有限数学示例

\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}

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

解题步骤 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}

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

解题步骤 2

求每一组的中点

M

$M$ 。

\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}

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

解题步骤 3

将每组频率乘以组中值。

\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}

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

解题步骤 4

化简

f \cdot M

$f \cdot M$ 列。

\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}

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

解题步骤 5

将

f \cdot M

$f \cdot M$ 列中的值相加。

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

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

解题步骤 6

将频率列中的值相加。

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

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

解题步骤 7

均值 (mu) 为

f \cdot M

$f \cdot M$ 的和除以

n

$n$ ，即为频率的和。

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

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

解题步骤 8

平均值是中点和频率的乘积之和除以频率总和。

μ = \frac{1530}{20}

$μ = \frac{1530}{20}$

解题步骤 9

化简

μ = \frac{1530}{20}

$μ = \frac{1530}{20}$ 的右边。

76.5

$76.5$

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有限数学 示例

有限数学示例