I am confused about something relating to histograms.
A histogram is a visual tool to display the distribution of numerical data.
One example is when we have the binomial distribution of a set of successes of spinning a coin:
Taking into account that the width of a bar in the histogram is 1 the use of the integral is used to show the probability that e.g. we get from 2 up to 5 heads
i.e.
$\int_{2}^{5}y_{x} dx$ where x is the number of heads.
Question: Is it because we can derive a curve that goes through all the midpoints of the top of each bar and we can use the integral approach or is that irrelevant?
So basically I am confused as to whether we can sum the values that each bar represent to get the value in that range or the histogram must describe an area that a graph covers
So if we have an histogram that is completely asymetric e.g.
that does not seem that we know of a formula for a graph that matches the boundaries of the histogram can we still use a summation of the histogram bars to get the value for that range of values?
For the first histogram it is exactly the CLT (or Laplace-De Moivre) theorem. You can see the full proof here, however the bottom line is that your original Binomial r.v can be approximated by the Normal distribution for large enough $n$, namely if you have $Bin(n,p)$ thus $$ P(X=k) \approx \frac{1}{\sqrt{2\pi n p q}}\exp\{-\frac{(k-np)^2}{2npq}\}. $$ So you can use integration to calculate the approximate approximate the probability of any number of success, in particular $$ P(2\le X\le 5)\approx \int_{2}^5 \frac{1}{\sqrt{2\pi n p q}}\exp\{-\frac{(k-np)^2}{2npq}\}\mathrm{d}k . $$