Timeline for Mean and variance after transformation
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2023 at 15:59 | answer | added | kjetil b halvorsen♦ | timeline score: 1 | |
| Sep 5, 2019 at 22:48 | history | edited | Glen_b | CC BY-SA 4.0 |
meaningful title
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| Sep 5, 2019 at 22:47 | comment | added | Glen_b | I have fixed the title and reopened but I would like the OP to undertake the other requested edits (i.e. to idenitfy a specific issue to ask about, something that will enable them to make some progress). | |
| Sep 5, 2019 at 22:45 | comment | added | whuber♦ | @Michael Glen reopened it quickly. I just wanted to thank you for your generous help in the comments you have posted. | |
| Sep 5, 2019 at 22:41 | history | reopened | Glen_b | ||
| Sep 5, 2019 at 18:12 | comment | added | Michael R. Chernick | @Glen_b Now that the OP has added the self study do you think the question should remain closed. If so what is your reason. I am thinking about voting to reopen it. | |
| Sep 5, 2019 at 18:10 | comment | added | Michael R. Chernick | @Glen_b I am not sure what the best advice to the OP should be. The question is very elementary & it wouldn't take much research to find the properties that explain how a linear transformation affects the mean and standard deviation of a random with a finite mean and standard deviation. I would have thought that this question is appropriate if the self study tag is used. Almost any hint that could be given as an answer would make it very obvious how to get the solution. I think I gave the hint that would serve as an answer but I felt too embarrassed to do more than make a comment. | |
| Sep 5, 2019 at 18:00 | comment | added | Michael R. Chernick | Yes in that case the the new mean would be (m+10)/2. | |
| Sep 5, 2019 at 13:16 | history | closed |
Sycorax♦ Glen_b |
Not suitable for this site | |
| Sep 5, 2019 at 13:15 | comment | added | Glen_b | Yes, I read that already; as I explained you have two choices in that case; you either need to ask a very specific question that would allow you to start, or you need to do more search and research. A very general "how do I do this" on a homework style question makes it not quite suitable for our site, even if you're seeking guidance. A more specific question, addressing the first thing you don't understand may well be suitable, though it's probably more productive to do more research yourself and move the specific question further along the process. | |
| Sep 5, 2019 at 10:50 | comment | added | user258093 | @MichaelChernick she added 10 points so will this mean that it would be (m+10)/2? | |
| Sep 5, 2019 at 10:48 | comment | added | user258093 | @Glen_b first of all the reason I did not show any attempt was because I did not know how to start working about the question. Second of all, I did not ask for someone to write or give me the answer but rather help me figure out how to start solving the question, as my textbook was not helpful and I could not find a website than answered my questions (I did do research) and I usually ask here when I am out of options meaning did my research, looked at my book and attempted to solve it. I won't ask for help just because I am lazy to solve this question | |
| Sep 4, 2019 at 23:36 | comment | added | Glen_b | ctd... reading your notes and textbook, looking at examples, checking Wikipedia, searching our site and the internet more generally). For example, see here and here. Please edit your question. | |
| Sep 4, 2019 at 23:36 | comment | added | Glen_b | 1. Every post here is seeking help with a stats question (otherwise it's off topic). As a result your title does nothing to help people figure out whether it's a question they want to try to help with. Please edit your title to reflect what the question is about. 2. Please read the help center on homework-style questions, which are treated differently. If you can't show any attempt you're going to need to ask something very specific about your difficulties. If you can't do that, you need to do some search and research (including ... ctd | |
| Sep 4, 2019 at 20:40 | review | Close votes | |||
| Sep 5, 2019 at 13:20 | |||||
| Sep 4, 2019 at 20:28 | comment | added | Michael R. Chernick | Since you subtract 10 & then divide by 2 the linearity of expectations tells you that for a sample with mean m the mean of the transformed data will be (m-10)/2. In this case plug in 156 for m. Regarding the standard deviation we know that the variance of a random variable multiplied by a constant is the square of the constant ties the original variance. The standard deviation is the square root of the variance. So in your case the standard deviation is 1/2 times the original standard deviation. In your case 24/2. | |
| Sep 4, 2019 at 20:25 | history | edited | whuber♦ |
edited tags
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| Sep 4, 2019 at 20:16 | review | First posts | |||
| Sep 4, 2019 at 20:29 | |||||
| Sep 4, 2019 at 20:08 | history | asked | user258093 | CC BY-SA 4.0 |