In truth, this is only one of many possible estimates for the mode, when the data are binned/grouped. You could construct a continuous probability distribution, and depending on how you discretize or "bin" the outcomes, you could get different modes.

Let us illustrate with an example. Suppose $$X \sim \operatorname{Gamma}(3, 1)$$ with density $$f_{X}(x) = \frac{x^2}{2} e^{-x}, \quad x > 0.$$ The true mode of this distribution is found by computing the derivative and looking for critical points: $$0 = f'(x) = -\frac{x^2}{2}(x-2) e^{-x},$$ hence $x = 2$ is the exact mode.

Now suppose we discretize the density into integer width bins, i.e., let $$Y = \lfloor X \rfloor,$$ so that $$\Pr[Y = y] = \Pr[y \le X < y+1] = \frac{1}{2} \int_{x=y}^{y+1} x^2 e^{-x} \, dx.$$ This is not difficult to compute exactly: $$\Pr[Y = y] = \frac{e^{-(y+1)}}{2} \left(-5 + 2e + 2(e-2)y + (e-1)y^2\right).$$ From this, we can compute $$\Pr[Y = 1] = \frac{5(e-2)}{2e^2} = f_0, \\ \Pr[Y = 2] = \frac{10e-17}{2e^3} = f_1, \\ \Pr[Y = 3] = \frac{17e-26}{2e^4} = f_2.$$ Using the formula provided, and with $l_0 = 2$, we have compute the mode as $$2 + \frac{\frac{10e-17}{2e^3} - \frac{5(e-2)}{2e^2}}{\frac{10e-17}{2e^3} - \frac{5(e-2)}{2e^2} + \frac{10e-17}{2e^3} - \frac{17e-26}{2e^4}} \cdot \frac{10e-17}{2e^3} \approx 2.0336342,$$ but we already knew that this calculation would yield a number strictly greater than $2$.

If, however, we binned the data differently, e.g. $$W = \lfloor X + 1/2 \rfloor,$$ we have $$\Pr[W = w] = \Pr[w - 1/2 \le X < w + 1/2] = \frac{e^{-(w+1/2)}}{8} \left(-13 + 5e + 4(e-3)w + 4(e-1) w^2\right),$$ and the resulting estimate for the mode is (calculations omitted) $1.67949$.

So what we can take away from this is that when data are binned from an underlying continuous distribution, you really can't tell where the mode is within the bin with the highest count, or even if the bin with the highest count actually contains the true mode.

They weren't derived, but rather DEFINED. Imagine you have a long list of numbers and you want to represent them all by one representative number. You want to be able to say what numbers in this list are generally like. The best way to do that will vary. In the scenario that all numbers are the same 3, 3, 3, 3 ,3 ... Obviously this number should be three. But what about other situation? If there's a a range of numbers, then you'll have some high and some low, so to represent it with one number you want to get the number "in the middle", but what number would that be? Well you could order all the numbers and see which number is then physically in the middle of the list. (This is a fine choice and called the median) Sometimes there are (for example) many high numbers and few low numbers, so the "middle number" (the median number) might end up disregarding the low numbers entirely. For example: 1, 2, 3, 1000, 1002, 1003, 1005 has a median of 1000, which doesn't really capture the start of the list You could instead share out the numbers evenly across the list until every number is the same. Then use that number. That way, every number in the list effects the outcome. ie: 0,2,2,5,6 There's a total of 15 here. Which can be shared amongst the entries to get a new list that is "similar" 3,3,3,3,3 This approach results in a number that we call the mean. It's trying to do the same job as the median, it just does it differently. Finally you might have a list where there are lots of repetitions. In this case, just counting the most common number might be most sensible (the mode)

Hint: To solve this question, we will first make 3 bar graphs of frequencies \\[{{f}_{1}},{{f}_{2}},{{f}_{3}}.\\] Mode is the value of the highest bar as that is of the maximum frequency. Finally, we will calculate the midpoint of the largest bar to get the value of the mode formula. Complete step-by-step answer:Let us first define the mode for grouped data. The mode of a list of data values is simply the most common values (or the values if any). When the data are grouped as in a histogram, we will normally talk only about the modal class (the class, or group with the greatest frequency) because we don’t know the individual values. The derivation of the mode formula is given by using the bar graph.\n \n \n \n \n Let the frequency of the modal class be \\[{{f}_{1}}.\\] The frequency of the class first after the modal class is \\[{{f}_{2}}.\\] From the above figure, we see that, triangle AEB is similar to triangle DEC. \\[\\Rightarrow \\Delta AEB\\sim \\Delta DEC\\]The relative side ratio is also equal. \\[\\Rightarrow \\Delta AEB\\sim \\Delta DEC\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{BE}{DE}\\]And BE is nothing but \\[{{f}_{1}}-{{f}_{0}}\\] and \\[DE={{f}_{1}}-{{f}_{2}}.\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{BE}{DE}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{{{f}_{1}}-{{f}_{2}}}\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{{{f}_{1}}-{{f}_{2}}}\\]Again we have \\[\\Delta BEF\\sim \\Delta BDC\\] from the figure.\\[\\Rightarrow \\dfrac{FE}{BC}=\\dfrac{BE}{BD}\\]Clearly, \\[BE={{f}_{1}}-{{f}_{0}}\\] and \\[BD=BE+ED\\]\\[\\Rightarrow BD=\\left( {{f}_{1}}-{{f}_{0}} \\right)+\\left( {{f}_{1}}-{{f}_{2}} \\right)\\]\\[\\Rightarrow BD={{f}_{1}}-{{f}_{0}}+{{f}_{1}}-{{f}_{2}}\\]\\[\\Rightarrow BD=2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}\\]Therefore, we have, \\[\\dfrac{FE}{BC}=\\dfrac{BE}{BD}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\]\\[\\Rightarrow \\dfrac{FE}{BC}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\]\\[\\Rightarrow FE=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times BC\\]We know that \\[BC={{f}_{1}},\\] so we can write\\[\\Rightarrow FE=\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]Let, FE be x.\\[\\Rightarrow x=\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]Therefore, the mode can be obtained by adding this value of x to \\[{{I}_{0}}.\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+x\\]Substituting the value of x as obtained from above, we get, \\[\\Rightarrow \\text{Mode}={{I}_{0}}+x\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\dfrac{\\left( {{f}_{1}}-{{f}_{0}} \\right)}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times {{f}_{1}}\\]Hence, the mode formula is determined. \\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\dfrac{\\left( {{f}_{1}}-{{f}_{0}} \\right)}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times {{f}_{1}}\\]Note: We have used the bar graph to determine the mode formula. So, \\[{{f}_{0}}\\] is considered a point after the first bar and the midpoint of the highest bar is the mode. The highest bar is in the middle. So, we have assumed x = midpoint of the largest bar and hence calculate \\[{{I}_{0}}+x\\] to get the mode value.