Timeline for Calculating Odds of Getting a Sample w/ a Specific Standard Deviation
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| S May 28, 2017 at 18:02 | history | bounty ended | EthanT | ||
| S May 28, 2017 at 18:02 | history | notice removed | EthanT | ||
| May 25, 2017 at 22:26 | vote | accept | EthanT | ||
| May 25, 2017 at 19:53 | answer | added | soakley | timeline score: 2 | |
| May 25, 2017 at 17:28 | history | tweeted | twitter.com/StackStats/status/867794617856659457 | ||
| May 25, 2017 at 16:31 | history | edited | EthanT | CC BY-SA 3.0 |
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| S May 25, 2017 at 16:31 | history | bounty started | EthanT | ||
| S May 25, 2017 at 16:31 | history | notice added | EthanT | Improve details | |
| May 23, 2017 at 17:38 | comment | added | EthanT | Thanks whuber, that's very interesting and something we hadn't thought of. Again, the masses are so tightly grouped (and pass a MATLAB chi2gof test for normality) I think they are "normal enough" for normality to be a reasonable assumption. Nonetheless, interesting point and I learned something new from it about distributions. again, would love to see the details spelled out as to how you reached 1 in 48 million above (wink, wink) | |
| May 23, 2017 at 16:01 | comment | added | whuber♦ | Ethan, concerning the distribution: Suppose, for the sake of imagining what might go on, that the dimensions of the explosive charge in the cannon have a joint Normal distribution. That implies (from geometry alone) that the volume of the charge itself varies like the cube of a Normal distribution. Assuming muzzle speed is proportional to the mass of the charge and the mass if proportional to its volume, that would yield a positively skewed, non-Normal distribution of speeds. Regardless, the answer might not depend too strongly on the shape of the distribution. | |
| May 23, 2017 at 15:55 | comment | added | EthanT | @whuber, I believe you've answered my question in your comment. Any chance of filling in the details in an "answer"? I'm not able to make the jump to calculating out the one in 48 million myself, given the amount of information given. Would be much appreciated ;-) | |
| May 23, 2017 at 15:46 | comment | added | EthanT | Remember, the numbers I put in the OP above are "made up" and don't represent the actual "gun" in question. I would say the problem might even get more pointed with match-grade rifle bullets. You're just not going to see a bullet come out at double the muzzle velocity. So, considering the type of performance I mention above, is there any reason to suspect something other than a normal distribution, supported with physical reasoning that justifies any (significant) departure from normality. I'm not saying there isn't, just having a hard time imagining it. | |
| May 23, 2017 at 15:46 | comment | added | EthanT | Hello Glen_b, I have access to large databases of the particular "gun" in question. I would have to say it is a near impossibility to ever see a muzzle velocity at double the mean for this particular "gun". Physically, where is that energy going to come from? All the prop charges are within 1% of each other in terms of weight, coming from a relatively consistent manufacturing process. In addition, plots of the muzzle velocity sure look like a normal distribution, with a tight standard deviation around the mean that is fractions of the mean velocity. | |
| May 23, 2017 at 10:53 | comment | added | Glen_b | There's obvious reasons to think it wouldn't be normal -- muzzle velocity is necessarily non-negative, and presuming you'll only consider projectiles that exit the barrel, actually necessarily positive. As a result you'd tend to expect somewhat skewed distributions (one can at least consider a possibility of a cannonball exiting with a bit more than double the mean velocity but it's impossible to be that below the mean, since that would be negative). It might be reasonable in some situations to use a normal approximation, but it's by no means obvious that one should automatically do so. | |
| May 23, 2017 at 3:48 | history | edited | EthanT | CC BY-SA 3.0 |
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| May 23, 2017 at 3:48 | comment | added | EthanT | Sorry, I assumed it was obvious it was a normal distribution. Although I guess that brings up another question: Is that assumption not as obvious as I thought? Are there reasons to think cannon (or any type of gun) muzzle velocities would not be represented by a normal distribution around some mean velocity? However, I guess for the sake of this problem, let's assume both the population and samples can be represented by normal distributions. | |
| May 23, 2017 at 3:42 | comment | added | whuber♦ | Your question has no unique answer for the information given. If you assume a specific distribution for the speeds, then the question can be answered, so please edit it to include that information. [There exist analytic answers for some distributions, such as a Normal distribution. For it, a known multiple of the sample standard deviation has a chi distribution. Note that the chance that the sample SD is exactly $2$ will be zero: you need to ask for the chance that the SD lies within some given range, such as $0-2$. For a Normal distribution, that chance is one in 48 million.] | |
| May 23, 2017 at 3:25 | history | asked | EthanT | CC BY-SA 3.0 |