Finite Math Examples

$6.1$ , $0.6$ , $4.9$ , $6.6$ , $5.9$ , $3.3$ , $5.1$ , $5.4$

Step 1

The number of classes can be estimated using the rounded output of Sturges' rule, $N = 1 + 3.322 \log (n)$ , where $N$ is the number of classes and $n$ is the number of items in the data set.

$1 + 3.322 \log (8) = 4.00006493$

Step 2

Select $4$ classes for this example.

$4$

Step 3

Find the data range by subtracting the minimum data value from the maximum data value. In this case, the data range is $6.6 - 0.6 = 6$ .

$6$

Step 4

Find the class width by dividing the data range by the desired number of groups. In this case, $\frac{6}{4} = 1.5$ .

$1.5$

Step 5

Round $1.5$ up to the nearest whole number. This will be the size of each group.

$2$

Step 6

Start with $0.6$ and create $4$ groups of size $2$ .

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0.6 - 1.6 \\ 2.6 - 3.6 \\ 4.6 - 5.6 \\ 6.6 - 7.6 \end{array}$

Step 7

Determine the class boundaries by subtracting $0.5$ from the lower class limit and by adding $0.5$ to the upper class limit.

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0.6 - 1.6 & 0.0\overline{9} - 2.1 \\ 2.6 - 3.6 & 2.1 - 4.1 \\ 4.6 - 5.6 & 4.1 - 6.1 \\ 6.6 - 7.6 & 6.1 - 8.1 \end{array}$

Step 8

Draw a tally mark next to each class for each value that is contained within that class.

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0.6 - 1.6 & 0.0\overline{9} - 2.1 & | \\ 2.6 - 3.6 & 2.1 - 4.1 & | \\ 4.6 - 5.6 & 4.1 - 6.1 & | | | | | \\ 6.6 - 7.6 & 6.1 - 8.1 & | | \end{array}$

Step 9

Count the tally marks to determine the frequency of each class.

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0.6 - 1.6 & 0.0\overline{9} - 2.1 & 1 \\ 2.6 - 3.6 & 2.1 - 4.1 & 1 \\ 4.6 - 5.6 & 4.1 - 6.1 & 5 \\ 6.6 - 7.6 & 6.1 - 8.1 & 2 \end{array}$