What is the Approximated Value of \Phi(x) for x>3.5 in a Normal Distribution?

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The approximated value of \Phi(x) for a normal distribution when x > 3.5 is very close to 1, but never actually reaches it, as \Phi(x) remains less than 1 for any finite x. For instance, \Phi(4) is approximately 0.9999683 and \Phi(5) is about 0.9999997. The values approach 1 closely enough that the difference is negligible in practical applications. Additionally, the claim that \Phi(x) could be 0.5 is incorrect, as this value only occurs at x = 0. Thus, for x > 3.5, \Phi(x) is effectively treated as 1 in most scenarios.
DamjanMk
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Hi,
I'd like to know the value that \Phi(x) of a normal distribution is approximated when x> 3,5.
I assume it is 1, since the value for x=3,49 is 0,9998...
But I got some answers at the university that it might be 0,5

Thanks
 
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There is no finite value of for which it is 1: that is, if you calculate precisely, it will always be the case that \Phi(x) < 1 for a finite number x. For example, \Phi(4) \approx 0.9999683 and \Phi(5) \approx 0.9999997 (calculation from R software). However, just like my two examples, the values essentially get so close to 1 that the difference is, for almost all situations, immaterial.

For your second comment ("But I got some answers at the university that it might be 0,5") - if you mean 0.5, that is definitely false: the only value that gives \Phi(x) = 0.5 is
x = 0.
 
Thank you :)
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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