Question

the last six days of August 2011 all time records for high temperatures in Austin Texas which of the following values that identifies the mean absolute deviation for these temperatures F. 4G. 107H. 103J. 3.3

Answer 1

We have to calculate the mean absolute deviation (MAD) for these temperatures.

We can express the MAD as:

`\text{MAD}=(1)/(n)\sum ^n_(i=1)|x_i-\barx`

So first we have to calculate the mean of the sample. We can do this as:

`\bar{x}=(1)/(n)\sum ^n_(i=1)x_i=(1)/(6)(103+110+112+109+105+103)=(1)/(6)\cdot642=107`

Now we can calculate the MAD as:

`\begin{gathered} \text{MAD}=(1)/(6)(|103-107|+|110-107|+|112-107|+|109-107|+|105-107|+|103-107|) \\ MAD=(1)/(6)(4+3+5+2+2+4) \\ \text{MAD}=(1)/(6)(20) \\ \text{MAD}\approx3.3 \end{gathered}`

Answer: The MAD is approximately 3.3.