Finite Math Examples

\begin{array}{c} Class & Frequency \\ 2 - 10 & 1 \\ 11 - 19 & 3 \\ 20 - 28 & 9 \end{array}

$\begin{array}{c} Class & Frequency \\ 2 - 10 & 1 \\ 11 - 19 & 3 \\ 20 - 28 & 9 \end{array}$

Step 1

Find the midpoint

M

$M$ for each class.

\begin{array}{c} Class & Frequency (f) & Midpoint (M) \\ 2 - 10 & 1 & 6 \\ 11 - 19 & 3 & 15 \\ 20 - 28 & 9 & 24 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) \\ 2 - 10 & 1 & 6 \\ 11 - 19 & 3 & 15 \\ 20 - 28 & 9 & 24 \end{array}$

Step 2

Multiply the frequency of each class by the class midpoint.

\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 2 - 10 & 1 & 6 & 1 \cdot 6 \\ 11 - 19 & 3 & 15 & 3 \cdot 15 \\ 20 - 28 & 9 & 24 & 9 \cdot 24 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 2 - 10 & 1 & 6 & 1 \cdot 6 \\ 11 - 19 & 3 & 15 & 3 \cdot 15 \\ 20 - 28 & 9 & 24 & 9 \cdot 24 \end{array}$

Step 3

Simplify the

f \cdot M

$f \cdot M$ column.

\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 2 - 10 & 1 & 6 & 6 \\ 11 - 19 & 3 & 15 & 45 \\ 20 - 28 & 9 & 24 & 216 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & f \cdot M \\ 2 - 10 & 1 & 6 & 6 \\ 11 - 19 & 3 & 15 & 45 \\ 20 - 28 & 9 & 24 & 216 \end{array}$

Step 4

Add the values in the

f \cdot M

$f \cdot M$ column.

6 + 45 + 216 = 267

$6 + 45 + 216 = 267$

Step 5

Add the values in the frequency column.

n = 1 + 3 + 9 = 13

$n = 1 + 3 + 9 = 13$

Step 6

The mean (mu) is the sum of

f \cdot M

$f \cdot M$ divided by

n

$n$ , which is the sum of frequencies.

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

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

Step 7

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

μ = \frac{267}{13}

$μ = \frac{267}{13}$

Step 8

Simplify the right side of

μ = \frac{267}{13}

$μ = \frac{267}{13}$ .

20.53846153

$20.53846153$

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