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ariana was looking at rent costs for 5 apartments. each rent was a diff…

Question

ariana was looking at rent costs for 5 apartments. each rent was a different amount between $1,000 and $1,400 per month, except for one apartment whose rent was $4,800 per month. ariana looked at the mean and median of the rents. then, she decided to remove the $4,800 rent from the set and recalculate the mean and median. how will removing this rent affect the mean and median? choose 1 answer: a both the mean and median will increase, but the median will increase by more than the mean. b both the mean and median will increase, but the mean will increase by more than the median. c both the mean and median will decrease, but the median will decrease by more than the mean. d both the mean and median will decrease, but the mean will decrease by more than the median.

Explanation:

Step1: Understand mean and median

The mean is the sum of all values divided by the number of values. The median is the middle - value when the data is arranged in ascending order.

Step2: Analyze the outlier effect

The value $4800$ is an outlier (much larger than the other values in the range of $1000 - 1400$). The mean is affected by outliers. Since $4800$ is much larger than the other values, it pulls the mean up.

Step3: Consider the median

For $n = 5$ data - points, the median is the 3rd value when arranged in ascending order. The outlier $4800$ is likely to be the largest value. Removing it will not change the position of the middle - value (if the other values remain in the same relative order), or may shift it to a value that is more representative of the non - outlier data.

Step4: Determine the change

When the outlier $4800$ is removed, the sum of the values decreases significantly for the mean calculation, and the number of values also decreases by 1. But the mean will decrease more than the median because the mean is more sensitive to extreme values.

Answer:

D. Both the mean and median will decrease, but the mean will decrease by more than the median.