Recent questions in Mode

Descriptive StatisticsAnswered question

vomi2xn 2023-03-06

Mean of the squares of the deviations from mean is called the:

Variance

Standard deviation

Quartile deviation

Mode

Variance

Standard deviation

Quartile deviation

Mode

Descriptive StatisticsAnswered question

Savanne1bo 2022-12-30

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

Descriptive StatisticsAnswered question

Annie French 2022-11-16

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

Descriptive StatisticsAnswered question

MUHAMAD LUQMANUL ARIEF ZULKIFLEE2022-08-06

At the Faculty of Computing, 53% of the students are male and 47% are female. 35% of the

males and 20% of the females from this group major in Computer Networking.

Find the probability that a student selected at random from this group is majoring in

Computer Networking.

Descriptive StatisticsAnswered question

Keenan Santos 2022-07-14

$X$ is a discrete uniform distribution on $1,2,\dots ,n$. 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 - $1,2,\dots ,n$?

Also, is every point mode as PDF has highest values there? So there are $n$ modes - $1,2,\dots ,n$?

Descriptive StatisticsAnswered question

glitinosim3 2022-07-09

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?

How do i show this? As this is a discrete R.V, can i be allowed to use Calculus?

Descriptive StatisticsAnswered question

Banguizb 2022-07-09

If we have $N$ sets, $\{{A}_{1},\dots ,{A}_{N}\}$, 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:

$\mathrm{Mode}(S)=\sum _{n=1}^{N}\mathrm{Mode}({A}_{n})$

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 $\mathrm{Mode}(S)\subseteq \sum _{n=1}^{N}\mathrm{Mode}({A}_{n})$, 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:

$\mathrm{Mode}(S)\subseteq \sum _{n=1}^{N}\mathrm{Mode}({A}_{n})=\{\sum _{n=1}^{N}{x}_{n}:{x}_{i}\in {A}_{i}\}$

This seems to be true, but I was wondering if we could say anything stronger?

Intuitively, I would like to think that we can simply take the sum of the modes, i.e:

$\mathrm{Mode}(S)=\sum _{n=1}^{N}\mathrm{Mode}({A}_{n})$

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 $\mathrm{Mode}(S)\subseteq \sum _{n=1}^{N}\mathrm{Mode}({A}_{n})$, 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:

$\mathrm{Mode}(S)\subseteq \sum _{n=1}^{N}\mathrm{Mode}({A}_{n})=\{\sum _{n=1}^{N}{x}_{n}:{x}_{i}\in {A}_{i}\}$

This seems to be true, but I was wondering if we could say anything stronger?

Descriptive StatisticsAnswered question

rjawbreakerca 2022-07-09

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".

$2,3,4,5,3,2$

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".

Descriptive StatisticsAnswered question

Araceli Clay 2022-07-08

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?

Descriptive StatisticsAnswered question

prirodnogbk 2022-07-07

Given the mean, median and mode of a function and have to find the probability density function.

mean: $\gamma -\beta {\mathrm{\Gamma}}_{1}$

median: $\gamma -\beta (ln2{)}^{1/\delta}$

mode: $\gamma -\beta (1-1/\delta {)}^{1/\delta}$

Also given that

${\mathrm{\Gamma}}_{k}=\mathrm{\Gamma}(1+k/\delta )$

$\mathrm{\Gamma}(z)={\int}_{0}^{\mathrm{\infty}}{t}^{z-1}dt$

$-\mathrm{\infty}<x<\gamma ,\beta >0,\gamma >0$

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 $\gamma -$. Does this mean essentially this is a typical Weibull distribution however shifted by $\gamma $ and therefore the pdf will be $\gamma -Weibullpdf"$

mean: $\gamma -\beta {\mathrm{\Gamma}}_{1}$

median: $\gamma -\beta (ln2{)}^{1/\delta}$

mode: $\gamma -\beta (1-1/\delta {)}^{1/\delta}$

Also given that

${\mathrm{\Gamma}}_{k}=\mathrm{\Gamma}(1+k/\delta )$

$\mathrm{\Gamma}(z)={\int}_{0}^{\mathrm{\infty}}{t}^{z-1}dt$

$-\mathrm{\infty}<x<\gamma ,\beta >0,\gamma >0$

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 $\gamma -$. Does this mean essentially this is a typical Weibull distribution however shifted by $\gamma $ and therefore the pdf will be $\gamma -Weibullpdf"$

Descriptive StatisticsAnswered question

spockmonkey40 2022-07-07

Consider random variable $Y$ with a Poisson distribution:

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

Mean and variance of $Y$ given $\theta $ are both equal to $\theta $. Assume that $\sum _{i=1}^{n}{y}_{i}>1$.

If we impose the prior $p\propto \frac{1}{\theta}$, 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:

$P(\theta |y)=likelihood\ast prior$

$P(\theta |y)\propto ({\theta}^{\sum _{i=1}^{n}{y}_{i}}{e}^{-n\theta})({\theta}^{-1})$

$P(\theta |y)\propto {\theta}^{(\sum _{i=1}^{n}{y}_{i})-1}{e}^{-n\theta}$

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

$P(y|\theta )=\frac{{\theta}^{y}{e}^{-\theta}}{y!},y=0,1,2,\dots ,\theta >0$

Mean and variance of $Y$ given $\theta $ are both equal to $\theta $. Assume that $\sum _{i=1}^{n}{y}_{i}>1$.

If we impose the prior $p\propto \frac{1}{\theta}$, 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:

$P(\theta |y)=likelihood\ast prior$

$P(\theta |y)\propto ({\theta}^{\sum _{i=1}^{n}{y}_{i}}{e}^{-n\theta})({\theta}^{-1})$

$P(\theta |y)\propto {\theta}^{(\sum _{i=1}^{n}{y}_{i})-1}{e}^{-n\theta}$

Descriptive StatisticsAnswered question

lilmoore11p8 2022-06-30

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.

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.

Descriptive StatisticsAnswered question

Bailee Short 2022-06-27

Suppose that ${X}_{1},{X}_{2},\dots ,{X}_{n}$ are IID Bernoulli random variables with success probability equal to an unknown parameter $\theta \in [0,1]$.

Let $A$ and $B$ be nonnegative constants. If we impose the prior $\pi (\theta )\propto ({\theta}^{a})((1-\theta {)}^{b})$, then what is the Bayesian posterior mode?

Let $A$ and $B$ be nonnegative constants. If we impose the prior $\pi (\theta )\propto ({\theta}^{a})((1-\theta {)}^{b})$, then what is the Bayesian posterior mode?

Descriptive StatisticsAnswered question

freakygirl838w 2022-06-24

A formula to calculate the mode for grouped data's is given:

Mode = $l+{\displaystyle \frac{({f}_{1}-{f}_{0})h}{2{f}_{1}-{f}_{0}-{f}_{2}}}$

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.

Mode = $l+{\displaystyle \frac{({f}_{1}-{f}_{0})h}{2{f}_{1}-{f}_{0}-{f}_{2}}}$

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.

Descriptive StatisticsAnswered question

flightwingsd2 2022-06-21

The distribution is left-skewed if meanThe distribution is right-skewed if mean>median>mode.

Can mode lie between mean and median?

Can mode lie between mean and median?

Descriptive StatisticsAnswered question

rose2904ks 2022-06-21

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 $\mathrm{log}(Y)$ 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?

Not sure how to solve this. A variable $Y$ has a lognormal distribution if $\mathrm{log}(Y)$ 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?

Descriptive StatisticsAnswered question

George Bray 2022-06-21

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:

$n(x)=2\ast ln(n(y))-(1/2)\ast ln(sd(y{)}^{2}+n(y{)}^{2})$

$sd(x)=-2\ast ln(n(y))+ln(sd(y{)}^{2}+n(y{)}^{2})$

$mode(y)=exp(n(y)-sd(y{)}^{2})$

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:

$mode(y)=exp(4\ast ln(n(y))-3/2\ast ln(n(y{)}^{2}-sd(y{)}^{2}))$

$mode(y{)}^{2/3}\ast sd(y{)}^{2}=n(y{)}^{2}\ast (n(y{)}^{2/3}-mode(y{)}^{2/3})$

Can anybody help me to complete this derivation?

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:

$n(x)=2\ast ln(n(y))-(1/2)\ast ln(sd(y{)}^{2}+n(y{)}^{2})$

$sd(x)=-2\ast ln(n(y))+ln(sd(y{)}^{2}+n(y{)}^{2})$

$mode(y)=exp(n(y)-sd(y{)}^{2})$

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:

$mode(y)=exp(4\ast ln(n(y))-3/2\ast ln(n(y{)}^{2}-sd(y{)}^{2}))$

$mode(y{)}^{2/3}\ast sd(y{)}^{2}=n(y{)}^{2}\ast (n(y{)}^{2/3}-mode(y{)}^{2/3})$

Can anybody help me to complete this derivation?

Descriptive StatisticsAnswered question

Sarai Davenport 2022-06-20

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.

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.

Descriptive StatisticsAnswered question

gledanju0 2022-06-20

Used the ECB mode (with block length $4$) to encrypt the message $m=1011000101001010$ into $c=0010011001001101$ using the key

$\pi =\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)$

and initialisation vector $IV=1010$. Now I to decrypt it we used the key

${\pi}^{-1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)$

however isn't the inverse of a permutation it written backwards? In this case shouldn't we have

${\pi}^{-1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 4& 3& 2\end{array}\right)$

Why is it the first one?

$\pi =\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)$

and initialisation vector $IV=1010$. Now I to decrypt it we used the key

${\pi}^{-1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)$

however isn't the inverse of a permutation it written backwards? In this case shouldn't we have

${\pi}^{-1}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 4& 3& 2\end{array}\right)$

Why is it the first one?

Descriptive StatisticsAnswered question

Amber Quinn 2022-06-17

Suppose

${f}_{X}(x)=\{\begin{array}{cc}0.5& x=0\\ x& 0<x\le 1\end{array}$

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

${f}_{X}(x)=\{\begin{array}{cc}x& 0<x\le 1\\ 1& 1<x\le 1.5\end{array}$

What would be the MODE in this case? $1?\text{}\text{}1.5?$ or any number in $[1,1.5]$

${f}_{X}(x)=\{\begin{array}{cc}0.5& x=0\\ x& 0<x\le 1\end{array}$

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

${f}_{X}(x)=\{\begin{array}{cc}x& 0<x\le 1\\ 1& 1<x\le 1.5\end{array}$

What would be the MODE in this case? $1?\text{}\text{}1.5?$ or any number in $[1,1.5]$

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These are usually used by investors, financial managers, banking experts, and specialists working in the investment area. The mode of a data set is exactly where the values can be encountered most often. Regardless of whether you are focusing on the equations or want to explore more than one data set, the use of mode examples will help you to determine the correct answers. If you feel uncertain and need more help with your descriptive statistics, the mode calculation should be used based on the specific questions that you see asked and resolved by others in the solutions provided below.