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Aug
14
comment Are there any other interesting functions such as $e^x$ whose derivative and integral are the same?
Well, there are close relatives, like $\frac{e^x+e^{-x}}{2}$ and $\frac{e^x-e^{-x}}{2}$, where the derivative bounces back and forth, and $\sin x$, $\cos x$ which exhibit similar but more complicated behaviour.
Aug
14
revised Range of variance
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Aug
14
answered Range of variance
Aug
14
comment Probability of X or Y
Not yet, soon. First of all, for the second quesition you need to subtract the $0.5832$ from the sum $\Pr(A)+\Pr(B)$.
Aug
14
comment Probability of X or Y
The labels on the bottom branches are wrong, There is no reason to think they are $0.65$, $0.35$. The $0,65$ refers to all respondents, not new motorway haters. By the way, $0,5832$ is about $58\%$, not $5.8\%$.
Aug
14
comment Median based on number of entries instead of values
There is no existing name that I know of used in Statistics. The median of the set of values is technically the correct name
Aug
14
revised How to solve? 1
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Aug
14
comment Counting the number of permutations of all sizes without repetition
Someone may write out a solution. I don't want to do it, too much typing.
Aug
14
revised How to solve? 1
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Aug
14
answered How to solve? 1
Aug
14
comment Counting the number of permutations of all sizes without repetition
There are messy expressions using Stirling numbers.
Aug
14
revised Unique Integer solution of a non-linear equation
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Aug
14
awarded  divisibility
Aug
14
awarded  optimization
Aug
14
answered Unique Integer solution of a non-linear equation
Aug
14
comment Counting the number of permutations of all sizes without repetition
Without repeated letters you would have $n+n(n-1)+n(n-1)(n-2)+\cdots +n!$. To get pleasant form, divide by $n!$. You get $n!(1+1+\frac{1}{2!}+\cdots+\frac{1}{(n-1)!})$, which unless $n$ is small is close to $n!e$, where $e$ is the base for natural logarithms.
Aug
14
comment Median based on number of entries instead of values
Discarding entries that happen to repeat would not make statistical sense. One could call the result $6$ the median of the set of values.
Aug
14
comment Median based on number of entries instead of values
Population median is defined for a probability distribution. In the case of "ambiguity" of the type we meet with samples, all "middle" values are declared to be medians. So in our case it would be all numbers between $0$ and $6$ inclusive. The top of this range of medians, and the bottom, have no special names. In the finite case, any point $p$ that minimizes the sum of the absolute values $\sum |x_i-p|$ is a median.
Aug
14
comment what distribution should I use for this question?
Thanks for spotting the typo. It should be $0.01$, the original standard deviation, divided by $\sqrt{72}$. Let $X_1,\dots,X_{72}$ be the sizes of our coins, assumed independent. Then the variance of the sum is the sum of the variances, so $(72)(0,01)^2$. When we divide by $72$ to get the sample mean, the variance gets divided by $72^2$. Now take the square root.
Aug
14
revised what distribution should I use for this question?
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