Timeline for How to calculate expected value and standard deviation if I have 100 values divided into 15 groups (normal distribution)?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Jun 10, 2012 at 12:10 | vote | accept | user1111261 | ||
| May 26, 2012 at 18:35 | answer | added | leonbloy | timeline score: 3 | |
| May 26, 2012 at 9:08 | comment | added | user1111261 | I think I finally understood the difference the grouping makes. The expected value is not calculated as an arithmetic mean of the values but I have to find the midpoint of each group and then the weighted average of all the midpoints will give me the expected value. However, the calculation of the stand.deviation is no different... | |
| May 25, 2012 at 23:16 | comment | added | Michael R. Chernick | @whuber my source was math.stackexchange.com/questions/148629/calculating-expected-value-standard-deviation-when-i-have-frequencies-and-intervals-in-percentage or math.stackexchange.com/questions/148629 for short | |
| May 25, 2012 at 23:14 | comment | added | Michael R. Chernick | @JoelW That seems like the right thing to do but the question was phrased in a way to make me think he was looking for something else based on the reference he was reading. He could do it either way with his data but your suggestion would gie a mor accurate answer. | |
| May 25, 2012 at 21:58 | comment | added | Joel W. | If you have the actual values, why not ignore the categories and just calculate the mean and standard deviation? Using categories makes the calculation less precise. Or is this a homework exercise to help you understand that? | |
| May 25, 2012 at 21:23 | comment | added | Michael R. Chernick | @whuber Those are both appropriate references but not the question I saw. I will look for my reference. | |
| May 25, 2012 at 21:15 | comment | added | whuber♦ | You're probably thinking of stats.stackexchange.com/questions/18797, @Michael. A related question is stats.stackexchange.com/questions/10433. | |
| May 25, 2012 at 21:13 | comment | added | Michael R. Chernick | My guess is that they are referring to grouped means and grouped standard deviations. A question like that came up once recently either here or on the mathematics site. The grouped mean is obtained by taking a weighted average of the midpoint of the bin where the weight is the number of cases falling in the bin. The same goes for the variance except the square differences are taken by square the difference of the midpoint of the bin with the weighted mean obtained previously as the grouped mean. I don't see where the normal distribution assumption comes into play. | |
| May 25, 2012 at 20:39 | comment | added | user1111261 | I have no idea what do they mean by that... Presumably, the exp.value is gonna be calculated the same as in the case of 15 values. However, the stand.deviation should probably be calculated for each group separately and then getting the stand.deviation by using those stand.deviation of each group. [The resuls are: exp.value: 25.37 and stand.deviation: 3.103] | |
| May 25, 2012 at 20:37 | comment | added | whuber♦ |
(1) Note that the "Stand.deviation" formula actually is for an estimator of a variance and that both formulas incorrectly refer to 15 values, whereas 100 values are in this dataset. (The use of "exp.value" in place of "mean" suggests these data should be treated as a population, whence the n-1 probably should be replaced by n.) (2) You ask about the difference between treating these data as 100 values and 15 values. Why don't you just compute the formulas you think should apply and compare what you get?
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| May 25, 2012 at 20:20 | comment | added | David LeBauer | is this homework? What is the rationale behind the grouping? What is the application / context? | |
| May 25, 2012 at 20:19 | history | edited | David LeBauer | CC BY-SA 3.0 |
added 196 characters in body
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| May 25, 2012 at 19:45 | history | asked | user1111261 | CC BY-SA 3.0 |