2 Tried to provide a lead in answering the question. edited Oct 24 '14 at 9:03 abaumann 1,6551111 silver badges99 bronze badges 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 $$Z+\omega$$. 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 $$[Z-\omega;Z+\omega]$$. 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 $$Z+\omega$$. 1 answered Oct 23 '14 at 14:21 abaumann 1,6551111 silver badges99 bronze badges 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 $$[Z-\omega;Z+\omega]$$.