Find the mean absolute deviation to the nearest tenth.

A fast food company polled a random sample of its customers to find how many times a month they eat out: 10, 3, 12, 15, 7, 8, 4, 12, 9, 14, 12. Find the mean absolute deviation of the data set to the nearest tenth.

The mean is equal to the sum of all data values divided by the number of data values.

The mean absolute deviation is the average of the absolute values of the differences between the mean and each value in the data set.

The given data set are :

10, 3, 12, 15, 7, 8, 4, 12, 9, 14, 12

First, find the mean of the data.

Mean$=\frac{10+3+12+15+7+8+4+12+9+14+12}{11}=\frac{106}{11}=9.63$

Next, find the absolute value of the difference between the mean and each value.

$\begin{array}{c}\left|10-9.63\right|=0.37,\left|3-9.63\right|=6.63,\left|12-9.63\right|=2.37,\left|15-9.63\right|=5.37,\\ \left|7-9.63\right|=2.63,\left|8-9.63\right|=1.63,\left|4-9.63\right|=5.63,\left|12-9.63\right|=2.37,\\ \left|9-9.63\right|=0.63,\left|14-9.63\right|=4.37,\left|12-9.63\right|=2.37\end{array}$

Now, find the mean of the differences.

$$\begin{gathered} =\frac{0.37+6.63+2.37+5.37+2.63+1.63+5.63+2.37+0.63+4.37+2.37}{11} \\ =\frac{34.37}{11} \\ =3.12 \\ \simeq 3.1 \end{gathered}$$

Therefore, the mean absolute deviation is 3.1.