Question

Taylor is a student in Ms. Garza's math class. Her scores on the last ten assignments are shown in the table below. Taylor's Scores 96 91 85 89 83 85 94 95 90 92 What is the mean absolute deviation of Taylor's scores? O A. 90 O B. 36 O C. 3.6 OD. none of these 1 2 3 4 5 6 7

Answer 1

First, calculate the mean of the scores. Since there are 10 scores, then the mean is given by:

$\begin{gathered} \bar{s}=\frac{96+91+85+89+83+85+94+95+90+92}{10} \\ =\frac{900}{10} \\ =90 \end{gathered}$

Calculate the absolute value of the difference between the mean and each of the scores:

$\begin{gathered} |s_1-\bar{s}|=|96-90|=6 \\ |s_2-\bar{s}|=|91-90|=1 \\ |s_3-\bar{s}|=|85-90|=5 \\ |s_4-\bar{s}|=|89-90|=1 \\ |s_5-\bar{s}|=|83-90|=7 \\ |s_6-\bar{s}|=|85-90|=5 \\ |s_7-\bar{s}|=|94-90|=4 \\ |s_8-\bar{s}|=|95-90|=5 \\ |s_9-\bar{s}|=|90-90|=0 \\ |s_{10}-\bar{s}|=|92-90|=2 \end{gathered}$

The mean absolute deviation is the average of the absolute deviations from each score to the mean. Calculate the mean from the deviations:

$\begin{gathered} \frac{6+1+5+1+7+5+4+5+0+2}{10} \\ =\frac{36}{10} \\ =3.6 \end{gathered}$

Therefore, the mean absolute deviation of Taylor's scores, is 3.6