Answer:
The correct choice is the third one.
Step-by-step explanation:
Answer: On average, a pet's weight during this vet visit is approximately 2.4 pounds away from 12.9 pounds.
If the MAD for another day's weights was 1.5, then that day’s weights would be less variable compared to the weights of pets seen today.
Step-by-step explanation:
Given: The data in the table outlines the weights of animals visiting a vet one day, in pounds.
Mean = 12.9
Median= 12.0
Mode = 12.0
Mean Absolute Deviation = 2.4
It’s understood that the mean absolute deviation (MAD) of a dataset indicates the average distance between each data point and the mean. It reflects the variation present in the dataset.
Therefore, the average weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.
Moreover, if another day had a MAD of 1.5, and since 1.5 < 2.4,[TAG_42]]
it implies that the weights on that day would be less variable compared to those of pets encountered on this day.
Assuming a maximum rental duration of 8 days, if we apply this to the equation, Happy Harry's Rentals costs $500, while Smilin' Sam's charges $600. Therefore, for the 7th and 8th days, Happy Harry's Rentals is a better option, while Smilin' Sam's is preferable for the first six days.
