many standard deviations a data point is from the mean.
Let's calculate the Z-score for Bob's pumpkin weight:
1. Calculate the deviation of Bob's pumpkin weight from the mean (average weight):
Deviation = Bob's pumpkin weight - Average weight
= 938.23 lbs - 1029.51 lbs
= -91.28 lbs
2. Calculate the Z-score using the following formula:
Z-score = Deviation / Standard deviation
= -91.28 lbs / 674.29 lbs
≈ -0.1352
The calculated Z-score for Bob's pumpkin weight is approximately -0.1352.
To determine if Bob's pumpkin will make it to the finals, we need to check if the Z-score falls within the range that corresponds to the percentage of entries making it to the finals. Typically, a Z-score between -1.96 and +1.96 corresponds to a range containing 95% of the data.
If the Z-score is within this range, Bob's pumpkin will make it to the finals. Otherwise, it will not.
Since the calculated Z-score of -0.1352 is within the range of -1.96 and +1.96, Bob's 938.23-pound pumpkin will make it to the finals.
Please note that the mentioned ranges and assumptions are based on common statistical practices and may vary in different contexts or competitions.