Q2. - Algebra 1

Question

The table shows the number of Calories in twelve different snacks. Which measure of central tendency would be most affected by the outlier 342 Calories?

 

Number of Calories in Snacks
122	87	149	121
64	138	342	72
179	105	99	114

 $\mathit{F}$ mean                       $\mathit{H}$ mode

$\mathit{G}$ median                  $\mathit{J}$ range

Step-by-Step Solution

Answer

The option $\mathit{F}$ is correct.

Step 1. State the concept.

Mean absolute deviation is given by the formula:

$\frac{|\overline{x}-x_1|+|\overline{x}-x_2|+\dots +|\overline{x}-x_n|}{n}$, where $\overline{x}$ is the mean and $x_i$ are the terms.

Step 2. Calculate the mean.

First calculate the mean number of calories in the snacks.

$\begin{array}{l}\overline{x}=\frac{\begin{array}{l}122+64+179+87+138+105+149\\ +342+99+121+72+114\end{array}}{12}\\ =\frac{1592}{12}\\ =132.667\\ \approx 133\end{array}$

Step 3. Calculate the mean absolute deviation.

In order to find mean absolute deviation, first find the absolute values of the differences.

 $\begin{array}{l}|\overline{x}-x_1|=|133-122|=11\\ |\overline{x}-x_2|=|133-64|=69\\ |\overline{x}-x_3|=|133-179|=46\\ |\overline{x}-x_4|=|133-87|=46\\ |\overline{x}-x_5|=|133-138|=5\\ |\overline{x}-x_6|=|133-105|=28\\ |\overline{x}-x_7|=|133-149|=16\\ |\overline{x}-x_8|=|133-342|=209\\ |\overline{x}-x_9|=|133-99|=34\\ |\overline{x}-x_{10}|=|133-121|=12\\ |\overline{x}-x_{11}|=|133-72|=61\\ |\overline{x}-x_{12}|=|133-114|=19\end{array}$

Now calculate mean absolute deviation.

$\frac{\begin{array}{l}11+69+46+46+5+28+\\ 16+209+34+12+61+19\end{array}}{12}=46.33$

Hence, mean will be that measure of central tendency that would be most affected by the outlier 342 Calories. Therefore, option $F$ is correct.