Can you bound exponentially the probability of a normal variable based on the z-score?

Question

That is, is there a bound $f(z_a)$ for $P(X > a)$ if the z-score of a is $z_a$?

Is there such a bound that decreases exponentially?

Something like $P(X > a) < c^{z_a}$?

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Here, a z-score is defined as:

z = (x-μ)/σ

where x is the raw value, μ is the population mean, and σ is the population standard deviation

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I know there are tables for calculating $P(X > a)$ given a z-score, but they don't help understand what is the function that links z-score and probability

Answer 1

This should be a comment but is too long due the links.

Maybe what you are looking for as bound for the probabilities are Concentration inequalities, which are many like the famous Chebyshev's or Markov's inequalities. But in the case of independent random variables, you can use the Chernoff bound which in the case of Normal distributed random variables $X\sim N(0,\sigma^2) $ (since you asked about z-scores), is indeed and exponential kind of bound with the shape: $$P(X<a)\leq e^{-\frac{a^2}{2\sigma^2}}$$