Following a reduction of Levenshtein distance between the original question and the current version, one can say that the total frequency of words that needed correction is merely the sum of all the $x$-values, let us call that the $L$ percentage. Assuming that this is actually exponentially distributed in $x$, which is not a good model for this (see below), the cumulative scaled distribution would be $L(1-e^{-\lambda x})$. However, the question is rather how would $L$ be distributed. For that, let us first suppose that we repeat the experiment $n$ fold to produce $n$ distributions, and add them up value for value, then we would be summing outcomes, which would be convolution, and that would produce an Erlang distribution, A.K.A. special case gamma distribution. However, I am not sure I would do that. Rather, this is a discrete distribution and the exponential distribution is continuous, i.e., $x$ cannot be $1/2$, for example. So the geometric distribution may be a better model.

Following a reduction of Levenshtein distance between the original question and the current version, one can say that the total frequency of words that needed correction is merely the sum of all the $x$-values, let us call that the $L$ percentage. Assuming that this is actually exponentially distributed in $x$, which is not a good model for this (see below), the cumulative scaled distribution would be $L(1-e^{-\lambda x})$. However, the question is rather how would $L$ be distributed. For that, let us first suppose that we repeat the experiment $n$ fold to produce $n$ distributions, and add them up value for value, then we would be summing outcomes, which would be convolution, and that would produce an Erlang distribution, A.K.A. special case gamma distribution. However, I am not sure I would do that. Rather, this is a discrete distribution and the exponential distribution is continuous, i.e., $x$ cannot be $1/2$, for example. So the geometric distribution may be a better model.