UC知道英语高手能帮我翻译下吗？跪求！！！
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a certain probability. It is based on a 99 % confidence interval for the unknown true
overall
resistance factor. The t5;0,995-quantile here indicates, that the 99 % confidence interval
6
5;0,995
k ± t × s will cover the unknown true overall resistance factor of connector i with a
probability of 99 %. It does not imply, however, that all mean resistance factors ki. will
be
covered by this confidence interval with a probability of 99 %.
For this quantity, the assumption can be made that the empirical means ki. for all
connectors i independently and identically follow a normal distribution.
Figure F.3 - Plot of estimated mean resistance factors ki. ,
estimated overall mean k and parameter a
Figure F.3 shows the estimated mean resistance factors ki. (plotted using a dotted line) of
the
six connectors (labelled as C1 to C6) as well as the estimated overall mean k (horizontal
solid
overall
resistance factor. The t5;0,995-quantile here indicates, that the 99 % confidence interval
6
5;0,995
k ± t × s will cover the unknown true overall resistance factor of connector i with a
probability of 99 %. It does not imply, however, that all mean resistance factors ki. will
be
covered by this confidence interval with a probability of 99 %.
For this quantity, the assumption can be made that the empirical means ki. for all
connectors i independently and identically follow a normal distribution.
Figure F.3 - Plot of estimated mean resistance factors ki. ,
estimated overall mean k and parameter a
Figure F.3 shows the estimated mean resistance factors ki. (plotted using a dotted line) of
the
six connectors (labelled as C1 to C6) as well as the estimated overall mean k (horizontal
solid
有一定的几率。它是基于一个99%的置信区间为未知的真实
整体
阻力的因素。这里显示了;0,995-quantile胸椎,99%的置信区间
6
0,995 5。
凯西±吨×将覆盖了未知电阻系数的真实全面连接器的一家
概率的99%。这并不意味着,然而,这意味着ki.阻力因子
被
被这个置信区间的可能性99%。
对这批数量大,我们可以用它假定的实证方法对所有ki.
我独立和同遵循连接器的正态分布。
图F.3 -块估计平均阻力因子,吻。
总体平均钾和参数估计
图F.3显示大约平均阻力因子的吻。用虚线(绘制的
这个
六接头(贴上碳到C6)以及估计整体意味着k(水平
固体
在一定的概率。它是根据99 ％置信区间为未知的真实
全部的
阻力系数。在T5荧光灯; 0995 -位数在这里表示，该99 ％置信区间
6
5 ; 0995
k ±吨× s将涵盖未知真正全面阻力系数的连接器一1
概率99 ％ 。这并不意味着，但是，所有的意思阻力因素棋。意志
是
涉及本置信区间的概率为99 ％ 。
这一数量，假设可以说，实证手段棋。所有
连接器相同的字母i独立和遵循正态分布。
图中三-图估计平均阻力因素棋。 ，
估计总的意思K和参数a
图中三显示估计平均阻力因素棋。 （策划使用虚线）的
那个
6连接器（称为的C1到的C6 ）以及估计总体平均度（水平
固体