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Lets assume that I have samples that fit to normal distribution. $f(x) = y$

How to calculate the probability that sum of values for $n$ next samples will equal $Z$.

$\mathbb P(f(x1) +f (x2) + f(x3) + f(xn) = Z) = ?$

Is it even possible?

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    $\begingroup$ This is a wrong question to ask. The probability of a normal variable equals to a particular value is 0. $\endgroup$ – Peter Oct 23 '14 at 14:50
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    $\begingroup$ The mathematical notation doesn't seem to match what you're asking. What's $f(.)$? What's $y$? I think you wanted to write $\Pr(X_1+ X_2 + X_3 \ldots + X_n = z)$. Note the convention of upper case letters just for the random variables. In any case the probability of the sum's being exactly any value is zero, as pointed out - this goes for any continuous random variable. A Poisson random variable can have a probability of being equal to $2$; a normal random variable only a probability of being $2 \pm 0.1$, say. $\endgroup$ – Scortchi - Reinstate Monica Oct 23 '14 at 17:38
  • $\begingroup$ The answers here may be helpful. $\endgroup$ – Scortchi - Reinstate Monica Oct 23 '14 at 17:52
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The sum of independent normally distributed variables will itself be normally distributed. This then implies that the probability of their sum being any particular value $Z$ is 0.

You might want to rethink your question in terms of putting an interval on Z - i.e., what is the probability that the result is within $[Z-\omega;Z+\omega]$.


Let $X_i \sim N(\mu_i,\sigma_i)$ and $X_j \sim N(\mu_j,\sigma_j)$ be independent normal random variables. Then their sum is given by $$X_{ij}=X_i+X_j \rightarrow X_{ij} \sim N(\mu_i+\mu_j,\sigma_i+\sigma_j)$$.

Given this, finding the probability that $X_{ij}$ is within some interval $[Z-\omega;Z+\omega]$ can be found by subtracting the CDF of $X_{ij}$ at $Z-\omega$ from the CDF at $Z+\omega$.

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  • $\begingroup$ abaumann and is there any formula that calculates probability you mentioned? $\endgroup$ – Robert.K Oct 24 '14 at 7:01
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    $\begingroup$ @Robert.K I've drawn up some of the considerations above. Can you do it from there? $\endgroup$ – abaumann Oct 24 '14 at 9:03
  • $\begingroup$ thanks a lot for your effort! I'm statistics noob but I'll try continue this way :) $\endgroup$ – Robert.K Oct 24 '14 at 9:40
  • $\begingroup$ Please feel free to ask again if you need more pointers :-) $\endgroup$ – abaumann Oct 24 '14 at 9:52

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