Probability that x samples from normal distribution sum to X - Cross Validated most recent 30 from stats.stackexchange.com 2019-10-22T01:27:34Z https://stats.stackexchange.com/feeds/question/121182 https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/121182 -1 Probability that x samples from normal distribution sum to X Robert.K https://stats.stackexchange.com/users/56495 2014-10-23T14:14:33Z 2014-10-24T09:03:31Z <p>Lets assume that I have samples that fit to normal distribution. $f(x) = y$</p> <p>How to calculate the probability that sum of values for $n$ next samples will equal $Z$.</p> <p>$\mathbb P(f(x1) +f (x2) + f(x3) + f(xn) = Z) = ?$</p> <p>Is it even possible?</p> https://stats.stackexchange.com/questions/121182/-/121183#121183 6 Answer by abaumann for Probability that x samples from normal distribution sum to X abaumann https://stats.stackexchange.com/users/11878 2014-10-23T14:21:12Z 2014-10-24T09:03:31Z <p>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.</p> <p>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 $[Z-\omega;Z+\omega]$.</p> <hr> <p>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 $$X_{ij}=X_i+X_j \rightarrow X_{ij} \sim N(\mu_i+\mu_j,\sigma_i+\sigma_j)$$.</p> <p>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 $Z+\omega$.</p>