[求助]一个关于不确定度的问题 [复制链接]

质量工程师的题目,请指教
请问06质量工程师专业知识的最后一条题怎么做的!
第一个分量,第二个分量都没问题,就是第三个分量,
题目是:分散区间的宽度为60正态分布,置信水平p=99.73%,问不确定度是?
99.73%:k=3
我们单位几个人都做了,有的是和答案中一样,用60/3=20
有的就认为答案错了,应该用半宽60/2=30,再用30/3=10
请问哪个对?大家请指教一下.

Normal distribution, also know as the "bell-shaped curve". A population is the entire set of observations from which the sample, a subset, is drawn.

Your question is a Norminal question,

(Hints: 50% at + 0.67 sigma, 68.3% at + 1 sigma, 99.7% at +3 sigma, > 99.999999% at + 6 sigma.

Therefore,

60/99.73% or 60/3 sigma = 20 (Ans.)

Remark: It is suggested to post your question to Six Sigma Room.

原帖由1楼楼主 nestor 于2008-06-17 20:30发表
Normal distribution, also know as the "bell-shaped curve". A population is the entire set of observations from which the sample, a subset, is drawn.
Your question is a Norminal question,
(Hints: 50% at + 0.67 sigma, 68.3% at + 1 sigma, 99.7% at +3 sigma, > 99.999999% at + 6 sigma.
.......

正常分配，也知道作为“钟形曲线”。 人口是样品，子集，得出的整个套观察。

您的问题是Norminal问题，
(提示： 50%在+ 0.67斯格码， 68.3%在+ 1个斯格码， 99.7%在+3斯格码， > 99.999999%在+ 6斯格码。)

所以，

60/99.73% 或 60/3斯格码= 20 (答复。)


陈述： 它被建议张贴您的问题对六斯格码室。