From a given z-score, what is the formula for calculating the probability of the deviation being due to chance?
e.g.
Given z = 1.2 standard deviations
The chance probability p = 0.11507
How did we derive p? (i.e. the formula, without having to refer to the normal distribution table)
The probability $p$ is given by $$p=\int_{1.2}^\infty \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\, dx.$$ Equivalently, it is $\dfrac{1}{2}$ minus the integral from $0$ to $1.2$ of the same thing.
The function we are integrating is the probability density function of the standard normal. Since $e^{-x^2/2}$ does not have an antiderivative that is expressible in closed form in terms of elementary functions, the definite integral has to be calculated in another way. The calculations are not simple, so an old-fashioned solution was to prepare a table.