有限数学 示例
组频率12-17318-23624-29430-352组频率12−17318−23624−29430−352
解题步骤 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)LowerLimitsUpperLimits12-173121718-236182324-294242930-3523035ClassFrequency(f)LowerLimitsUpperLimits12−173121718−236182324−294242930−3523035
解题步骤 2
组界是用来分隔不同组的数字。组与组之间的差距大小是一个组的上界限和下一个组的下界限之间的差。在本例中,即 间隔=18-17=1间隔=18−17=1。
间隔=1间隔=1
解题步骤 3
每组的下界可以通过从该组下限中减去间隙值 12=0.512=0.5 的二分一计算得出。另外,每组的上界可以通过该组上限加上间隙值 12=0.512=0.5 的二分一计算得出。
ClassFrequency(f)LowerLimitsLowerBoundariesUpperLimitsUpperBoundaries12-1731212-0.51717+0.518-2361818-0.52323+0.524-2942424-0.52929+0.530-3523030-0.53535+0.5ClassFrequency(f)LowerLimitsLowerBoundariesUpperLimitsUpperBoundaries12−1731212−0.51717+0.518−2361818−0.52323+0.524−2942424−0.52929+0.530−3523030−0.53535+0.5
解题步骤 4
化简下界列和上界列。
ClassFrequency(f)LowerLimitsLowerBoundariesUpperLimitsUpperBoundaries12-1731211.51717.518-2361817.52323.524-2942423.52929.530-3523029.53535.5ClassFrequency(f)LowerLimitsLowerBoundariesUpperLimitsUpperBoundaries12−1731211.51717.518−2361817.52323.524−2942423.52929.530−3523029.53535.5
解题步骤 5
在原始表格中增加两列,列出每一类的上限和下限。
ClassFrequency(f)LowerBoundariesUpperBoundaries12-17311.517.518-23617.523.524-29423.529.530-35229.535.5ClassFrequency(f)LowerBoundariesUpperBoundaries12−17311.517.518−23617.523.524−29423.529.530−35229.535.5