Question

caden is backing up his computer. the sizes of the 8 files that were most recently backed up are: 62.8 mb, 62.1 mb, 62.3 mb, 19.6 mb, 37.1 mb, 82.0 mb, 26.9 mb, 20.8 mb. what is the mean absolute deviation of the file sizes? if the answer is a decimal, round it to the nearest tenth. mean absolute deviation (mad): mb

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{62.8 + 62.1+62.3 + 19.6+37.1+82.0+26.9+20.8}{8}=\frac{373.6}{8}=46.7$ MB.

Step2: Calculate the absolute - deviations

$|62.8 - 46.7|=16.1$, $|62.1 - 46.7| = 15.4$, $|62.3 - 46.7|=15.6$, $|19.6 - 46.7| = 27.1$, $|37.1 - 46.7|=9.6$, $|82.0 - 46.7| = 35.3$, $|26.9 - 46.7|=19.8$, $|20.8 - 46.7| = 25.9$.

Step3: Calculate the mean of the absolute - deviations

The mean absolute deviation (MAD)$=\frac{16.1+15.4 + 15.6+27.1+9.6+35.3+19.8+25.9}{8}=\frac{164.8}{8}=20.6$ MB.

Answer:

20.6