Statistics Examples

$0$ , $1$ , $6$ , $9$ , $8$ , $1$ , $4$ , $3$ , $6$

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.

$1 + 3.322 \log (7) = 3.80741568$

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 $9 - 0 = 9$ .

$9$

Step 4

Find the class width by dividing the data range by the desired number of groups. In this case, $\frac{9}{4} = 2.25$ .

$2.25$

Step 5

Round $2.25$ up to the nearest whole number. This will be the size of each group.

$3$

Step 6

Start with $0$ and create $4$ groups of size $3$ .

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 \\ 3 - 5 \\ 6 - 8 \\ 9 - 11 \end{array}$

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.

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 \\ 3 - 5 & 2.5 - 5.5 \\ 6 - 8 & 5.5 - 8.5 \\ 9 - 11 & 8.5 - 11.5 \end{array}$

Step 8

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

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & | | | \\ 3 - 5 & 2.5 - 5.5 & | | \\ 6 - 8 & 5.5 - 8.5 & | | | \\ 9 - 11 & 8.5 - 11.5 & | \end{array}$

Step 9

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

$\begin{array}{c} Class & Class Boundaries & Frequency \\ 0 - 2 & - 0.5 - 2.5 & 3 \\ 3 - 5 & 2.5 - 5.5 & 2 \\ 6 - 8 & 5.5 - 8.5 & 3 \\ 9 - 11 & 8.5 - 11.5 & 1 \end{array}$

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, $n = 3 + 2 + 3 + 1 = 9$ .

$n = 9$

Step 12

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

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 3 & \frac{3}{9} \\ 3 - 5 & 2.5 - 5.5 & 2 & \frac{2}{9} \\ 6 - 8 & 5.5 - 8.5 & 3 & \frac{3}{9} \\ 9 - 11 & 8.5 - 11.5 & 1 & \frac{1}{9} \end{array}$

Step 13

Simplify the relative frequency column.

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} \\ 0 - 2 & - 0.5 - 2.5 & 3 & 0.\overline{3} \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{2} \\ 6 - 8 & 5.5 - 8.5 & 3 & 0.\overline{3} \\ 9 - 11 & 8.5 - 11.5 & 1 & 0.\overline{1} \end{array}$

Step 14

Multiply every relative frequency by $100$ to get the percentage frequency.

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 3 & 0.\overline{3} & 0.\overline{3} \cdot 100 % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{2} & 0.\overline{2} \cdot 100 % \\ 6 - 8 & 5.5 - 8.5 & 3 & 0.\overline{3} & 0.\overline{3} \cdot 100 % \\ 9 - 11 & 8.5 - 11.5 & 1 & 0.\overline{1} & 0.\overline{1} \cdot 100 % \end{array}$

Step 15

Simplify the Percentage column.

$\begin{array}{c} Class & Class Boundaries & Frequency (f) & f_{i} & Percent \\ 0 - 2 & - 0.5 - 2.5 & 3 & 0.\overline{3} & 33.\overline{3} % \\ 3 - 5 & 2.5 - 5.5 & 2 & 0.\overline{2} & 22.\overline{2} % \\ 6 - 8 & 5.5 - 8.5 & 3 & 0.\overline{3} & 33.\overline{3} % \\ 9 - 11 & 8.5 - 11.5 & 1 & 0.\overline{1} & 11.\overline{1} % \end{array}$

Enter YOUR Problem