Consider the distribution function of the normal distribution, $$ F(x)=\int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi}} e^{-y^{2} / 2} d y $$ We want to calculate the $F(1.96)$ with a two decimal precision. Devise a method for doing so, and find the number of divisions needed using the Trapezoidal method.
$f(y)= \frac{1}{\sqrt{2 \pi}} e^{-y^{2} / 2}$ is symmetric, so from $-\infty$ to $0$ the integral is $1/2$. So we can just calculate $\int_0^xf(y)dy$. I now want to find the $n$ such that $|E|≤ 0.01$. Observe that:
$$\max_{0\leq x\leq 1.96} |f''(y)| = |f''(0)| =\left| (0^2-1)\frac{1}{\sqrt{2\pi}}e^{\frac{-0^2}{2}}\right| = \frac{1}{\sqrt{2\pi}}$$
So the upper bound for the error
$$E=-\frac{(b-a) h^{2}}{12} f^{\prime \prime}(\xi), \quad \xi\in(a,b)$$
is
$$|E| \leq -\frac{(1.96) h^{2}}{12} f''(0), \quad \xi\in(0,1.96)$$
I get that $n\geq 1$. Is my method correct?
Kind regards,
