# Mode problems and answers

Recent questions in Mode
Amber Quinn 2022-06-17 Answered

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

Emanuel Keith 2022-06-13 Answered

### 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.

Jenna Beasley 2022-06-04 Answered

### for the frequency distribution given below $v=-\frac{1}{6},-\frac{1}{4},0,\frac{1}{2},\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

Yasmin Camacho 2022-05-28 Answered

### 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\left(X=x\right)=\left(\left(x-1\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(x-r\right)}\right)$$\left(\left(t-2\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(t-r-1\right)}\right)=\left(\left(t-1\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(t-r\right)}\right)$$\left(\left(t-2\right)choose\left(r-1\right)\right)\left({q}^{-1}\right)=\left(\left(t-1\right)choose\left(r-1\right)\right)$$\left(\left(t-2\right)\left(t-3\right)...\left(t-r\right)\right)/\left(r-1\right)!x\left({q}^{-1}\right)=\left(\left(t-1\right)\left(t-2\right)...\left(t-r+1\right)\right)/\left(r-1\right)!$$\left(t-r\right)/q=t-1$ [I am stuck here]

Matilda Webb 2022-05-09 Answered

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

Karissa Sosa 2022-05-07 Answered

### Supposewhere $x$ is normal with mean $\mu$ and variance $\sigma$. Then I see how to derive mode of $f\left(y\right)$ (distribution of $y$), as we need to find the value y that makes${f}^{\prime }\left(y\right)==0$However, why is mode not simply${e}^{\mu }$?$y$ 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 ($\mu$) hence $y$'s mode should be${e}^{\mu }$what mistake have I made?

Taliyah Spencer 2022-05-03 Answered

### 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 ?

Maeve Bowers 2022-05-03 Answered