Bootstrapping is a *nonparametric* approach that might help, if you don't know if or cannot assume that your data are normal.

Draw a sample of size *n*. Get the *k*-th percentile via `quantile()`

or similar. Draw another sample of size *n*, *sampling with replacement*. Get the *k*-th percentile. Repeat some *B* number of times, *B* being the number of bootstrapped samples.

After *B* samples, you get a set of *B* measurements that estimate the *k*-th percentile. Get the 2.5th and 97.5th percentile of this set of measurements. This the 95% confidence interval around your original *k*-th percentile measurement.

Example code in R:

```
bootstrap_sample <- function(x, n, B, p) {
bstrap <- vector(mode = "numeric", length = B)
for (i in 1:B) {
s <- sample(x, n, replace = T)
bstrap[i] <- quantile(s, p)
}
bstrap
}
# x = some vector of signal
# n = length of vector
# B = number of bootstrap samples
# kth = k-th percentile of interest
x <- unlist(read.table(...))
n <- length(x)
B <- 10000
kth <- 0.95
# bootstrap samples
x.95thPercentile_95pctCI_samples <- bootstrap_sample(x, n, B, kth)
# k-th percentile (k = 95%)
x.95thPercentile = quantile(x, kth)
# 95th percentile's 95% confidence interval
x.95thPercentile_95pctCI <- quantile(x.95thPercentile_95pctCI_samples, c(0.025, 0.975))
# bounds (asymmetric)
x.95thPercentile_95pctCI_lower_bound = x.95thPercentile - x.95thPercentile_95pctCI[1]
x.95thPercentile_95pctCI_upper_bound = x.95thPercentile + x.95thPercentile_95pctCI[2]
```

I am not sure I understand the problem. If you can draw a random sample from your population then you can estimate the distribution and its parameters. With your example, draw a large sample and find out the proportion of values that are greater than 35. You can also use the estimated mean and standard deviation if you can assume the population has a normal distribution.

I don't think biostars is the best place to find an answer. Statistics-related questions are more likely to get an answer in CrossValidated