有限数学 示例

求频数分布表的方差
ClassFrequency360-3692370-3793380-3895390-3997400-4095410-4194420-4294430-4391
解题步骤 1
求每一组的中点 M
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解题步骤 1.1
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)LowerLimitsUpperLimits360-3692360369370-3793370379380-3895380389390-3997390399400-4095400409410-4194410419420-4294420429430-4391430439
解题步骤 1.2
组中点为组下限加组上限除以 2
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)360-3692360369360+3692370-3793370379370+3792380-3895380389380+3892390-3997390399390+3992400-4095400409400+4092410-4194410419410+4192420-4294420429420+4292430-4391430439430+4392
解题步骤 1.3
化简所有中点列。
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)360-3692360369364.5370-3793370379374.5380-3895380389384.5390-3997390399394.5400-4095400409404.5410-4194410419414.5420-4294420429424.5430-4391430439434.5
解题步骤 1.4
在原始表格中增加一列,列出中值。
ClassFrequency(f)Midpoint(M)360-3692364.5370-3793374.5380-3895384.5390-3997394.5400-4095404.5410-4194414.5420-4294424.5430-4391434.5
ClassFrequency(f)Midpoint(M)360-3692364.5370-3793374.5380-3895384.5390-3997394.5400-4095404.5410-4194414.5420-4294424.5430-4391434.5
解题步骤 2
计算各组中点的平方 M2
ClassFrequency(f)Midpoint(M)M2360-3692364.5364.52370-3793374.5374.52380-3895384.5384.52390-3997394.5394.52400-4095404.5404.52410-4194414.5414.52420-4294424.5424.52430-4391434.5434.52
解题步骤 3
化简 M2 列。
ClassFrequency(f)Midpoint(M)M2360-3692364.5132860.25370-3793374.5140250.25380-3895384.5147840.25390-3997394.5155630.25400-4095404.5163620.25410-4194414.5171810.25420-4294424.5180200.25430-4391434.5188790.25
解题步骤 4
将每一中点的平方乘以其频率 f
ClassFrequency(f)Midpoint(M)M2fM2360-3692364.5132860.252132860.25370-3793374.5140250.253140250.25380-3895384.5147840.255147840.25390-3997394.5155630.257155630.25400-4095404.5163620.255163620.25410-4194414.5171810.254171810.25420-4294424.5180200.254180200.25430-4391434.5188790.251188790.25
解题步骤 5
化简 fM2 列。
ClassFrequency(f)Midpoint(M)M2fM2360-3692364.5132860.25265720.5370-3793374.5140250.25420750.75380-3895384.5147840.25739201.25390-3997394.5155630.251089411.75400-4095404.5163620.25818101.25410-4194414.5171810.25687241420-4294424.5180200.25720801430-4391434.5188790.25188790.25
解题步骤 6
求所有频率的和。在本例中,所有频率的和为 n=2,3,5,7,5,4,4,1=31
f=n=31
解题步骤 7
fM2 列的和。在本例中,即 265720.5+420750.75+739201.25+1089411.75+818101.25+687241+720801+188790.25=4930017.75
fM2=4930017.75
解题步骤 8
求平均值 μ
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解题步骤 8.1
求每一组的中点 M
ClassFrequency(f)Midpoint(M)360-3692364.5370-3793374.5380-3895384.5390-3997394.5400-4095404.5410-4194414.5420-4294424.5430-4391434.5
解题步骤 8.2
将每组频率乘以组中值。
ClassFrequency(f)Midpoint(M)fM360-3692364.52364.5370-3793374.53374.5380-3895384.55384.5390-3997394.57394.5400-4095404.55404.5410-4194414.54414.5420-4294424.54424.5430-4391434.51434.5
解题步骤 8.3
化简 fM 列。
ClassFrequency(f)Midpoint(M)fM360-3692364.5729370-3793374.51123.5380-3895384.51922.5390-3997394.52761.5400-4095404.52022.5410-4194414.51658420-4294424.51698430-4391434.5434.5
解题步骤 8.4
fM 列中的值相加。
729+1123.5+1922.5+2761.5+2022.5+1658+1698+434.5=12349.5
解题步骤 8.5
将频率列中的值相加。
n=2+3+5+7+5+4+4+1=31
解题步骤 8.6
均值 (mu) 为 fM 的和除以 n,即为频率的和。
μ=fMf
解题步骤 8.7
平均值是中点和频率的乘积之和除以频率总和。
μ=12349.531
解题步骤 8.8
化简 μ=12349.531 的右边。
398.37096774
398.37096774
解题步骤 9
标准差方程为 S2=fM2-n(μ)2n-1
S2=fM2-n(μ)2n-1
解题步骤 10
将计算值代入 S2=fM2-n(μ)2n-1
S2=4930017.75-31(398.37096774)231-1
解题步骤 11
化简 S2=4930017.75-31(398.37096774)231-1 的右边以得出方差 S2=344.51612903
344.51612903
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