Finite Math Examples

$11$ , $200 {(1 + 0.08)}^{2}$

Step 1

The class width can be found by finding the difference of the maximum data value and the minimum data value (range of data) divided by the number of classes.

$\frac{(Maximum Data Value - Minimum Data Value)}{(Number of Classes)} = \frac{(Range of Data)}{(Number of Classes)}$

Step 2

Since the number of items in the data $2$ is too small to be arranged in a set of classes, it is easier to list the items with their related frequencies.

$\begin{array}{c} Item & Frequency \\ 11 & 1 \\ 233.28 & 1 \end{array}$