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5

Here is a back-of-the-envelope calculation. Start with the average value of $ab$, which is $(n+1)^2/4$. Consider powers of $2$ only. Three-quarters of the $(a,b)$ pairs lack $2$ as a common factor. 3/16 have $2^1$ as a common factor, $3/64$ have $4$ as a common factor and so on. So, on average, dealing with powers of $2$ reduces the average by a factor ...

4

The histogram represents $2 + 0 + 4 + 1 + 2 + 0 + 3 = 12$ apples in total. Two of them have three seeds, none with four, and so forth. Thus, to find the average, we must evaluate the sum of the seeds divided by the number of apples. Can you do the rest?

3

Theorem 6.3 of Olivier Bordelles, Mean values of generalized gcd-sum and lcm-sum functions, Journal of Integer Sequences, Vol. 10 (2007), Article 07.9.2, says, for any real number $x$ sufficiently large, $$\sum_{n\le x}\sum_{j=1}^n[n,j]={\zeta(3)\over8\zeta(2)}x^4+O\bigl(x^3(\log x)^{2/3}(\log\log x)^{4/3}\bigr)$$

3

\begin{align} & \sum_x \sum_y x P_{XY}(x,y) = \sum_x \sum_y x \Pr(X=x\ \&\ Y=y) \\[10pt] = {} & \sum_x \left( x \sum_y \Pr(X=x\ \&\ Y=y) \right) \tag 1 \end{align} The justification of the last step above is that as (lower-case) $y$ runs through the list of all possible values of (capital) $Y$, the index $x$ does not change. It can therefore ...

2

My numeric investigations show that for a given $n$ the list values all tend toward $$r=\frac{2n+1}3$$ Therefore, if you are given the repeated value $r$, you can find $n$ from $$n=\frac{3r-1}2$$ The complete proof escapes me so far, however. So far I see that the transformation is linear, and it appears that the only real eigenvalue is $1$ and the only ...

2

The average is the total divided by the count. The average rate of change is the total distance traveled divided by the amount of time. When you take calculus you will see that average rate over an interval is different from instantaneous rate at a point in time.

1

You might be interested in reading about Simpson's paradox (easily found online). It can happen, for example, that A has a better average every single year than player B does, yet player B has the better combined average. For example: $$A:\;\;\;\left[\frac23,\frac{5}{10},\frac{5}{10}\right]$$ $$B:\;\;\;\left[\frac{12}{20},\frac13,\frac25\right]$$ Then, ...

1

The difference between the two results is a consequence of the fact that he played different number of games in different years. If every year he would play the same number of games, the two results would be the same. Now to understand the issue, think about a simplified model. Lets say that the player played 100 games the first year, and 200 games the ...

1

Your proposed method gives too much weight to the most recent test. If you want each test (of the most recent 5) to have the same weight the formula you want is: $$New\;Average=\frac{4(Old\; Average) + New\; Score}{5}$$ In the first example you mention (old average = 3, most recent test = 8) you get 4.

1

You cannot find the average attendance over a month with only a single week's information. This cannot be done. You might guess that the attendance of that one week was the same as during the whole month --- and that may or may not be accurate. (Probably not, especially if it was only one meeting).

1

There are a variety of functions that map $(0,+\infty)$ to a finite interval. For example, $\tan^{-1} x$ maps $(0,+\infty)$ to $(0,\frac{\pi}{2})$.

1

The two methods are fundamentally different, and so in general will produce different results. The row average of row $i$ is: $$\dfrac{\sum_{j=1}^{N_c} {e_{ij}\times w_{ij}}}{\sum_{j=1}^{N_c} w_{ij}}$$ where $N_c$ is the number of columns, $e_{ij}$ is the entry in the $j$-th column of the $i$-th row in the data table, and $w_{ij}$ is the entry in the ...

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