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What is the mean absolute deviation of the temperatures?
First Day Temperatures




Month
Temperature 
^(@)^(@) F)


January
9


February
15


March
20


April
18


May
19


June
27

What is the mean absolute deviation of the temperatures?\newlineFirst Day Temperatures\newline\begin{tabular}{|c|c|}\newline\hline Month & Temperature ^{\circ}{ }^{\circ} F) \\\newline\hline January & 99 \\\newline\hline February & 1515 \\\newline\hline March & 2020 \\\newline\hline April & 1818 \\\newline\hline May & 1919 \\\newline\hline June & 2727 \\\newline\hline\newline\end{tabular}

Full solution

Q. What is the mean absolute deviation of the temperatures?\newlineFirst Day Temperatures\newline\begin{tabular}{|c|c|}\newline\hline Month & Temperature ^{\circ}{ }^{\circ} F) \\\newline\hline January & 99 \\\newline\hline February & 1515 \\\newline\hline March & 2020 \\\newline\hline April & 1818 \\\newline\hline May & 1919 \\\newline\hline June & 2727 \\\newline\hline\newline\end{tabular}
  1. Calculate Mean: Find the mean of the temperatures.\newline(9+15+20+18+19+27)/6=108/6=18(9 + 15 + 20 + 18 + 19 + 27) / 6 = 108 / 6 = 18\newlineMean temperature: 1818
  2. Absolute Deviations: Calculate the absolute deviations from the mean.\newline918=9|9 - 18| = 9\newline1518=3|15 - 18| = 3\newline2018=2|20 - 18| = 2\newline1818=0|18 - 18| = 0\newline1918=1|19 - 18| = 1\newline2718=9|27 - 18| = 9
  3. Mean of Deviations: Find the mean of these absolute deviations.\newline(9+3+2+0+1+9)/6=24/6=4(9 + 3 + 2 + 0 + 1 + 9) / 6 = 24 / 6 = 4\newlineMean absolute deviation: 44

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