Finite Math Examples

$\begin{array}{c} Class & Frequency \\ 0 - 9 & 2 \\ 10 - 19 & 4 \\ 20 - 29 & 9 \\ 30 - 39 & 8 \\ 40 - 49 & 2 \\ 50 - 59 & 1 \end{array}$

Step 1

Find the midpoint

$M$ for each class.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) \\ 0 - 9 & 2 & 4.5 \\ 10 - 19 & 4 & 14.5 \\ 20 - 29 & 9 & 24.5 \\ 30 - 39 & 8 & 34.5 \\ 40 - 49 & 2 & 44.5 \\ 50 - 59 & 1 & 54.5 \end{array}$

Step 2

Multiply the frequency of each class by the class midpoint.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 0 - 9 & 2 & 4.5 & 2 \cdot 4.5 \\ 10 - 19 & 4 & 14.5 & 4 \cdot 14.5 \\ 20 - 29 & 9 & 24.5 & 9 \cdot 24.5 \\ 30 - 39 & 8 & 34.5 & 8 \cdot 34.5 \\ 40 - 49 & 2 & 44.5 & 2 \cdot 44.5 \\ 50 - 59 & 1 & 54.5 & 1 \cdot 54.5 \end{array}$

Step 3

Simplify the

$f \cdot M$ column.

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 0 - 9 & 2 & 4.5 & 9 \\ 10 - 19 & 4 & 14.5 & 58 \\ 20 - 29 & 9 & 24.5 & 220.5 \\ 30 - 39 & 8 & 34.5 & 276 \\ 40 - 49 & 2 & 44.5 & 89 \\ 50 - 59 & 1 & 54.5 & 54.5 \end{array}$

Step 4

Add the values in the

$f \cdot M$ column.

$9 + 58 + 220.5 + 276 + 89 + 54.5 = 707$

Step 5

Add the values in the frequency column.

$n = 2 + 4 + 9 + 8 + 2 + 1 = 26$

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.

$μ = \frac{707}{26}$

Step 8

Simplify the right side of

$μ = \frac{707}{26}$ .

$27.19230769$