MATH SOLVE

2 months ago

Q:
# The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {41, 50, 58, 62, 72, 83} What does the mean absolute deviation (MAD) of the data sets tell you about the temperatures over the 6-month period? Select from the drop-down menus to correctly answer the question. The MAD for City 2 is _______ the MAD for City 1, which means the average monthly temperatures of City 2 vary _______ the average monthly temperatures for City 1.

Accepted Solution

A:

Answer:The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1. Step-by-step explanation:For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,So for city 1:[tex]Mean = x1 = \frac{20+24+40+63+76+89}{6}[/tex][tex]x1 = \frac{312}{6}[/tex][tex]x1 = 52[/tex]Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)[tex]20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37[/tex]The deviations will be added then.[tex]Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24[/tex]So the mean absolute deviation for city 1 is 24 ..For city 2:[tex]Mean = x2 = \frac{41+50+58+62+72+83}{6}[/tex][tex]x2 = \frac{366}{6}[/tex][tex]x2 = 61[/tex]Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)[tex]41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22[/tex]The deviations will be added then.[tex]Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33[/tex]So the MAD for city 2 is 11.33 ..So,The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.