Estimate probability that random variable is smaller than given value I am not an expert in statistics but I am facing the following problem: I have a random variable that basically arrises as difference of count data. I have 67 points and they don't seem to be normally distributed. Nevertheless I would like to estimate the probability that the underlying random variable takes values larger than, say, 400.
Would it be statistically sound to estimate a PDF using a kernel density estimation and then compute the probability $P(X>400)$ through that estimation? Or could it be a way to go to fit a normal distribution? However, I have to say that testing on normality fails. Please also see the attached figure.
Thanks. :) 

 A: The probability $P\left(X>t\right)$ you are estimating is the complement of the cumulative distribution function $F(x)=P\left(X\leq t\right)$. which could be nonparametrically estimated by the empirical cumulative distribution function $F_n\left(t\right) = \frac1{n}\sum\limits_{i=1}^n I_{[X_i\leq t]}$. It means that the point estimate of $P\left(X>400\right)$ would just be the number of observations above $400$, which seems to be zero here.
This could be easily supplemented by the confidence interval from the  Dvoretzky–Kiefer–Wolfowitz inequality:
$$P\left(\sup\limits_t\left|F_n(t)-F(t)\right|>\varepsilon\right)\leq 2e^{-2n\varepsilon^2}$$
Taking the right side to be equal to the desired confidence level $\alpha$, it's easy to show that for any given $t$ the interval $$\left[ 1 - F_n(t) - \sqrt{-\frac1{2n}\log\frac{\alpha}{2}}; 1 - F_n(t) + \sqrt{-\frac1{2n}\log\frac{\alpha}{2}}\right]$$ would cover $P\left(X>t\right)$ with confidence $1-\alpha$ (and it should be trimmed if its boundaries are below $0$ or above $1$).
In your case, with $n=67$ and $F_n(400)=1$ with $95$% confidence we could estimate that the probability of $P(X>400)$ is between $0$ and $\approx0.166$.
