motion2020 wrote:
What is the resultant Inter-Quartile range for a set of unique positive values, if the values in the top 50-percentile are decreased by 3 and then divided by 5, while the values in the bottom 50 percentile (median included) are divided by 5 and then reduced by 3, given the original Inter-Quartile range was 48?
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
Source:
Scholardenp.s. I have spent 4-5 minutes trying to figure out and solve this question. It's an awesome one
Refer to the figure below;
Old IQR = \((x - y) = 48\)
New IQR = \(\frac{(x - 3)}{5} - (\frac{y}{5} - 3)\)
= \(\frac{x}{5} - \frac{y}{5} + 3 - \frac{3}{5}\)
= \(\frac{{(x - y) + 12}}{5} = \frac{{48 + 12}}{5} = \frac{60}{5} = 12\)
Hence, option C
Attachments
What is the resultant Inter-Quartile range.png [ 4.08 KiB | Viewed 2020 times ]