有限数学 示例

求频率表的方差 table[[Class,Frequency],[90-99,4],[80-89,6],[70-79,4],[60-69,3],[50-59,2],[40-49,1]]
ClassFrequency90-99480-89670-79460-69350-59240-491ClassFrequency909948089670794606935059240491
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)404915059260693707948089690994
解题步骤 2
求每一组的中点 MM
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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。
ClassFrequency(f)LowerLimitsUpperLimits40-491404950-592505960-693606970-794707980-896808990-9949099ClassFrequency(f)LowerLimitsUpperLimits404914049505925059606936069707947079808968089909949099
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)40491404940+49250592505950+59260693606960+69270794707970+79280896808980+89290994909990+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)40491404944.550592505954.560693606964.570794707974.580896808984.590994909994.5
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)4049144.55059254.56069364.57079474.58089684.59099494.5
ClassFrequency(f)Midpoint(M)40-49144.550-59254.560-69364.570-79474.580-89684.590-99494.5ClassFrequency(f)Midpoint(M)4049144.55059254.56069364.57079474.58089684.59099494.5
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)M24049144.544.525059254.554.526069364.564.527079474.574.528089684.584.529099494.594.52
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)M24049144.51980.255059254.52970.256069364.54160.257079474.55550.258089684.57140.259099494.58930.25
解题步骤 5
将每一中点的平方乘以其频率 ff
ClassFrequency(f)Midpoint(M)M2fM240-49144.51980.2511980.2550-59254.52970.2522970.2560-69364.54160.2534160.2570-79474.55550.2545550.2580-89684.57140.2567140.2590-99494.58930.2548930.25ClassFrequency(f)Midpoint(M)M2fM24049144.51980.2511980.255059254.52970.2522970.256069364.54160.2534160.257079474.55550.2545550.258089684.57140.2567140.259099494.58930.2548930.25
解题步骤 6
化简 fM2fM2 列。
ClassFrequency(f)Midpoint(M)M2fM240-49144.51980.251980.2550-59254.52970.255940.560-69364.54160.2512480.7570-79474.55550.252220180-89684.57140.2542841.590-99494.58930.2535721ClassFrequency(f)Midpoint(M)M2fM24049144.51980.251980.255059254.52970.255940.56069364.54160.2512480.757079474.55550.25222018089684.57140.2542841.59099494.58930.2535721
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=20f=n=20
解题步骤 8
fM2fM2 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
fM2=121165fM2=121165
解题步骤 9
求平均值 μμ
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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.
ClassFrequency(f)40-49150-59260-69370-79480-89690-994ClassFrequency(f)404915059260693707948089690994
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)4049144.55059254.56069364.57079474.58089684.59099494.5
解题步骤 9.3
将每组频率乘以组中值。
ClassFrequency(f)Midpoint(M)fM40-49144.5144.550-59254.5254.560-69364.5364.570-79474.5474.580-89684.5684.590-99494.5494.5ClassFrequency(f)Midpoint(M)fM4049144.5144.55059254.5254.56069364.5364.57079474.5474.58089684.5684.59099494.5494.5
解题步骤 9.4
化简 fMfM 列。
ClassFrequency(f)Midpoint(M)fM40-49144.544.550-59254.510960-69364.5193.570-79474.529880-89684.550790-99494.5378ClassFrequency(f)Midpoint(M)fM4049144.544.55059254.51096069364.5193.57079474.52988089684.55079099494.5378
解题步骤 9.5
fM 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 fM 的和除以 n,即为频率的和。
μ=fMf
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=fM2-n(μ)2n-1
S2=fM2-n(μ)2n-1
解题步骤 11
将计算值代入 S2=fM2-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
 [x2  12  π  xdx ]