统计学 示例

求频数分布表的方差
ClassFrequency19.55-21.82321.83-24.1524.11-26.38926.39-28.66628.67-30.942
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)LowerLimitsUpperLimits19.55-21.82319.5521.8221.83-24.1521.8324.124.11-26.38924.1126.3826.39-28.66626.3928.6628.67-30.94228.6730.94
解题步骤 1.2
组中点为组下限加组上限除以 2
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)19.55-21.82319.5521.8219.55+21.82221.83-24.1521.8324.121.83+24.1224.11-26.38924.1126.3824.11+26.38226.39-28.66626.3928.6626.39+28.66228.67-30.94228.6730.9428.67+30.942
解题步骤 1.3
化简所有中点列。
ClassFrequency(f)LowerLimitsUpperLimitsMidpoint(M)19.55-21.82319.5521.8220.68521.83-24.1521.8324.122.96524.11-26.38924.1126.3825.24526.39-28.66626.3928.6627.52528.67-30.94228.6730.9429.805
解题步骤 1.4
在原始表格中增加一列,列出中值。
ClassFrequency(f)Midpoint(M)19.55-21.82320.68521.83-24.1522.96524.11-26.38925.24526.39-28.66627.52528.67-30.94229.805
ClassFrequency(f)Midpoint(M)19.55-21.82320.68521.83-24.1522.96524.11-26.38925.24526.39-28.66627.52528.67-30.94229.805
解题步骤 2
计算各组中点的平方 M2
ClassFrequency(f)Midpoint(M)M219.55-21.82320.68520.685221.83-24.1522.96522.965224.11-26.38925.24525.245226.39-28.66627.52527.525228.67-30.94229.80529.8052
解题步骤 3
化简 M2 列。
ClassFrequency(f)Midpoint(M)M219.55-21.82320.685427.86922521.83-24.1522.965527.39122524.11-26.38925.245637.31002526.39-28.66627.525757.625624928.67-30.94229.805888.338025
解题步骤 4
将每一中点的平方乘以其频率 f
ClassFrequency(f)Midpoint(M)M2fM219.55-21.82320.685427.8692253427.86922521.83-24.1522.965527.3912255527.39122524.11-26.38925.245637.3100259637.31002526.39-28.66627.525757.62562496757.625624928.67-30.94229.805888.3380252888.338025
解题步骤 5
化简 fM2 列。
ClassFrequency(f)Midpoint(M)M2fM219.55-21.82320.685427.8692251283.60767521.83-24.1522.965527.3912252636.956124924.11-26.38925.245637.3100255735.79022526.39-28.66627.525757.62562494545.7537528.67-30.94229.805888.3380251776.67605
解题步骤 6
求所有频率的和。在本例中,所有频率的和为 n=3,5,9,6,2=25
f=n=25
解题步骤 7
fM2 列的和。在本例中,即 1283.607675+2636.9561249+5735.790225+4545.75375+1776.67605=15978.783825
fM2=15978.783825
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)19.55-21.82320.68521.83-24.1522.96524.11-26.38925.24526.39-28.66627.52528.67-30.94229.805
解题步骤 8.2
将每组频率乘以组中值。
ClassFrequency(f)Midpoint(M)fM19.55-21.82320.685320.68521.83-24.1522.965522.96524.11-26.38925.245925.24526.39-28.66627.525627.52528.67-30.94229.805229.805
解题步骤 8.3
化简 fM 列。
ClassFrequency(f)Midpoint(M)fM19.55-21.82320.68562.054921.83-24.1522.965114.82524.11-26.38925.245227.20526.39-28.66627.525165.14928.67-30.94229.80559.61
解题步骤 8.4
fM 列中的值相加。
62.0549+114.825+227.205+165.149+59.61=628.845
解题步骤 8.5
将频率列中的值相加。
n=3+5+9+6+2=25
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=628.84525
解题步骤 8.8
化简 μ=628.84525 的右边。
25.1538
25.1538
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=15978.783825-25(25.1538)225-1
解题步骤 11
化简 S2=15978.783825-25(25.1538)225-1 的右边以得出方差 S2=6.7059359
6.7059359
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