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有限数学 示例
ClassFrequency90-99480-89670-79460-69350-59240-491ClassFrequency90−99480−89670−79460−69350−59240−491
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。
ClassFrequency(f)40-49150-59260-69370-79480-89690-994ClassFrequency(f)40−49150−59260−69370−79480−89690−994
解题步骤 2
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。
ClassFrequency(f)LowerLimitsUpperLimits40-491404950-592505960-693606970-794707980-896808990-9949099ClassFrequency(f)LowerLimitsUpperLimits40−491404950−592505960−693606970−794707980−896808990−9949099
解题步骤 2.2
组中点为组下限加组上限除以 22。
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40-491404940+49250-592505950+59260-693606960+69270-794707970+79280-896808980+89290-994909990+992ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40−491404940+49250−592505950+59260−693606960+69270−794707970+79280−896808980+89290−994909990+992
解题步骤 2.3
化简所有中点列。
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40-491404944.550-592505954.560-693606964.570-794707974.580-896808984.590-994909994.5ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)40−491404944.550−592505954.560−693606964.570−794707974.580−896808984.590−994909994.5
解题步骤 2.4
在原始表格中增加一列,列出中值。
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)40−49144.550−59254.560−69364.570−79474.580−89684.590−99494.5
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)40−49144.550−59254.560−69364.570−79474.580−89684.590−99494.5
解题步骤 3
计算各组中点的平方 M2M2。
ClassFrequency(f)Midpoint(M)M240-49144.544.5250-59254.554.5260-69364.564.5270-79474.574.5280-89684.584.5290-99494.594.52ClassFrequency(f)Midpoint(M)M240−49144.544.5250−59254.554.5260−69364.564.5270−79474.574.5280−89684.584.5290−99494.594.52
解题步骤 4
化简 M2M2 列。
ClassFrequency(f)Midpoint(M)M240-49144.51980.2550-59254.52970.2560-69364.54160.2570-79474.55550.2580-89684.57140.2590-99494.58930.25ClassFrequency(f)Midpoint(M)M240−49144.51980.2550−59254.52970.2560−69364.54160.2570−79474.55550.2580−89684.57140.2590−99494.58930.25
解题步骤 5
将每一中点的平方乘以其频率 ff。
ClassFrequency(f)Midpoint(M)M2f⋅M240-49144.51980.251⋅1980.2550-59254.52970.252⋅2970.2560-69364.54160.253⋅4160.2570-79474.55550.254⋅5550.2580-89684.57140.256⋅7140.2590-99494.58930.254⋅8930.25ClassFrequency(f)Midpoint(M)M2f⋅M240−49144.51980.251⋅1980.2550−59254.52970.252⋅2970.2560−69364.54160.253⋅4160.2570−79474.55550.254⋅5550.2580−89684.57140.256⋅7140.2590−99494.58930.254⋅8930.25
解题步骤 6
化简 f⋅M2f⋅M2 列。
ClassFrequency(f)Midpoint(M)M2f⋅M240-49144.51980.251980.2550-59254.52970.255940.560-69364.54160.2512480.7570-79474.55550.252220180-89684.57140.2542841.590-99494.58930.2535721ClassFrequency(f)Midpoint(M)M2f⋅M240−49144.51980.251980.2550−59254.52970.255940.560−69364.54160.2512480.7570−79474.55550.252220180−89684.57140.2542841.590−99494.58930.2535721
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=20n=1,2,3,4,6,4=20。
∑f=n=20∑f=n=20
解题步骤 8
求 f⋅M2f⋅M2 列的和。在本例中,即 1980.25+5940.5+12480.75+22201+42841.5+35721=1211651980.25+5940.5+12480.75+22201+42841.5+35721=121165。
∑f⋅M2=121165∑f⋅M2=121165
解题步骤 9
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.
ClassFrequency(f)40-49150-59260-69370-79480-89690-994ClassFrequency(f)40−49150−59260−69370−79480−89690−994
解题步骤 9.2
求每一组的中点 MM。
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)40−49144.550−59254.560−69364.570−79474.580−89684.590−99494.5
解题步骤 9.3
将每组频率乘以组中值。
ClassFrequency(f)Midpoint(M)f⋅M40-49144.51⋅44.550-59254.52⋅54.560-69364.53⋅64.570-79474.54⋅74.580-89684.56⋅84.590-99494.54⋅94.5ClassFrequency(f)Midpoint(M)f⋅M40−49144.51⋅44.550−59254.52⋅54.560−69364.53⋅64.570−79474.54⋅74.580−89684.56⋅84.590−99494.54⋅94.5
解题步骤 9.4
化简 f⋅Mf⋅M 列。
ClassFrequency(f)Midpoint(M)f⋅M40-49144.544.550-59254.510960-69364.5193.570-79474.529880-89684.550790-99494.5378ClassFrequency(f)Midpoint(M)f⋅M40−49144.544.550−59254.510960−69364.5193.570−79474.529880−89684.550790−99494.5378
解题步骤 9.5
将 f⋅M 列中的值相加。
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
解题步骤 9.7
均值 (mu) 为 f⋅M 的和除以 n,即为频率的和。
μ=∑f⋅M∑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=153020
解题步骤 9.9
化简 μ=153020 的右边。
76.5
76.5
解题步骤 10
标准差方程为 S2=∑f⋅M2-n(μ)2n-1。
S2=∑f⋅M2-n(μ)2n-1
解题步骤 11
将计算值代入 S2=∑f⋅M2-n(μ)2n-1。
S2=121165-20(76.5)220-1
解题步骤 12
化简 S2=121165-20(76.5)220-1 的右边以得出方差 S2=216.84210526。
216.84210526