bisection method - quantiles of the normal distribution

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I got stuck on this question, I can't understand it.

I need to solve $$ \Phi(a)=\int_{-\infty}^a\frac1{\sqrt{2\pi}}\exp\left(-\frac12x^2\right)dx=\alpha $$ by the bisection method, the purpose of the question is to calculate the 30th, 70th, and 95th percentages of a standard normal distribution.

In other words, in order to calculate the appropriate fracture the equation must be solved: $$ f(a)=\Phi(a)-\alpha=0 $$

I know the algorithm of the method, but I don't succeed to do it. Any explanation, or any help, will be great.

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Your teacher will not appreciate this answer, which is more appropriate than bisection.

See the problem as a differential equation

$$\frac{d\Phi}{dx}=\frac{e^{-x^2/2}}{\sqrt{2\pi}}$$ or

$$\frac{dx}{d\Phi}=\sqrt{2\pi}\,e^{x^2/2}=f(x,\Phi)$$ and integrate by Runge-Kutta (or similar) from $\Phi=\frac12$ to $\alpha$. (Starting from $\frac12$ is numerically safer, and we know that $\Phi(0)=\frac12$.)