I have this exercise where 
is the answer. Using this annex(text at the top means tenths of the x), 
the value becomes 
. What I don't understand is how it becomes 2,58. Our first answer is 0,495, so I'm looking on the left searching the 0,4 table. After that, I'm looking at the tenths of our answer, but there is no number equivalent to 2,58. Is the exercise wrong or am I not understanding the annex right?
Not fully understanding gaussian-laplace table
laplace-distributionmathematical-statisticstables
Related Solutions
The aspect emphasised is that the relevance of a term or a document does not increase proportionally with term (or document) frequency. Using a sub-linear function therefore helps dampen down this effect. To that extent, the influence of very large or very small values (e.g. very rare words) is also amortised. Finally, as most people intuitively perceive scoring functions to be somewhat additive, using logarithms will make probability of different independent terms from $P(A, B) = P(A) \, P(B)$ to look more like $\log(P(A,B)) = \log(P(A)) + \log(P(B))$.
As the Wikipedia article you link notes the justification of TF-IDF is still not well-established; it is/was a heuristic that we want to make rigorous, not a rigorous concept we want to transfer to the real world. As mentioned by @Anony-Mousse as a very good read on the matter is Robertson's Understanding Inverse Document Frequency: On theoretical arguments for IDF. It gives a broad overview of the whole framework and attempts to ground TF-IDF methodology to the relevance weighting of search terms.
You almost got there.
Your 0.32 would more accurately be 0.315 (for number of sd's below the mean). This gives a central area of about 25% and a left tail area of about 38%. More accurately, (100-24.8)/2=37.6 (Your 24.8 is 100-?-? = 24.8 and you want to figure out what that ? value would be)
Best Answer
Note to readers: On this forum, $\Phi(x)$ usually denotes the cumulative distribution function of the standard normal random variable, and thus $\Phi(x) = \Phi_{\text{CDF}}(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \mathrm dx$, which takes on values increasing from $0$ at $-\infty$ to $1$ at $\infty$ as $x$ increases from $-\infty$ to $\infty$. In contrast, the OP's notation uses $\Phi(x)$ (I will call it $\Phi_{\text{OP}}(x))$ to mean $\int_0^x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \mathrm dx$, which takes on values increasing from $0$ at $x=0$ to $0.5$ at $x=\infty$ as $x$ increases from $0$ to $\infty$. Thus, $\Phi_{\text{CDF}}(x) = 0.5+ \Phi_{\text{OP}}(x)$, The table shows the values of $\Phi_{\text{OP}}(x)$.
You need to find the value of $x$ for which the "table entry" is $0.4950$. Unfortunately, there is no such entry; the closest that we can come is $x=2.57$ for which the table entry is $0.4949$ or $x=2.58$ for which the table entry is $0.4951$. What is really needed is the smallest value of $x$ for which we are guaranteed that the "table entry" is $0.495$ or larger. By the way, that's an exact number, no round-off etc: if you write the number as a decimal, all the digits after that $5$ are 0; $0.495000000000000\cdots$. Now, since $x=2.58$ clearly satisfies the desired criterion $(0.4951 > 0.495)$, we choose that value as the answer.
An alternative would be to do linear interpolation and say that $x=2.575$, midway between $x=2.57$ and $x=2.58$, is the right answer, but this would be incorrect. The value of $x$ at which $\Phi_{\text{CDF}}(x) = 0.9950$, which value of $x$ is denoted by $x_{0.9950}$, is just a little smaller than $2.58$. According to the tables on pages 968-871 of Abramowitz and Stegun, Handbook of Mathematical Functions, $\Phi_{\text{CDF}}(2.58) = 0.9950599642\cdots$ (and so $\Phi_{\text{OP}}(2.58) = 0.4950599642\cdots$), and so the exact value of $x_{0.9950}$ is definitely larger than the $2.575$ value obtained by linear interpolation.