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 Author: lakhtin Time: 03/20/05 15:05 PST This is a reply to message no. 15592 by cessna Reply | Original Message | New Topic | List Topics | List Messages on This Topic
 Current Topic:Quality: -- Calculating Sigma from Mean and Std Deviations sir, you might have understood the basics given by cessna. when you see a normal distribution curve, it is a bell shaped curve you see two things: mean, which lies at the centre and the value you are looking for (i think you have called that value load). let mean be M and load be L then the "number of standard deviations" you are (the load) from M is z=[M-L]/sd, where sd is standard deviation. now that you have got the value of z, you can see in any normal distribution table to know the percentage of area that lies between M to z or less than or greater than z. now coming to 6 sigma, 6 sigma means that whatever you produce the chances that you would go wrong are just 3.34 in 1 million. the normal curve is divided into 6 equal parts, each part is 1 sd. the area of curve = area of 6 sd's or 6 sigma = 99.9997% this means if your process lies with in 6 sd's you would give 99.9997% accracy. regards lakhtin...
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