PPRuNe Forums > Other Aircrew Forums > Flight Testing > CEP assessment PDA View Full Version : CEP assessment peeush14th Apr 2011, 08:10This is the CEP related doubt that I have. Going by the definition of CEP i.e. if CEP is n meters, 50% of rounds land within n meters of the target, 43% between n and 2n, and 7 % between 2n and 3n meters, and the proportion of rounds that land farther than three times the CEP from the target is less than 0.2%. Now to say for example, if the min and max error in a measurement is 0% and 2% respectively, then value of 3n=2% (ignoring the area for 0.2% of the rounds discussed above). If that be so then n=0.67%, and hence it may be correct to say that in 50% of the measurements made there is a likelihood of 0.67% error, for 93% of the measurements made (within '2n' metres: i.e. including the n and the 2n areas ;50%+43%) there is a likelihood of 1.34% error (2n value) and for 100% of the measurements a probability of 2% error (within 3n area) exists. Well this is where I could reach given my very basic understanding in mathematics. Would request the august audience to guide me for corrections. Thanks Matthew Parsons15th Apr 2011, 16:43peeush, I don't see the flight test relevance to your question. It seems like you're trying to use a normal distribution on measurement errors, by assuming that the maximum range of measurement errors is equivalent to three standard deviations, and from there deducing that half of your measurements have a measurement error of one third of the maximum. For that to work you need to identify all sources of measurement error and confirm that each is error is idendependant of all others. Once you've done that, you can claim a probability that 50% of the measurements are within n/3, but you cannot state that it is so. On average 50% of coin flips will turn up heads, that doesn't mean you won't ever see 10 heads in 10 flips. Matthew. peeush16th Apr 2011, 04:23Matthew, Thanks for the comments, I would like to confirm the following:- 1. The errors taken into calculations are independent of each other. 2. It is understood that there would only be a probability of n/3 errors occuring in 50% of the cases and 2n/3 in 93% of the cases. Given the above statements, it would imply that if I have a max error of 2% in system, for 50% of the cases I may expect a probability of 0.67% error, for 93% of the cases 1.34% error and so on (in accordance with the CEP definition stated in the original post) Thats what I finally intended to confirm. Thanks