How far from the North Pole

Idea. The distance to the pole depends only on the height of the point, and the height of a uniform point on a sphere is uniform. So the answer is a one-variable integral.

Step 1: the distance. For a point $(x,y,z)$ on the unit sphere, the squared distance to $(0,0,1)$ is $$\begin{array}{rl}{x}^2+y^2+(z-1{)}^2& =1-z^2+z^2-2z+1\\ & =2-2z.\end{array}$$

Step 2: the height is uniform. The band of the sphere between heights $z$ and $z+dz$ has area $2\pi dz$, whatever $z$. So $z$ is uniform on $[-1,1]$.

Step 3. $$\begin{array}{rl}\mathbb{E}\left[\sqrt{2-2z}\right]& =\frac{1}{2}{\int }_{-1}^1\sqrt{2-2z}dz\\ & =\frac{1}{4}{\int }_0^4\sqrt{u}du=\frac{1}{4}\cdot \frac{16}{3}=\frac{4}{3}.\end{array}$$ (We substituted $u=2-2z$, so $dz=-\frac{du}{2}$.)

So the expected distance is $\frac{4}{3}\approx 1.333$. It is a little less than $\sqrt{2}\approx 1.414$, the distance from the pole to the equator, where the average height $z=0$ lies: the distance $\sqrt{2-2z}$ is a concave function of $z$, so its average is below its value at the average height (Jensen's inequality).