Finite Math Examples

$\begin{array}{c} Class & Frequency \\ 90 - 99 & 4 \\ 80 - 89 & 6 \\ 70 - 79 & 4 \\ 60 - 69 & 3 \\ 50 - 59 & 2 \\ 40 - 49 & 1 \end{array}$

Step 1

Reorder the classes with their related frequencies (ƒ) in an ascending order (lowest number to highest), which is the most common.

$\begin{array}{c} Class & Frequency (f) \\ 40 - 49 & 1 \\ 50 - 59 & 2 \\ 60 - 69 & 3 \\ 70 - 79 & 4 \\ 80 - 89 & 6 \\ 90 - 99 & 4 \end{array}$

Step 2

Find the midpoint $M$ for each class.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) \\ 40 - 49 & 1 & 44.5 \\ 50 - 59 & 2 & 54.5 \\ 60 - 69 & 3 & 64.5 \\ 70 - 79 & 4 & 74.5 \\ 80 - 89 & 6 & 84.5 \\ 90 - 99 & 4 & 94.5 \end{array}$

Step 3

Multiply the frequency of each class by the class midpoint.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 40 - 49 & 1 & 44.5 & 1 \cdot 44.5 \\ 50 - 59 & 2 & 54.5 & 2 \cdot 54.5 \\ 60 - 69 & 3 & 64.5 & 3 \cdot 64.5 \\ 70 - 79 & 4 & 74.5 & 4 \cdot 74.5 \\ 80 - 89 & 6 & 84.5 & 6 \cdot 84.5 \\ 90 - 99 & 4 & 94.5 & 4 \cdot 94.5 \end{array}$

Step 4

Simplify the $f \cdot M$ column.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 40 - 49 & 1 & 44.5 & 44.5 \\ 50 - 59 & 2 & 54.5 & 109 \\ 60 - 69 & 3 & 64.5 & 193.5 \\ 70 - 79 & 4 & 74.5 & 298 \\ 80 - 89 & 6 & 84.5 & 507 \\ 90 - 99 & 4 & 94.5 & 378 \end{array}$

Step 5

Add the values in the $f \cdot M$ column.

$44.5 + 109 + 193.5 + 298 + 507 + 378 = 1530$

Step 6

Add the values in the frequency column.

$n = 1 + 2 + 3 + 4 + 6 + 4 = 20$

Step 7

The mean (mu) is the sum of $f \cdot M$ divided by $n$ , which is the sum of frequencies.

$μ = \frac{\sum_{}^{} f \cdot M}{\sum_{}^{} f}$

Step 8

The mean is the sum of the product of the midpoints and frequencies divided by the total of frequencies.

$μ = \frac{1530}{20}$

Step 9

Simplify the right side of $μ = \frac{1530}{20}$ .

$76.5$