Finite Math Examples

$\begin{array}{c} x & y \\ 0.5 - 2.5 & 5 \\ 2.5 - 4.5 & 8 \\ 4.5 - 6.5 & 2 \\ 6.5 - 8.5 & 8 \end{array}$

Step 1

Find the midpoint $M$ for each class.

$\begin{array}{c} x & y & Midpoint (M) \\ 0.5 - 2.5 & 5 & 1.5 \\ 2.5 - 4.5 & 8 & 3.5 \\ 4.5 - 6.5 & 2 & 5.5 \\ 6.5 - 8.5 & 8 & 7.5 \end{array}$

Step 2

Multiply the frequency of each class by the class midpoint.

$\begin{array}{c} x & y & Midpoint (M) & f \cdot M \\ 0.5 - 2.5 & 5 & 1.5 & 5 \cdot 1.5 \\ 2.5 - 4.5 & 8 & 3.5 & 8 \cdot 3.5 \\ 4.5 - 6.5 & 2 & 5.5 & 2 \cdot 5.5 \\ 6.5 - 8.5 & 8 & 7.5 & 8 \cdot 7.5 \end{array}$

Step 3

Simplify the $f \cdot M$ column.

$\begin{array}{c} x & y & Midpoint (M) & f \cdot M \\ 0.5 - 2.5 & 5 & 1.5 & 7.5 \\ 2.5 - 4.5 & 8 & 3.5 & 28 \\ 4.5 - 6.5 & 2 & 5.5 & 11 \\ 6.5 - 8.5 & 8 & 7.5 & 60 \end{array}$

Step 4

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

$7.5 + 28 + 11 + 60 = 106.5$

Step 5

Add the values in the frequency column.

$n = 5 + 8 + 2 + 8 = 23$

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.

$μ = \frac{106.5}{23}$

Step 8

Simplify the right side of $μ = \frac{106.5}{23}$ .

$4.63043478$