The gross-error bound corresponding to $b$ is
$$
c(b)=\frac{b}{\mathbb P(|Z|\leq b)}
=\frac{b}{2\Phi(b)-1}.
$$
It is strictly increasing because
$$
2\Phi(b)-1=\int_{-b}^b\phi(x)\,dx
>2b\phi(b),
$$
so the numerator of $c'(b)$ is positive. Furthermore
$$
\lim_{b\downarrow0}c(b)
=\frac1{2\phi(0)}=\sqrt{\frac\pi2},
\qquad
\lim_{b\to\infty}c(b)=\infty.
$$
Thus $b\in(0,\infty)$ corresponds exactly to $c\in(\sqrt{\pi/2},\infty)$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 223 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-223/2/b/solution