Stepwise Descriptive Statistics Learning

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:
$P(\mathrm{\theta <!--\; \theta \; -->}|y)=likelihood\ast <!--\; \ast \; -->prior$
$P(\mathrm{\theta <!--\; \theta \; -->}|y)\propto <!--\; \propto \; -->({\mathrm{\theta <!--\; \theta \; -->}}^{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i}{e}^{-<!--\; -\; -->n\mathrm{\theta <!--\; \theta \; -->}})({\mathrm{\theta <!--\; \theta \; -->}}^{-<!--\; -\; -->1})$
$P(\mathrm{\theta <!--\; \theta \; -->}|y)\propto <!--\; \propto \; -->{\mathrm{\theta <!--\; \theta \; -->}}^{(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{y}_i)-<!--\; -\; -->1}{e}^{-<!--\; -\; -->n\mathrm{\theta <!--\; \theta \; -->}}$

I had a question about an instantaneous z score. The question reads as
$P(z=2)$
I think that this is 0, as z-score is the area under the curve and could be found by integrating the PDF over the range. As the range is 0, the area is also zero. Is that correct?

In a given population, a score of X= 88 corresponds to z= +2.00 and a score of X= 79 corresponds to z= -1.00. Find the mean and standard deviation for the population.

I know how to find X and z-scores as well as how to plug things into the z-score formula, but I'm not sure how to solve this one. It says to sketch out a distribution table and find where the mean and SD fall on it, but I'm not sure how to do that.

The 2006 Statistical Abstract of the United States reports on a survey that asked a national sample of 80,000 American households about pet ownership. Suppose that one-third of all American households own a pet cat. The survey discovered that 31.6% of all the households sampled owned a pet cat. What is the z-score of this?

From the standard deviation formula for a sample proportion, I found that standard deviation is 0.0016. From there, I plugged that into the z-score formula, and got (0.316-0.333)/0.0016 = -10.625. However, a z-score that high baffles me and I cannot imagine getting a z-score that high. Where did I go wrong?

I'm new to studying z-scores and I've been told that for a gaussian statistic, around 95% of the values lie within the area two standard deviations above and below the mean, which (in accordance to my interpretation) would imply,
${\int <!--\; \int \; -->}_{\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->2\mathrm{\sigma <!--\; \sigma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}+2\mathrm{\sigma <!--\; \sigma \; -->}}A{e}^{-<!--\; -\; -->((x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}{)}^2}\mathrm{d}x=0.95\ast <!--\; \ast \; -->{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{+\mathrm{\infty <!--\; \infty \; -->}}A{e}^{-<!--\; -\; -->((x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}{)}^2}\mathrm{d}x$
Firstly, am I correct in my presumption? and secondly, is there any way to calculate the integral on the left to prove this point mathematically?

I have been reading a Statistics book and they show the following formula for z-score:
$z=\frac{x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}$
And then they explain about the $\mathrm{\mu <!--\; \mu \; -->}$ and the $\mathrm{\sigma <!--\; \sigma \; -->}$ in the formula like this:

These are the mean and standard deviation of data containing the value x.

But is that so ? I don't think the mean and standard deviation should contain the value x.

On the first periodic exam in Statistics, the population mean was 70 and the population standard deviation was 9. How to determine the standard score of a student who got a score of 88 assuming that the scores are normally distribtued?

I need to figure out how to find what proportion of scores fall above a z score of 1.65. I'm having trouble and this hasn't been explained much in class. We have to use a z-table to get the correct scores so I'll supply them here.
In column b (area between mean and the z) the score is 0.4505 for a z score of 1.65. And in column c the score is .0495 for a z score of 1.65. I hope you guys can help me.
How do I figure out what proportion fall above 1.65?

I am comparing 2 variables and have an outlier in x of 82.27. I want to find the z score of this outlier. When using the z score equation x-mean/standard deviation, do I use the mean and standard deviation of my x values or y values?

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.

How can we use Cepheid variables and supernovae to determine stellar distances?

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?

What type of data should a chi-squared test be applied to?

I have learned that z-scores are "Measures of Relative Standing" and that they can be used to compare two scores that are from different normal distributions.

For example if I wanted to compare how I performed

with a 85% in a math class with mean 70 and std dev 15;
and
with a 70% in a english class with mean 50 and std dev 15;

I would use the z-score formula to calculate a relative value for two situation and use it to compare them.

However why is it not enough just to compare each classes with the mean and use this as a basis for comparison? For example

85% -70% = 15%
70% - 50% = 20%
20 > 15 therefore I must have performed better in english

I know this is incorrect but I don't understand why. thx for the help

I was given a question where I was supposed to find the probability of obtaining y between two scores, however when I input my answer it tells me that I'm wrong, the question is given below along with my answer to the question:

Question
For a normal distribution with sample mean =−19 and standard deviation =6.85 find $p(-<!--\; -\; -->16.15\le <!--\; \le \; -->y\le <!--\; \le \; -->-<!--\; -\; -->15.27)$, where y is a random draw from the normal distribution. Round to 4 decimal places.

My answer
I obtained the z scores for both the y values −15.27 and −16.15 and their respective z scores are 0.5445 and 0.4161. The probability under the curve with a z score of 0.5445 is 0.7054 and the probability under the curve with a z score of 0.4161 is 0.6628. With those two probabilities, I obtained the difference (getting 0.0426) and I put in that answer and was told that I was wrong. Can someone give me insight as to what I'm doing wrong?

What are stars that pulsate in and out called?

Having some issue with the concept of Z score.
When exactly do I use $Z=\frac{\stackrel{¯<!--\; ¯\; -->}{X}-<!--\; -\; -->u}{\mathrm{\sigma <!--\; \sigma \; -->}}$
and when do I use $Z=\frac{\stackrel{¯<!--\; ¯\; -->}{X}-<!--\; -\; -->u}{\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}}$
I get very confused with in what situation should be using which Z calculation. Really appreciate it to have someone explain the concept