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Given the particular normal distribution specified below, what is the probability that a random observation falls within the specified range .004 greater and less than the average?

original Lower limit is .496

original upper limit is .504

$$\mu = .500$$ defective equipment cause the average to change to $$\mu = .449$$

$$\sigma = .002$$

new Lower limit is .445

new upper limit is .453

So how do you compensate for your average changing because of defective equipment?


OLD info When I use my ti 89 normal cdf function I get 2.24 which isn't even a probability. Is there something special I need to do because my average changes or is there something special about these numbers?

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  • $\begingroup$ Are the numbers accurate? With the mean this small, even your lower limit is effectively impossible. 20 sigma event and all. $\endgroup$ – lulu Jul 7 '15 at 20:30
  • $\begingroup$ Well, maybe I am misunderstanding your question. I thought you were asking "given a mean of .449 and a standard deviation of .002 what is the probability that an observation falls between .496 and .504 ?". As I say, the answer to that is 0 to any sensible precision level. But maybe I have the wrong question in mind, or maybe there's a typo in the numbers. $\endgroup$ – lulu Jul 7 '15 at 20:37
  • $\begingroup$ @lulu Yes the numbers are accurate, but I did a poor job of explaining the question after reading out loud to myself several times. I am working on editing it to make more sense. Thank you for pointing it out. $\endgroup$ – rocker Jul 7 '15 at 20:50
  • $\begingroup$ I see the edit, but the sentence "I have a normal distribution that I am trying to calculate the probability it stays between my upper and lower limit." is still a bit unclear. do you mean "Given the particular normal distribution specified below, what is the probability that a random observation falls within the specified range?" $\endgroup$ – lulu Jul 7 '15 at 21:08
  • $\begingroup$ @lulu Yes :). Thank you. $\endgroup$ – rocker Jul 7 '15 at 21:13
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If $\mu=0.5$ and $\sigma=0.002$, $P(0.496<X<0.504)=P(\frac{0.496-0.5}{0.002}<Z<\frac{0.504-0.5}{0.002})=\Phi(2)-\Phi(-2)\approx 0.9545$, where $Z$ is the standard normal r.v. and $\Phi(\cdot)$ its cumulative density function.

If $\mu=0.499$ and $\sigma=0.002$, $P(0.496<X<0.504)=P(\frac{0.495-0.499}{0.002}<Z<\frac{0.503-0.499}{0.002})=\Phi(2)-\Phi(-2)\approx 0.9545$.

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