Talk:Three-point estimation

This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Statistics Low‑importance This article is within the scope of WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.StatisticsWikipedia:WikiProject StatisticsTemplate:WikiProject StatisticsStatistics Low This article has been rated as Low-importance on the importance scale.

Typos everywhere. That is all [dated 20 May 2013]

 "Confidence level in E value +/- SD is approximately 68%... etc"

Technically, what is being referred to here is not the 'confidence level in E', the following is also technically incorrect:

 "Calculates the SD value for the total project work as SD (Project Work) = √(Σ SD (Task) 2)"

In the first case, the SD referred to {"SD = (b − a) / 6"} is the SD of the triangular distribution not the the standard error of the mean (which is unknown in this example). In the second case, this summation only works if all tasks times are independent, identically distributed, and normally distributed - plainly not often the case, otherwise you wouldn't be using 3pt estimates.

Needs updating. [dated 20 May 2013]

The formula ${\displaystyle \operatorname {SD} ({\text{project}})={\sqrt {\sum {\operatorname {SD} ({\text{task}})^{2}}}}}$ makes no sense; square root of a square is the value itself: ${\displaystyle {\sqrt {\sum {\operatorname {SD} ({\text{task}})^{2}}}}=\sum {\operatorname {SD} ({\text{task}})}}$. It's probably supposed to be the root mean square: ${\displaystyle \operatorname {SD} ({\text{project}})={\sqrt {{1 \over n}\left(\operatorname {SD} ({\text{task}}_{1})^{2}+\operatorname {SD} ({\text{task}}_{2})^{2}+\cdots +\operatorname {SD} ({\text{task}}_{n})^{2}\right)}}}$ 91.153.55.81 (talk) 21:44, 25 September 2022 (UTC)Reply