Statistics Examples

$2$ , $3$ , $4$ , $5$ , $3$ , $6$ , $8$ , $6$ , $4$ , $2$

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.

$1 + 3.322 \log (6) = 3.58501845$

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 $8 - 2 = 6$ .

$6$

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 $2$ and create $4$ groups of size $2$ .

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 2 - 3 \\ 4 - 5 \\ 6 - 7 \\ 8 - 9 \end{array}$

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.

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 2 - 3 & 1.5 - 3.5 \\ 4 - 5 & 3.5 - 5.5 \\ 6 - 7 & 5.5 - 7.5 \\ 8 - 9 & 7.5 - 9.5 \end{array}$

Step 8

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

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 2 - 3 & 1.5 - 3.5 & | | | | \\ 4 - 5 & 3.5 - 5.5 & | | | \\ 6 - 7 & 5.5 - 7.5 & | | \\ 8 - 9 & 7.5 - 9.5 & | \end{array}$

Step 9

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

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 2 - 3 & 1.5 - 3.5 & 4 \\ 4 - 5 & 3.5 - 5.5 & 3 \\ 6 - 7 & 5.5 - 7.5 & 2 \\ 8 - 9 & 7.5 - 9.5 & 1 \end{array}$

Step 10

The relative frequency of a data class is the percentage of data elements in that class. The relative frequency can be calculated using the formula $f_{i} = \frac{f}{n}$ , where $f$ is the absolute frequency and $n$ is the sum of all frequencies.

$f_{i} = \frac{f}{n}$

Step 11

$n$ is the sum of all frequencies. In this case, $n = 4 + 3 + 2 + 1 = 10$ .

$n = 10$

Step 12

The relative frequency can be calculated using the formula $f_{i} = \frac{f}{n}$ .

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 2 - 3 & 1.5 - 3.5 & 4 & \frac{4}{10} \\ 4 - 5 & 3.5 - 5.5 & 3 & \frac{3}{10} \\ 6 - 7 & 5.5 - 7.5 & 2 & \frac{2}{10} \\ 8 - 9 & 7.5 - 9.5 & 1 & \frac{1}{10} \end{array}$

Step 13

Simplify the relative frequency column.

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 2 - 3 & 1.5 - 3.5 & 4 & 0.4 \\ 4 - 5 & 3.5 - 5.5 & 3 & 0.3 \\ 6 - 7 & 5.5 - 7.5 & 2 & 0.2 \\ 8 - 9 & 7.5 - 9.5 & 1 & 0.1 \end{array}$

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