# Tag Info

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It is a strange assignment, but: Poisson have mean=variance. Calculate both from data! Uniform has mean midways between min and max, and a variance easily found from min and max Normal has no relation between mean and variance, so if not passes first two ... besides normal is symmetric around mean, so just look at some quantiles symmetrically around the ...

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Mean: You answered yourself Standard deviation: Find first a pooled variance, see How to calculate pooled variance of two or more groups given known group variances, means, and sample sizes? Median: This is more difficult, see Is it possible to calculate Q1, Median, Q3, StDev from already aggregated data?, Weighted median, Calculate one median for data ...

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I would rather go for a histogram (with unequal bin length), since that can be made from your data without approximations. Box-plots are not made for your format of data, so you would need approximations in addition to those already made in binning.

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First, your formula can be written in several different ways: $$\left| \frac{A}{B}-1\right|=\left|\frac{A-B}{B} \right|=\frac{\left|A-B \right|}{|B|}$$ with the last being the form for one definition of (absolute) relative difference. Second, that form places a particular importance on Condition B as the reference. If that's the case your formula might ...

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That histogram is clearly bimodal, so with 50000 observations that cannot be a fluke, and your data is not normal, so you do not need a test. See also Is normality testing 'essentially useless'? But rather tell us why you think you need normality? What is your goal with your data?

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One possible explanation: it depends what you mean by adding data because: shapley value shows the individual contribution of a feature that is above or below the average contribution or in other words: This is the predicted value for the data point x minus the average predicted value or in other words efficiency of a shap-value https://christophm.github....

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If $f$ is a probability density, what is conserved is the elementary probability $f\ dx$: $$f_X dx = f_Y dy$$ This shows that the densities $f_X$ and $f_Y$ don't necessarily have the same dimensions.

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A lot of people don't understand the PDF. Some people think it's a probability, somehow. The term itself has a clue: a density. Think of a density of of a physical body. For instance, a density of gold in liquid form is 17.3 g/cc or0.63 lb/cubic inches. It doesn't surprise you that the density is different when we change the units of measure? PDF is the ...

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First, you hardly need a formal test with $n = 50\,000$ observations and such an obviously bimodal histogram. There are various kinds of tests of normality. The Shapiro-Wilk test, as implemented in R by shapiro.test, will take up to 5000 observations. The Kolmogorov-Smirnov test, as implemented in R by ks.test, will test whether a sample follows a ...

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The statistics that you mention could help you to prove your point that the temperature rarely changes - things like the inter-quartile range would provide evidence of how stable it is. Instead of providing evidence that you don't need these statistics, you should see these statistics as tools for understanding whether your hypothesis is correct.

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The fact that an event happens rarely doesn't automatically mean that statistics shouldn't be tracked. You need to understand the business reason for the request. Also, tracking can be done on conditional and unconditional base. That's why I'm afraid you will not get any help from statisticians here.

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You should look into Kaplan Meier analysis, which is a way of handling this type of "right censored" data. In these cases, you want to leverage every data point as much as you can, so you'll still want to use the fact that Person 5's phone hasn't broken in 3 years, even though we don't know how long it will actually last (but we know it's at least ...

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The False Negative Rate is FN/(FN+TP) and not FN/(TN+FN) which is the complement of the False Omission Rate. Your colleague's formula is just the number of false negatives divided by the total number of observations. The confusion matrix itself seems more meaningful. What does representative mean here? When it comes to fallacies, you can come up with ...

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The benefit of distributional assumptions is that you can do additional inference rather than just prediction. Using the mean as a prediction of some population level parameter is often good, the mean has lots of very desirable properties and is well studied. That being said, if the underlying distribution is long tailed and you choose to ignore that, then ...

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A common task in statistics is to perform parameter estimation. The parameter estimator is a function of the sample. For example, consider the following two parameter estimator for the mean of normal distribution: $\hat{\mu}_1=\frac{X_1+X_2}{2}$ $\hat{\mu}_2=X_1$ We have $Var(\hat{\mu}_1)=\frac{\sigma^2}{2}$ but $Var(\hat{\mu}_2)=\sigma^2$. Even though ...

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