The following is a question which I personally find badly constructed, and would like to gather some opinion on:
Suppose you select a coin from a pile that is thought to have probability of landing heads $p\sim\mathcal{N}(\frac12, \frac1{100})$. The probability of landing heads three times in a row lies within $(\frac1{N+1},\frac1N)$, where $N\in\mathbb{Z}^+$. Find $N$.
This makes absolutely zero sense to me. First of all, now $p$ has a range spanning $\mathbb{R}$ which is unacceptable for a probability. Furthermore, even if you just assumed that the probability matched the mean exactly, that would be $N=8$. Where would the $N+1=9$ come into play? Am I missing something completely, or is this a badly constructed question?
Essentially the question is asking you to find $\mathbb E[p^3]$, which is not exactly $\frac{1}{8}$.
You do not need to do this precisely, as it wants the answer as the integer part of its reciprocal. This leaves room for some interpretation of the "normal distribution" for $p$. As stated, the probability $p$ is in $[0,1]$ would be $\Phi(5)-\Phi(-5) \approx 0.9999994266$, where $\Phi()$ is the cumulative distribution function of a standard normal distribution.
So it will not be too dramatic if we ignore this point and just find $\int\limits_{-\infty}^\infty p^3 \,f(p)\,dp$. If you do the calculus, this turns out to be $\dfrac{7}{50}=0.14$.
Alternatively, we could condition on $p$ being in $[0,1]$ and find $\frac{\int\limits_{0}^1 p^3 \,f(p)\,dp}{\int\limits_{0}^1 f(p)\,dp}=\dfrac{7}{50} - \dfrac{3}{20\sqrt{2\pi}\exp(25/2)(\Phi(5)-\Phi(-5))} \approx 0.139999777$.
Both of these are about $\dfrac{1}{7.143}$ so between $\dfrac{1}{7}$ and $\dfrac{1}{7+1}$.