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To understand this, you need to see what triggers a nuclear decay. The answer is a beautiful example of quantum mechanical behavior. Nothing triggers it. It is just that the world is fundamentally quantum mechanical, and probabilistic. All the other answers that "no, there is no triggering event, it just happens, quantum mechanics is like that" ...


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I see that people on this site mostly seem to think you can just multiply numbers together to get probabilities, and thus the answer is that the probability is something of order $10^{-10^{25}}$. The trouble with this is that the decay events are not entirely independent events, so this method of calculation is wrong. It is ok as a first very VERY rough ...


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@Nihar has an excellent answer: It's possible but with a chance of 1 in $10^{1.94\times10^{25}}$ That is a truly large number. When you use exponents that need to be represented with their own exponents, it can sometimes be difficult to think about what they actually mean. for some perspective: There are about $5\times10^{19}$ atoms in a grain of sand ...


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The uncertainty specified by a manufacturer is not evaluated by considering a single sensor but by considering the spread across the whole production, determined both from sampling the production and by considering design parameters. Therefore, you should not interpret the specifications as "95 % of the time the uncertainty is within this range" ...


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In order for that to happen in the real world you need to start with about 3.8 million kilograms of that material. Here is how you come up with that number. You start from the the formula connecting the half-life to the number of particles over time $$ N(t) = N_0 \left(\frac{1}{2}\right)^\frac{t}{t_{1/2}} $$ Now you replace $N(t)$ with what you would like to ...


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The answer is 'no'. This 'no' is on the same level like: Can it happen that you float for 15 minutes in the middle of your room. (Statistical mechanics tells technically yes, but again with a for all practical purposes zero probability) Can you put a monkey in front of a typewriter and get Shakespeare novels out of it? Can you walk through a solid wall (non ...


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One thing to keep in mind is that this is not only a statistics question and the analogy of atoms decaying and flipping coins can be misleading. For example, uranium 235 has a half life of more than 700 million years, but when brought in the right configuration (close packed) and in the right amount (above critical mass), it decays practically in an instant.....


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TLDR: statistical models are models, and thus by definition not a perfect reflection of reality. Nihar's answer is good but I'm going to tackle it from a different direction. First off, if we only look at statistical mechanics you can run through the math and of course you will find an extremely small probability. You might stop there. But statistical ...


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With a single input variable $x$ the covariance "matrix" of the estimator is a single number: the variance of the estimator. If you are really interested in this case, I did not understand the question. I understand your question, if we consider the model $f(x_1, x_2) = k_1 x_1 + k_2 x_2$ with two input variables. In this case, the (least square) ...


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The short answer is yes. No matter how many atoms there are, there is always a (sometimes vanishingly small) chance that all of them decay in the next minute. The fun answer is actually seeing how small this probability gets for large numbers of atoms. Let's take iodine-131, which I chose because it has the reasonable half-life of around $8$ days = $\text{...


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That can be done using a Dirac delta when integrating over all the possible values of the momentum $$\Omega\left(E_{0}\right)=\int\Pi_{i=1}^{N}\dfrac{dp_{ix}dp_{iy}}{h^{2}}dx_idy_i\delta\left(E_{0}-\sum_i E_{i}(\vec{r}_{i},\vec{p}_{i})\right),$$ where $\Omega\left(E_{0}\right)$ is the number of states with energy $E_0$ (I am not completely sure about the ...


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The general error propagation for a function $f(x_1, x_2,...,x_n)$ is $$ \Delta f = \sqrt{\sum_{i=1}^n \left(\frac{\partial f}{\partial x_i} \Delta x_i \right)^2}. $$ If $f(x) = 1/x$, we can say that $$ \frac{\partial f}{\partial x} = -\frac{1}{x^2} $$ and so $$ \Delta f = \frac{\Delta x}{x^2}. $$


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This answer is a comment really. The standard deviation has a strict statistical meaning The root-mean-square deviation of x from its average is called the standard deviation. For a set of discrete measurements, the standard deviation takes the form for continuous: .... Determining the average or mean in the above expression involves the distribution ...


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The idea is that if a particle can be anywhere in the sphere of radius $\Delta R$, then $\sigma_x\sim\Delta R$ (you can try to calculate the standard deviation of the position of a particle that can be anywhere on a sphere of radius $\Delta R$ and it will be proportional to $\Delta R$). It is a bit like the Fermi Problem (https://en.wikipedia.org/wiki/...


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The first formula is more precise. For a gaussian wave package the standard deviations in x and p are related by this expression. A wave function that is uniform over a spherical volume and zero outside it has high momentum components due to its sharp edges. This causes he product of x and p standard deviations to be larger than $\hbar/2 = h/4\pi$.


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Yes! Suppose we could entropy as a function of two state variables like temperature and volume. Eg:Consider for example an ideal gas: $$ S = nC_v \ln T + nR \ln V$$ So, for a cyclic process consider a shift of state variables in the form: $$ (T_1 , V_1) \to (T_2 , V_2) \to (T_3 , V_3) \to (T_1,V_1)$$ And corresponding entropy change $$\Delta S = \sum_{i=1}^3 ...


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In the case of a bell curve representing the entire population, or the mathematics of a phenomenon… — any applicable bell curve — the concept of a standard deviation is the point at which the curve changes from getting more and more steep to getting less and less steep. In other words… starting at the middle, and moving steadily away from the centre… the ...


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The standard deviation is just one measure of uncertainty. Sometimes results are summarized as a mean (best estimate) and a standard deviation. If all you have is a set of data, the mean and standard deviation of the data can be easily calculated without any assumptions as to the underlying probability distribution. This does not imply that the mean plus/...


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If you study probability theory, you will rarely find references to the standard deviation $s$ . The reason is that the standard deviation is a biased estimator: On average the samples' standard deviation $s$ is smaller than the populations' standard deviation $\sigma$, see example below. Therefore, mathematicians usually prefer the sample variance $s^2$, ...


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The 68% confidence level is chosen because it corresponds to 1$\sigma$ of a Gaussian distribution. Therefore, if the distribution is Gaussian, converting from the 68% confidence level to the 95% confidence level is done by simply multiplying by 2 (more on this). In metrology, distributions are often assumed to be Gaussian since the central limit theorem ...


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Are there good reasons to use the coverage factor of $\pm 1\sigma$ as the standard representation of uncertainties or is it only a matter of convention/tradition? The reason is convention however with full respect to the statistically rigorous (correct) way that we should use the reported values in comparisons. Consider first that, when we establish a ...


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We talk in terms of standard deviation because this is traditionally the quantity you use to specify the variance of a Gaußian distribution specifically and any random distribution more generally. You seem to be misinterpreting the recommendation that all uncertainties be reported as standard deviations as a guideline that this should also always be what ...


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