Question

there were 6 rain - storms last summer. the amount of rainfall during each storm was 6 centimeters 3 centimeters 1 centimeter 8 centimeters 7 centimeters 9 centimeters. what was the mean absolute deviation of the rainfall amounts? if the answer is a decimal, round it to the nearest tenth. mean absolute deviation (mad): centimeters

Explanation:

Step1: Calculate the mean

Let the rainfall amounts be \(x_1 = 6\), \(x_2=3\), \(x_3 = 1\), \(x_4=8\), \(x_5=7\), \(x_6 = 9\). The mean \(\bar{x}=\frac{6 + 3+1+8+7+9}{6}=\frac{34}{6}=\frac{17}{3}\approx5.67\)

Step2: Calculate the absolute - deviations

\(\vert x_1-\bar{x}\vert=\vert6 - \frac{17}{3}\vert=\vert\frac{18 - 17}{3}\vert=\frac{1}{3}\approx0.33\)
\(\vert x_2-\bar{x}\vert=\vert3-\frac{17}{3}\vert=\vert\frac{9 - 17}{3}\vert=\frac{8}{3}\approx2.67\)
\(\vert x_3-\bar{x}\vert=\vert1-\frac{17}{3}\vert=\vert\frac{3 - 17}{3}\vert=\frac{14}{3}\approx4.67\)
\(\vert x_4-\bar{x}\vert=\vert8-\frac{17}{3}\vert=\vert\frac{24 - 17}{3}\vert=\frac{7}{3}\approx2.33\)
\(\vert x_5-\bar{x}\vert=\vert7-\frac{17}{3}\vert=\vert\frac{21 - 17}{3}\vert=\frac{4}{3}\approx1.33\)
\(\vert x_6-\bar{x}\vert=\vert9-\frac{17}{3}\vert=\vert\frac{27 - 17}{3}\vert=\frac{10}{3}\approx3.33\)

Step3: Calculate the mean absolute deviation

\(MAD=\frac{\frac{1}{3}+\frac{8}{3}+\frac{14}{3}+\frac{7}{3}+\frac{4}{3}+\frac{10}{3}}{6}=\frac{\frac{1 + 8+14+7+4+10}{3}}{6}=\frac{\frac{44}{3}}{6}=\frac{44}{18}=\frac{22}{9}\approx2.4\)

Answer:

\(2.4\)