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

按其相关频率的升序（从低到高）将各组重新排序，这是最常见的排序方式。

\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

$M$ 。

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解题步骤 2.1

每组的下限为该组的最小值。另外，每组的上限为该组的最大值。

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

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

解题步骤 2.2

组中点为组下限加组上限除以

2

$2$ 。

\begin{array}{c} Class & Frequency (f) & Lower Limits & Upper Limits & Midpoint (M) \\ 40 - 49 & 1 & 40 & 49 & \frac{40 + 49}{2} \\ 50 - 59 & 2 & 50 & 59 & \frac{50 + 59}{2} \\ 60 - 69 & 3 & 60 & 69 & \frac{60 + 69}{2} \\ 70 - 79 & 4 & 70 & 79 & \frac{70 + 79}{2} \\ 80 - 89 & 6 & 80 & 89 & \frac{80 + 89}{2} \\ 90 - 99 & 4 & 90 & 99 & \frac{90 + 99}{2} \end{array}

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

解题步骤 2.3

化简所有中点列。

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

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

解题步骤 2.4

在原始表格中增加一列，列出中值。

\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}

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

\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

计算各组中点的平方

M^{2}

$M^{2}$ 。

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

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

解题步骤 4

化简

M^{2}

$M^{2}$ 列。

\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 \\ 60 - 69 & 3 & 64.5 & 4160.25 \\ 70 - 79 & 4 & 74.5 & 5550.25 \\ 80 - 89 & 6 & 84.5 & 7140.25 \\ 90 - 99 & 4 & 94.5 & 8930.25 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 \\ 60 - 69 & 3 & 64.5 & 4160.25 \\ 70 - 79 & 4 & 74.5 & 5550.25 \\ 80 - 89 & 6 & 84.5 & 7140.25 \\ 90 - 99 & 4 & 94.5 & 8930.25 \end{array}$

解题步骤 5

将每一中点的平方乘以其频率

f

$f$ 。

\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} & f \cdot M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 & 1 \cdot 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 & 2 \cdot 2970.25 \\ 60 - 69 & 3 & 64.5 & 4160.25 & 3 \cdot 4160.25 \\ 70 - 79 & 4 & 74.5 & 5550.25 & 4 \cdot 5550.25 \\ 80 - 89 & 6 & 84.5 & 7140.25 & 6 \cdot 7140.25 \\ 90 - 99 & 4 & 94.5 & 8930.25 & 4 \cdot 8930.25 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} & f \cdot M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 & 1 \cdot 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 & 2 \cdot 2970.25 \\ 60 - 69 & 3 & 64.5 & 4160.25 & 3 \cdot 4160.25 \\ 70 - 79 & 4 & 74.5 & 5550.25 & 4 \cdot 5550.25 \\ 80 - 89 & 6 & 84.5 & 7140.25 & 6 \cdot 7140.25 \\ 90 - 99 & 4 & 94.5 & 8930.25 & 4 \cdot 8930.25 \end{array}$

解题步骤 6

化简

f \cdot M^{2}

$f \cdot M^{2}$ 列。

\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} & f \cdot M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 & 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 & 5940.5 \\ 60 - 69 & 3 & 64.5 & 4160.25 & 12480.75 \\ 70 - 79 & 4 & 74.5 & 5550.25 & 22201 \\ 80 - 89 & 6 & 84.5 & 7140.25 & 42841.5 \\ 90 - 99 & 4 & 94.5 & 8930.25 & 35721 \end{array}

$\begin{array}{c} Class & Frequency (f) & Midpoint (M) & M^{2} & f \cdot M^{2} \\ 40 - 49 & 1 & 44.5 & 1980.25 & 1980.25 \\ 50 - 59 & 2 & 54.5 & 2970.25 & 5940.5 \\ 60 - 69 & 3 & 64.5 & 4160.25 & 12480.75 \\ 70 - 79 & 4 & 74.5 & 5550.25 & 22201 \\ 80 - 89 & 6 & 84.5 & 7140.25 & 42841.5 \\ 90 - 99 & 4 & 94.5 & 8930.25 & 35721 \end{array}$

解题步骤 7

求所有频率的和。在本例中，所有频率的和为

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

$n = 1, 2, 3, 4, 6, 4 = 20$ 。

\sum_{}^{} f = n = 20

$\sum_{}^{} f = n = 20$

解题步骤 8

求

f \cdot M^{2}

$f \cdot M^{2}$ 列的和。在本例中，即

1980.25 + 5940.5 + 12480.75 + 22201 + 42841.5 + 35721 = 121165

$1980.25 + 5940.5 + 12480.75 + 22201 + 42841.5 + 35721 = 121165$ 。

\sum_{}^{} f \cdot M^{2} = 121165

$\sum_{}^{} f \cdot M^{2} = 121165$

解题步骤 9

求平均值

μ

$μ$ 。

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解题步骤 9.1

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 9.2

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 9.3

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 9.4

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 9.5

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 9.6

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 9.7

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 9.8

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 9.9

化简

μ = \frac{1530}{20}

$μ = \frac{1530}{20}$ 的右边。

76.5

$76.5$

76.5

$76.5$

解题步骤 10

标准差方程为

S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}

$S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}$ 。

S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}

$S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}$

解题步骤 11

将计算值代入

S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}

$S^{2} = \frac{\sum_{}^{} f \cdot M^{2} - n {(μ)}^{2}}{n - 1}$ 。

S^{2} = \frac{121165 - 20 {(76.5)}^{2}}{20 - 1}

$S^{2} = \frac{121165 - 20 {(76.5)}^{2}}{20 - 1}$

解题步骤 12

化简

S^{2} = \frac{121165 - 20 {(76.5)}^{2}}{20 - 1}

$S^{2} = \frac{121165 - 20 {(76.5)}^{2}}{20 - 1}$ 的右边以得出方差

S^{2} = 216.84210526

$S^{2} = 216.84210526$ 。

216.84210526

$216.84210526$

有限数学 示例

有限数学示例