Step by Step Mode Calculation to Boost Your Math Skills

For $NB(r,p)$ where $r$ is the number of successes and the $p$ is the probability of success. Also we have:
$f(x)=(\genfrac{}{}{0ex}{}{r+x-<!--\; -\; -->1}{r-<!--\; -\; -->1})p^r{q}^x$
Shown that $f(x)=\frac{(r+x-<!--\; -\; -->1)q}{x}f(x-<!--\; -\; -->1)$. So:
$f(x)>f(x-<!--\; -\; -->1)⟺<!--\; ⟺\; -->\frac{(r-<!--\; -\; -->1)q}{p}>x$
$f(x)<f(x-<!--\; -\; -->1)⟺<!--\; ⟺\; -->\frac{(r-<!--\; -\; -->1)q}{p}<x$
And after a long calculation, I've found that the mode is the integer part of $\frac{(r-<!--\; -\; -->1)q}{p}$. Where I did wrong?

The mode of a binomial $B(n,p)$ distribution is equal to $k=\lfloor <!--\; \lfloor \; -->(n+1)p\rfloor <!--\; \rfloor \; -->$. I am wondering how to estimate the probability at this value, i.e. $(\genfrac{}{}{0ex}{}{n}{k})p^k(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->k}$, like what's the order of this value in terms of $n$. Is it exp($-<!--\; -\; -->\mathrm{\Omega <!--\; \Omega \; -->}(n)$)?

Find the mode, we need to find the value of $x$ for which $f^{\prime }(x)=0$ and $f^{″}(x)<0$. After applying derivative to the pdf, how do we proceed?

A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?
From the information, I got the following information:
11 integers with a mean of 10 means a total must be 110.
A unique mode means that there must be at least two 8's.
if there are two 8's, and the median is 9, there must be at least two numbers greater than 9 also.

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Calculator mode. ENG mode?? (SCI vs. FLO vs. ENG)
When I type 8 ∗ 6
FLO = $48$
SCI = $4.8\ast <!--\; \ast \; -->{10}^1$
ENG = $48\ast <!--\; \ast \; -->{10}^0$
1) What is the point of ENG mode??
2) What are situations where you want answers in SCIentific notation?

for the frequency distribution given below $v=-<!--\; -\; -->{\displaystyle \frac{1}{6}},-<!--\; -\; -->{\displaystyle \frac{1}{4}},0,{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}}$
$f=12,16,21,8,27$
find mean ,median mode what percent of the population is non negative what percent of the population is negative valued

Consider a "discrete" random variable $X$. A mode of $X$ is just a maximizer of $P(X=x)$. This is obviously useful, and we can easily see that a mode is a "most likely" value for $X$.
If, instead, we have a "continuous" real-valued random variable $X$ with a PDF $f_X$, I think we usually define a mode of $X$ to be a maximizer of $f_X$. I have two questions:
1. How can we interpret the mode of a continuous random variable? In other words, why is the mode of a continuous random variable useful to probability theory?
2. Is there a more general definition of mode, removing the assumptions above that $X$ is real-valued and has a PDF?

Given a sample (the scope is 72 elements) with mode=54 mean=55,7 median=54,5. The 73th value of the extended sample is 56. What can I say about the mode, median and mean of the extended sample?

Mentioned that the mode of negative binomial distribution can be found by: t = 1 + ((r-1)/p) where t is some number, r is the number of successes and p is the probability of success If t is an integer, there will be 2 modes at t and t-1 If t is not an integer, the integer part of t is the mode.
This is what I have worked on so far: ((r-1)/p) is the expected number of attempts to achieve all required successes excluding the final success The expected number of attempts + the definite final success = the total number of expected attempts.The total number of expected attempts is the mode.
Why could there be two modes of t-1 and t?
$P(X=x)=((x-<!--\; -\; -->1)choose(r-<!--\; -\; -->1))(p^r)(q^{(x-<!--\; -\; -->r)})$
$((t-<!--\; -\; -->2)choose(r-<!--\; -\; -->1))(p^r)(q^{(t-<!--\; -\; -->r-<!--\; -\; -->1)})=((t-<!--\; -\; -->1)choose(r-<!--\; -\; -->1))(p^r)(q^{(t-<!--\; -\; -->r)})$
$((t-<!--\; -\; -->2)choose(r-<!--\; -\; -->1))(q^{-<!--\; -\; -->1})=((t-<!--\; -\; -->1)choose(r-<!--\; -\; -->1))$
$((t-<!--\; -\; -->2)(t-<!--\; -\; -->3)...(t-<!--\; -\; -->r))/(r-<!--\; -\; -->1)!x(q^{-<!--\; -\; -->1})=((t-<!--\; -\; -->1)(t-<!--\; -\; -->2)...(t-<!--\; -\; -->r+1))/(r-<!--\; -\; -->1)!$
$(t-<!--\; -\; -->r)/q=t-<!--\; -\; -->1$ [I am stuck here]

The probability density function of the random variable $X$ is defined by
$f(x)=\{\begin{array}{ll}4(x-<!--\; -\; -->x^3),& 0\le <!--\; \le \; -->x\le <!--\; \le \; -->1\\ 0,& \text{elsewhere}\end{array}$
What is the probability that three independent observations from the distribution of $X$ are all less than the mode of $X$?
This is a question I got incorrect. I have very little experience with mode and I understand that it's the value of the random variable with the highest probability. The solution for finding the mode is as follows:
The max point occurs when $f^{\prime }(x)=0$
$f^{\prime }(x)=4-<!--\; -\; -->12x^2=0$
$x=\sqrt{\frac{1}{3}}$
This is the max point or mode because $f^{″}(x)$ is a negative number.

How can I calculate the mode in a grouped frequency distribution when the largest frequency occurs in two or more classes?

Suppose
where $x$ is normal with mean $\mathrm{\mu <!--\; \mu \; -->}$ and variance $\mathrm{\sigma <!--\; \sigma \; -->}$. Then I see how to derive mode of $f(y)$ (distribution of $y$), as we need to find the value y that makes
$f^{\prime }(y)==0$
However, why is mode not simply
$e^{\mathrm{\mu <!--\; \mu \; -->}}$?
$y$ is a monotonic function of $x$, and so when $x$ reaches its mode, then y should also reach its mode. The mode of $x$ is its mean ($\mathrm{\mu <!--\; \mu \; -->}$) hence $y$'s mode should be
$e^{\mathrm{\mu <!--\; \mu \; -->}}$
what mistake have I made?

Looking for some real world examples for mode in Statistics involving topics which students like say Football or Social networks. Also they need to clearly identify differences in the usefulness of mode and mean. For example which player to pick for a football match depending on scores against a particular team while playing against that team. Mean doesnt make sense here. Any thoughts ?

Suppose that $X$ has the Binomial distribution with parameters $n,p$ . How can I show that if $(n+1)p$ is integer then $X$ has two mode that is $(n+1)p$ or $(n+1)p-<!--\; -\; -->1?$?