Step by Step Mode Calculation to Boost Your Math Skills

Mean of the squares of the deviations from mean is called the:
Variance
Standard deviation
Quartile deviation
Mode

Tell whether the statement is true or false : A data always has a mode.

Find the mean, median, and mode for the following sample of scores: 6, 2, 4, 1, 2, 2, 3, 4, 3, 2. Clearly show all steps to get answers. You need to type the formula for the mean and then calculate it. For the median and mode you need to explain how you got the answers

$X$ is a discrete uniform distribution on $1,2,\dots <!--\; \dots \; -->,n$. I know that the median is $\frac{n+1}{2}$ for odd $n$. I need to find median when $n$ is even. Would it be $\frac{n}{2}$ or $\frac{n}{2}+1$, whichever is greater?
Also, is every point mode as PDF has highest values there? So there are $n$ modes - $1,2,\dots <!--\; \dots \; -->,n$?

A random variable, X, is defined as X ~ Geo(p). I know the mode is 1 as it is the value of X with highest probability.
How do i show this? As this is a discrete R.V, can i be allowed to use Calculus?

If we have $N$ sets, $\{A_1,\dots <!--\; \dots \; -->,A_N\}$, and we form a set $S$ by taking the sum of each element in the set with each element in the other sets, what can we say about the mode of $S$?
Intuitively, I would like to think that we can simply take the sum of the modes, i.e:
$\mathrm{Mode}<!--\; \; -->(S)=\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}\mathrm{Mode}<!--\; \; -->(A_n)$
However, this seems unlikely, especially as we would expect that $\mathrm{Mode}<!--\; \; -->(A_n)$ could potentially be a set of values, rather than a single value.
So I was wondering if we'd be able to relax this condition to state that $\mathrm{Mode}<!--\; \; -->(S)\subseteq <!--\; \subseteq \; -->\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}\mathrm{Mode}<!--\; \; -->(A_n)$, where we define $\mathrm{Mode}<!--\; \; -->(A)+\mathrm{Mode}<!--\; \; -->(B)$ as the set formed by taking the sum of each element in $\mathrm{Mode}<!--\; \; -->(A)$ with each element in $\mathrm{Mode}<!--\; \; -->(B)$, formally:
$\mathrm{Mode}<!--\; \; -->(S)\subseteq <!--\; \subseteq \; -->\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}\mathrm{Mode}<!--\; \; -->(A_n)=\{\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}{x}_n:x_i\in <!--\; \in \; -->A_i\}$
This seems to be true, but I was wondering if we could say anything stronger?

According to the definition I read, it came to my notice that the number with highest frequency has to be a mode for a given data set, but then what if I have all the numbers as distinct... In that scenario we won't have a particular number having a frequency more than other elements in the data set... Now if I consider a case when we have 2 numbers in a dataset with same max number of occurrences like:
$2,3,4,5,3,2$
Here 2, 3 both happen to have same maximum frequency and thus we say there are 2 modes... The above is stated similar in case we have 3 modes or multi modes ... So if there are all distinct numbers then we would have each number having the same maximum frequency as 1 ..so we can say all the numbers are modes...for that dataset...But then I have seen on some websites claiming that such data sets have "NO MODE".

Five test scores have a mean of 91, a median of 93, and a mode of 95. The possible scores on the tests are from 0 to 100. a) What is the sum of the lowest two test scores? b) What are the possible values of the lowest two test scores?

Given the mean, median and mode of a function and have to find the probability density function.
mean: $\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}{\mathrm{\Gamma <!--\; \Gamma \; -->}}_1$
median: $\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}(ln2{)}^{1/\mathrm{\delta <!--\; \delta \; -->}}$
mode: $\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}(1-<!--\; -\; -->1/\mathrm{\delta <!--\; \delta \; -->}{)}^{1/\mathrm{\delta <!--\; \delta \; -->}}$
Also given that
${\mathrm{\Gamma <!--\; \Gamma \; -->}}_k=\mathrm{\Gamma <!--\; \Gamma \; -->}(1+k/\mathrm{\delta <!--\; \delta \; -->})$
$\mathrm{\Gamma <!--\; \Gamma \; -->}(z)={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{t}^{z-<!--\; -\; -->1}dt$
$-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}<x<\mathrm{\gamma <!--\; \gamma \; -->},\mathrm{\beta <!--\; \beta \; -->}>0,\mathrm{\gamma <!--\; \gamma \; -->}>0$
Now I understand how to calculate the mean, mode and median when given a probability density function. However I'm struggling to go backwards. I initially tried to "reverse" the process by differentiating the mean or median however I know this is skipping the substitution over the given limit.
I then looked for patterns with known distributions and realised they are from Weibull distribution however $\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->$. Does this mean essentially this is a typical Weibull distribution however shifted by $\mathrm{\gamma <!--\; \gamma \; -->}$ and therefore the pdf will be $\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->Weibullpdf"$

Consider random variable $Y$ with a Poisson distribution:
$P(y|\mathrm{\theta <!--\; \theta \; -->})=\frac{{\mathrm{\theta <!--\; \theta \; -->}}^y{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}}}{y!},y=0,1,2,\dots <!--\; \dots \; -->,\mathrm{\theta <!--\; \theta \; -->}>0$
$P(y|\mathrm{\theta <!--\; \theta \; -->})=\frac{{\mathrm{\theta <!--\; \theta \; -->}}^y{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}}}{y!},y=0,1,2,\dots <!--\; \dots \; -->,\mathrm{\theta <!--\; \theta \; -->}>0$
Mean and variance of $Y$ given $\mathrm{\theta <!--\; \theta \; -->}$ are both equal to $\mathrm{\theta <!--\; \theta \; -->}$. Assume that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i>1$.
If we impose the prior $p\propto <!--\; \propto \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}$, then what is the Bayesian posterior mode?
I was able to calculate the likelihood and the posterior, but I'm having trouble calculating the mode so I'm wondering if I got the right posterior:
$P(\mathrm{\theta <!--\; \theta \; -->}|y)=likelihood\ast <!--\; \ast \; -->prior$
$P(\mathrm{\theta <!--\; \theta \; -->}|y)\propto <!--\; \propto \; -->({\mathrm{\theta <!--\; \theta \; -->}}^{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i}{e}^{-<!--\; -\; -->n\mathrm{\theta <!--\; \theta \; -->}})({\mathrm{\theta <!--\; \theta \; -->}}^{-<!--\; -\; -->1})$
$P(\mathrm{\theta <!--\; \theta \; -->}|y)\propto <!--\; \propto \; -->{\mathrm{\theta <!--\; \theta \; -->}}^{(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i)-<!--\; -\; -->1}{e}^{-<!--\; -\; -->n\mathrm{\theta <!--\; \theta \; -->}}$

The operation is of two step.
1. Bin a data in 10 bins. (the distribution is unimodal) and
2. Then find the bin with maximum density.
In other words finding the mode of a distribution.

Suppose that $X_1,X_2,\dots <!--\; \dots \; -->,X_n$ are IID Bernoulli random variables with success probability equal to an unknown parameter $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->[0,1]$.
Let $A$ and $B$ be nonnegative constants. If we impose the prior $\mathrm{\pi <!--\; \pi \; -->}(\mathrm{\theta <!--\; \theta \; -->})\propto <!--\; \propto \; -->({\mathrm{\theta <!--\; \theta \; -->}}^a)((1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^b)$, then what is the Bayesian posterior mode?

A formula to calculate the mode for grouped data's is given:
Mode = $l+{\displaystyle \frac{(f_1-<!--\; -\; -->f_0)h}{2f_1-<!--\; -\; -->f_0-<!--\; -\; -->f_2}}$
Where, $l=$ lower limit of the modal class,
$h=$ size of the class interval,
$f_1=$ frequency of the modal class,
$f_0=$ frequency of the class preceding the modal class,
$f_2=$ frequency of the class succeeding the modal class.
Explain the derivation of this formula.

The distribution is left-skewed if meanThe distribution is right-skewed if mean>median>mode.
Can mode lie between mean and median?

Given is a lognormal distribution with median $e$ and mode $\sqrt{e}$. What is the variance of the lognormal distribution?
Not sure how to solve this. A variable $Y$ has a lognormal distribution if $\log <!--\; \; -->(Y)$ has a normal distribution. So I'm thinking you can solve the question by finding the mean and standard deviation of the associated normal distribution by using the given median and mode. But I don't know how to. For a normal distribution, the median and mode equal the mean, but for a lognormal distribution they evidently do not. How to use these values to find the variance?

need to generate some random data from lognormal distribution, where I set the mode and standart deviation of that lognormal distribution. For this purpose I choose to use random numbers generator from lognormal distribution. This generator takes two numbers, that are mean and sd of underlying normal distribution.
So far its clear I need to derive mean and sd of normal distribution, which is underlaying for lognormal distribution where I know mode and sd. I know the equations for derivation of mean and sd:
NOTATION:
$n(x)=$ mean of normal distribution
$sd(x)=sd$ of normal distribution
$n(y)=$ mean of lognormal distribution
$sd(y)=$ sd of lognormal distribution
$mode(y)=$ mode of lognormal distribution
EQUATIONS:
$n(x)=2\ast <!--\; \ast \; -->ln(n(y))-<!--\; -\; -->(1/2)\ast <!--\; \ast \; -->ln(sd(y{)}^2+n(y{)}^2)$
$sd(x)=-<!--\; -\; -->2\ast <!--\; \ast \; -->ln(n(y))+ln(sd(y{)}^2+n(y{)}^2)$
$mode(y)=exp(n(y)-<!--\; -\; -->sd(y{)}^2)$
Here I stuck because I cant get the equation for $n(y)$ from these equations, that I need to compute $n(x)$. So far I ended:
$mode(y)=exp(4\ast <!--\; \ast \; -->ln(n(y))-<!--\; -\; -->3/2\ast <!--\; \ast \; -->ln(n(y{)}^2-<!--\; -\; -->sd(y{)}^2))$
$mode(y{)}^{2/3}\ast <!--\; \ast \; -->sd(y{)}^2=n(y{)}^2\ast <!--\; \ast \; -->(n(y{)}^{2/3}-<!--\; -\; -->mode(y{)}^{2/3})$
Can anybody help me to complete this derivation?

How mean i.e. average be equal to mode? What I understand is that median will be the point that is equidistant from the two extremes so here the peak in the bell curve will be the median.
Mode is the item that has occurred frequently - so the peak point has occurred twice - also , the extremes has occurred twice. but I take pairs (x, y)then probably I can say that the peak is the mode.
What I can't understand is the average? How the average will be the highest point. Basically, If I just consider three points - two extreme and the highest then the average will not be the highest point. So, I couldn't understand that the bell curve has the median, mode and mean equal.

Used the ECB mode (with block length $4$) to encrypt the message $m=1011000101001010$ into $c=0010011001001101$ using the key
$\mathrm{\pi <!--\; \pi \; -->}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)$
and initialisation vector $IV=1010$. Now I to decrypt it we used the key
${\mathrm{\pi <!--\; \pi \; -->}}^{-<!--\; -\; -->1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)$
however isn't the inverse of a permutation it written backwards? In this case shouldn't we have
${\mathrm{\pi <!--\; \pi \; -->}}^{-<!--\; -\; -->1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 4& 3& 2\end{array}\right)$
Why is it the first one?

Suppose
$f_X(x)=\{\begin{array}{cc}0.5& x=0\\ x& 0<x\le <!--\; \le \; -->1\end{array}$
What is the MODE of this distribution? I think it should be $0$ because it has the highest mass, even though $1$ has the highest density.
Another one
$f_X(x)=\{\begin{array}{cc}x& 0<x\le <!--\; \le \; -->1\\ 1& 1<x\le <!--\; \le \; -->1.5\end{array}$
What would be the MODE in this case? $1?1.5?$ or any number in $[1,1.5]$