Stepwise Descriptive Statistics Learning

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?

What is a numerical variable and what is a categorical variable?

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($-<!--\; -\; -->\mathrm{\Omega <!--\; \Omega \; -->}(n)$)?

A distribution has a standard deviation of o=10. Find the z-score for each of the following locations in the distribution.
a) Above the mean by 5 points b) Above the mean by 2 points c) Below the mean by 20 points d) Below the mean by 15 points
Please don't give me the answer, but explain how I would work this out.

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?

Prove that if diam(G)=3 for a regular graph G, then $diam(\stackrel{¯<!--\; ¯\; -->}{G})=2.$.
I know that without using regularity one can show that if $diam(G)\ge <!--\; \ge \; -->3,\text{then }diam(\stackrel{¯<!--\; ¯\; -->}{G})\le <!--\; \le \; -->3$. How do I use regularity to show that it must be exactly 2?

Boxplots and bar graphs
I'm studying for the GRE and came across several questions that I was unable to answer in a practice booklet, even after looking at the answer and trying to work backwards, and searching google and other sites for helpful hints. I think I am missing a fundamental understanding or useful heuristic for solving many of these problems; any advice would be greatly appreciated as my exam is Monday (Aug 1st).
1. Eight hundred insects were weighed, and the resulting measurements, in milligrams, are summarized in the boxplot below.
If the 80th percentile of the measurements is 130 milligrams, about how many measurements are between 126 milligrams and 130 milligrams?
I calculated the range (41), the quartiles(Q1=114, Q2=118, Q3=126), and the IQR (12), but I'm confused about the question. If the 80th percentile (so 80% of the measurements?) is at 130, then 640 are within this percentile. I'm not sure if this is true and even if it is, where to go from here. Each quartile is 25% of the data, correct? So from 126 to 146 must contain 25%, or 200 measurements? (1-.8)(200) = 40, but conceptually I'm lacking what that means.
2.This question refers to the following graph:

(a) In 2003 the family used a total of 49 percent of its gross annual income for two of the categories listed. What was the total amount of the family’s income used for those same categories in 2004 ?
My confusion lies with the fact that I can't seem to find any combination of two categories that add up to 49%. Plus, it seems that the total % expenditure in each year is 101%. The chart is not 100% accurately drawn, but even being very liberal in measuring there seems to be a discrepancy.

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?

I saw this somewhere. Is this right?
$P(z<-<!--\; -\; -->4.3333)=0.00003$
Most z-score table stops at 3.4. If this is right, kindly explain how it was gotten. Thanks

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?

From a given z-score, what is the formula for calculating the probability of the deviation being due to chance?
e.g.
Given z = 1.2 standard deviations
The chance probability p = 0.11507
How did we derive p? (i.e. the formula, without having to refer to the normal distribution table)

$X\sim <!--\; \sim \; -->N(15,4)$
Find P( X > 18.7 | X > 11.7 ).
Attempt: Rewrite as P ( x > 18.7) - P( x > 11.7)
Using the phi function
$\mathrm{\varphi <!--\; \varphi \; -->}(1.85)-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}(-<!--\; -\; -->1.65)=(0.9678)-<!--\; -\; -->(0.0495)=0.9183$
Where did I go wrong?

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]