Your question is not entirely clear on whether it is stated that the current is uniformly distributed on $[0,25]$. Your title says "normal" distribution; the statement of the problem does not explicitly state it is uniform (just continuous), and then your answer assumes a uniform distribution. So I have also made the assumption that $X$ is uniform.
You already calculated the cumulative distribution function $$F_X(x) = \begin{cases} 0, & x < 0 \\ \frac{x}{25}, & 0 \le x \le 25 \\ 1, & 25 < x \end{cases}$$ and you also calculated the expectation $$ \operatorname{E}[X] = 12.5 = \mu.$$ These are correct. The variance is defined by $$\operatorname{Var}[X] = \operatorname{E}[(X-\mu)^2] = \int_{x=0}^{25} (x - 25)^2 f_X(x) \, dx,$$ where $f_X(x) = \frac{1}{25}$ is the density on $x \in [0,25]$. So what you need to calculate is $$\int_{x=0}^{25} \frac{(x - 25)^2}{25} \, dx.$$ Can you do this?
Seeing the number of similar questions on the site, let us try to delineate some clear and automatic methods.
First step: Write correctly the joint PDF. Here, the PDF $f$ of $(X,Y)$ is defined on the whole plane $\mathbb R^2$ (as every other joint PDF) by the formula $$f(x,y)=\mathbf 1_{0<x<1}\,\mathbf 1_{0<y<1}$$
Again:
First motto: The joint PDF of any $d$-dimensional random vector should be defined on the whole space $\mathbb R^d$, not only on a subset of $\mathbb R^d$.
Second step: Deduce the joint PDF of a transformed couple. Here, one can consider $(Z,T)=(X/Y,Y)$ (but other choices for the second coordinate $T$ are possible).
Then $(X,Y)=(ZT,T)$ hence $dxdy=tdzdt$ and the change of variable formula based on the Jacobian of the transformation $(x,y)\to(z,t)=(x/y,y)$ yields the joint PDF $g$ of $(Z,T)$ as $$g(z,t)=|t|\,f(x,y)=|t|\,f(zt,t)=t\,\mathbf 1_{0<zt<1}\,\mathbf 1_{0<t<1}$$
Second motto: Get yourself familiar with the change of variables formula.
Third step: Marginalize the joint PDF to deduce the desired PDF. Here, the PDF $f_Z$ of $Z$ is simply $$f_Z(z)=\int_\mathbb Rg(z,t)dzdt=\int_0^1t\,\mathbf 1_{0<t<1/z}dt=\mathbf 1_{z>0}\int_0^{\min\{1,1/z\}}tdt=\tfrac12\min\{1,1/z\}^2\mathbf 1_{z>0}$$ An equivalent formulation is $$f_Z(z)=\tfrac12\mathbf 1_{0<z<1}+\tfrac1{2z^2}\mathbf 1_{z>1}$$
Third motto: Well, no real motto for this last step, simply, one can hope you get the automaticity of the approach...
Best Answer
Actually, by the definition of probability density function, $p(x)$ only needs to be piecewise continuous because what we really care is its integral, so it is not a problem that $p(x)$ is discontinuous at $x=1$. You can refer to your textbook to see the definition.