Timeline for Order Statistics Problem: Wackerly/Mendenhall/Scheaffer, 5th Ed., Problem 6.58
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2021 at 15:28 | vote | accept | Adrian Keister | ||
| Mar 16, 2021 at 15:27 | comment | added | Adrian Keister | Great, thanks! My answer certainly seems to be correct by that means. | |
| Mar 16, 2021 at 15:09 | comment | added | whuber♦ |
You can check it with a quick simulation, such as this one in R: mean(apply(matrix(runif(4e5, 0, 2), 4), 2, max)).
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| Mar 16, 2021 at 14:44 | comment | added | Adrian Keister | Is my (second) solution correct, as claimed? | |
| Mar 16, 2021 at 14:43 | history | edited | Adrian Keister | CC BY-SA 4.0 |
added 1242 characters in body
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| Mar 15, 2021 at 22:40 | comment | added | whuber♦ | The four events must be uniformly distributed within the first two minutes. | |
| Mar 15, 2021 at 22:06 | comment | added | Adrian Keister | @whuber Ok, that makes sense. How would I compute the expected waiting time in part 2? | |
| Mar 10, 2021 at 19:51 | comment | added | whuber♦ | Consider four independent uniform variables in an interval. The chance that any value lies in the first half of the interval is (by definition of uniform!) exactly $1/2.$ Therefore the chance all four lie in the first half is $1/2^4 = 1/16.$ So, however you choose to perform the calculation, you should obtain $1/16$ as your answer! | |
| Mar 10, 2021 at 19:35 | comment | added | Adrian Keister | Yeah, I wondered about that; but where are the errors? Is it the $n$ value? | |
| Mar 10, 2021 at 19:29 | comment | added | whuber♦ | The value $1\times 10^{-11}$ is so obviously wrong, you should sit back and rethink your interpretation of the question. | |
| Mar 10, 2021 at 18:59 | history | answered | Adrian Keister | CC BY-SA 4.0 |