Finite Math Examples

$\begin{array}{c} x & y \\ 1 - 10 & 8 \\ 11 - 20 & 7 \\ 21 - 30 & 9 \\ 31 - 40 & 6 \end{array}$

Step 1

Find the midpoint $M$ for each class.

$\begin{array}{c} x & y & Midpoint (M) \\ 1 - 10 & 8 & 5.5 \\ 11 - 20 & 7 & 15.5 \\ 21 - 30 & 9 & 25.5 \\ 31 - 40 & 6 & 35.5 \end{array}$

Step 2

Multiply the frequency of each class by the class midpoint.

$\begin{array}{c} x & y & Midpoint (M) & f \cdot M \\ 1 - 10 & 8 & 5.5 & 8 \cdot 5.5 \\ 11 - 20 & 7 & 15.5 & 7 \cdot 15.5 \\ 21 - 30 & 9 & 25.5 & 9 \cdot 25.5 \\ 31 - 40 & 6 & 35.5 & 6 \cdot 35.5 \end{array}$

Step 3

Simplify the $f \cdot M$ column.

$\begin{array}{c} x & y & Midpoint (M) & f \cdot M \\ 1 - 10 & 8 & 5.5 & 44 \\ 11 - 20 & 7 & 15.5 & 108.5 \\ 21 - 30 & 9 & 25.5 & 229.5 \\ 31 - 40 & 6 & 35.5 & 213 \end{array}$

Step 4

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

$44 + 108.5 + 229.5 + 213 = 595$

Step 5

Add the values in the frequency column.

$n = 8 + 7 + 9 + 6 = 30$

Step 6

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 7

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

$μ = \frac{595}{30}$

Step 8

Simplify the right side of $μ = \frac{595}{30}$ .

$19.8\overline{3}$