The distribution function defines the probability that the random variable is less than or equal to a given x: $P(n\leq x) = F(x)$. In this case, x is the distance of a single jump.
What we need is the expected value V of the distribution function. To get this, we want the derivative of the distribution function, $f(x) = F'(x)$, which produces a function known as the "probability density function" of x. This function will tell us how likely any given jump is to be x units.
Now, given that, the "expected value" of the random variable is equal to:
This "expected value", much like the term states, is the long-term average value of X given a sufficiently large number of trials. For a normal distribution, it is usually the point at which $f(x)$ is at its peak; however many probability functions are not normal (for instance, the probability of an n-sided die rolled once showing a particular face x is 1/n regardless of x). Now, that equation is the general form. We know that the frog always jumps forward, and thus the jump distance is never negative; therefore, we should actually solve the integral for a lower bound of 0:
$$E[x] = \int_0^\infty xf(x)dx$$
$E[x]$ will be the length of the average jump; it has a roughly 50-50 shot to be higher or lower, and thus given enough jumps, the sum of any n jumps should be $E[x] * n$. Now it's arithmetic; solve $E[x] * n >= d$ for n.