4
$\begingroup$

The formula for finding the mode in grouped data is given by:

$$ mode = l + \frac {f_1 - f_0}{2 f_1 - f_0 - f_2} X h $$

where,

l = the lower limit of the modal class,

$ f_1 $ = the frequency of the modal class, $ f_0 $ = the frequency of the class preceding the modal class, $ f_2 $ = the frequency of the class succeeding the modal class, h = class width.

There's already a good answer here; an excerpt:

"Now, observe that: $$ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} + \frac{f_1 - f_2}{2f_1 - f_0 - f_2} = \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} + \frac{f_1 - f_2}{(f_1 - f_0) + (f_1 - f_2)} = 1 $$ So if we want to divide an interval of width h into two pieces, where the ratio of sizes of those two pieces is $ (f_1 - f_0) : (f_1 - f_2) $, ), the first piece will have width $ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} h $. This is what the formula for estimating the mode does. It splits the width of the modal bar into two pieces whose ratio of widths is $ (f_1 - f_0) : (f_1 - f_2) $, , and it says the mode is at the line separating those two pieces, that is, at a distance $ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} h $, from the left edge of that bar, $ l $."

The answer does a very good job of explaining what the formula is, but it doesn't touch on:

why we'd expect the mode to be at the line separating the two pieces. Why can't the mode be somewhere else?

I understand that this is approximating, but why do we use this particular approximation?

Further, why do we use the differences between $ f_1 $ and $ f_0 $ & $ f_1 $ and $ f_2 $:

why do we care how much the frequency of the modal class is higher or lower than the frequencies of the classes preceding or succeeding it?

$\endgroup$
0
$\begingroup$

This formula works exactly if the distribution is continuous with pdf $p(u)=au^2+bu+c$. Furthermore, that's the simplest pdf which yields an easy mode and can be fitted exactly to the data.

We solve for $a,b,c$ in terms of $f_0, f_1, f_2$ in the equations

$$f_0 =\! \int_{L-h}^{L} p(u)du, \ \ f_1 =\! \int_{L}^{L+h} p(u)du, \ \ f_2 =\! \int_{L+h}^{L+2h} p(u)du$$

This gives $$a= \frac{f_0-2 f_1+f_2}{2 h^3},\ b= \frac{(f_1-f_0)h- (f_0-2 f_1+f_2)L}{h^3}$$

Substituting those values gives the desired formula: $$mode = \frac{-b}{2a} = L + \frac{f_1-f_0}{2f_1-f_0-f_2}h.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.