Expert Assistance for Population Data: Comprehensive Resources and Practice Problems

Minimum of the variance of a data set given the variances of subsets
Suppose we have a population data set X which is partitioned into two subsets A and B, with population variance 3 and 4, respectively. Is it true that the population variance of A is at least 3 (i.e. min{3,4})?

Question regarding the definition of confident interval
Someone calculates the 95% confidence interval for the average weight of teenagers by collecting data from a simple random sample of 100 teenagers. Circle the correct interpretation(s) of the confidence interval (there may be more than one correct answer):
1) There is a 95% chance that the average weight of all teenagers falls in this range.
2) There is a 95% chance that the interval created includes the average weight of all teenagers.
Solution given says that
1) Almost correct, but no: the average weight of all teenagers is not random
2) Correct
Now, I've been reading and rereading the two questions and I can't differentiate how the two questions are different.
To me, the second question basically restates the same scenario as the first question.
What I know about confidence interval is that it's for the population. The interval gives a range and how confident the range covers the TRUE value.
So in the first case, it sounds exactly like what the definition suggests?
Is the solution given correct or did I missed some nuances between the two questions?

Modelling wealth with a Pareto distribution: how do I estimate the parameters?
I wish to create a function that will estimate the wealth of a person in the United States. It would be used to make a table with each decile and their estimated wealth.
This estimate will be based on very rudimentary data, and is only for personal interest. The data is:
- The total wealth of the bottom 90% is equal to the total wealth of the top 0.1%.
- Both proportions have 22% of the total wealth.
- The total wealth under the distribution is $80 trillion.
- The total population is 160 million households.
Given this data, how would I create parameter estimates for the exponent and scale of a pareto distribution? What would be f(x) where x is from (0,1), and the solution is the wealth of someone richer than that proportion of people? For example f(0.1) is someone richer than or equal to exactly 10% of the least wealthy, and could equal 1,000 dollars. F(0.5) is the median wealth, and could be 200,000 dollars. F(0.9999) is richer than 99.99% and would be somewhere in the tens or hundreds of millions of dollars.

differential equation of a population growth and change
I want to formulate a system of equations and initial conditions of the following data: Each year the population1 grows by 4% and population2 by 2%. Also each year 3% of population1 leaves it and go to population2 and 1% vice versa, leaves population2 and goes to population1.
The initial populations are: $population1=20000000,population2=5000000$.
I thought of the next formalism of this:
$$x(t+1)=1.04\ast <!--\; \ast \; -->x(t)-<!--\; -\; -->0.03\ast <!--\; \ast \; -->x(t)+0.01\ast <!--\; \ast \; -->y(t)$$
$$y(t+1)=1.02\ast <!--\; \ast \; -->y(t)-<!--\; -\; -->0.01\ast <!--\; \ast \; -->y(t)+0.03\ast <!--\; \ast \; -->x(t)$$
where x(t) is the size of population1 and y(t) is the size of population2. Is this is correct in those term or maybe I should reformulate it in another way ?
I think that I need some how to get something like this (ignore the concrete number and function):
$$\stackrel{¯<!--\; ¯\; -->}{X^{\prime }}=\left(\begin{array}{cc}1& 1\\ -<!--\; -\; -->2& -<!--\; -\; -->2\end{array}\right)\stackrel{¯<!--\; ¯\; -->}{X}+\left(\begin{array}{c}{e}^{-<!--\; -\; -->t}\\ -<!--\; -\; -->2e^{-<!--\; -\; -->t}\end{array}\right)$$
with an initial conditions like that:
$$\stackrel{¯<!--\; ¯\; -->}{X}(0)=\left(\begin{array}{c}-<!--\; -\; -->1\\ 0\end{array}\right)$$
but I don't know how to do it, so will be grateful for some help.
An idea I had is:
$$x^{\prime }\left(t\right)=\frac{x(t+1)-x\left(t\right)}{(t+1-t)}=(1.04-0.03-1)x\left(t\right)+0.01y\left(t\right)$$
is it correct ?

Sample Standard Deviation vs. Population Standard Deviation
I have an HP 50g graphing calculator and I am using it to calculate the standard deviation of some data. In the statistics calculation there is a type which can have two values:
Sample Population
I didn't change it, but I kept getting the wrong results for the standard deviation. When I changed it to "Population" type, I started getting correct results!
Why is that? As far as I know, there is only one type of standard deviation which is to calculate the root-mean-square of the values!
Did I miss something?

The exercise statement (roughly): Assume there is a terrorist prevention system that has a 99% chance of correctly identifying a future terrorist and 99.9% chance of correctly identifying someone that is not a future terrorist. If there are 1000 future terrorists among the 300 million people population, and one individual is chosen randomly from the population, then processed by the system and deemed a terrorist. What is the chance that the individual is a future terrorist?
Attempted exercise solution:
I use the following event labels:
A -> The person is a future terrorist
B -> The person is identified as a terrorist
Then, some other data:
$$P(A)=\frac{{10}^3}{3\cdot <!--\; \cdot \; -->{10}^8}=\frac{1}{3\cdot <!--\; \cdot \; -->{10}^5}$$
$$P(\stackrel{¯<!--\; ¯\; -->}{A})=1-<!--\; -\; -->P(A)$$
$$P(B\mid <!--\; \mid \; -->A)=0.99$$
$$P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->A)=1-<!--\; -\; -->P(B\mid <!--\; \mid \; -->A)$$
$$P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})=0.999$$
$$P(B\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})=1-<!--\; -\; -->P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})$$
What I need to find is the chance that someone identified as a terrorist, is actually a terrorist. I express that through P(A | B) and use Bayes Theorem to find its value.
$$P(A\mid <!--\; \mid \; -->B)=\frac{P(A\cap <!--\; \cap \; -->B)}{P(B)}=\frac{P(B\mid <!--\; \mid \; -->A)\cdot <!--\; \cdot \; -->P(A)}{P(B\mid <!--\; \mid \; -->A)\cdot <!--\; \cdot \; -->P(A)+P(B\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})\cdot <!--\; \cdot \; -->P(\stackrel{¯<!--\; ¯\; -->}{A})}$$
The answer I get after plugging-in all the values is: $3.29\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}$, the book's answer is $3.29\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->4}$.
Can someone help me identify what I'm doing wrong? Also, in either case, I find that it is very unintuitive that the probability of success is so small. If someone could explain it to me in more intuitive terms I'd be very grateful.

Normalizing sample data to unit variance
Suppose you have data points $X_1,...,X_n$ drawn from some distribution P (regard them as iid random variables) and let the distribution be "well-behaved" so that the mean is $\mathrm{\mu <!--\; \mu \; -->}$ and variance is ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$. Now suppose that we want to normalize our data, i.e., find a mapping $X_i\mapsto <!--\; \mapsto \; -->Y_i$ such that $Y_i$ are iid drawn from a distribution with mean $=0$ and variance 1. Suppose that we don't know the true values $\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2$, then it's not too hard to make the mean zero by using the map
$$X_i\mapsto <!--\; \mapsto \; -->Y_i\equiv <!--\; \equiv \; -->X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}},\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}\equiv <!--\; \equiv \; -->\frac{1}{n}\underset{1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i$$
Indeed, $\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}$ is the sample mean and it's quite easy to check that
$$\mathbb{E}{Y}_i=0$$
Now if I would want to do the same for the variance, the natural method would be something like
$$X_i\mapsto <!--\; \mapsto \; -->Y_i\equiv <!--\; \equiv \; -->\frac{X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}}{\stackrel{^<!--\; ^\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}},{\stackrel{^<!--\; ^\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}}^2\equiv <!--\; \equiv \; -->\frac{1}{n-<!--\; -\; -->1}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}{)}^2$$
Now we know by the convergence in distribution, that when $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, we see that $Y_i\to <!--\; \to \; -->(X_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}$ in distribution, but the variance doesn't seem to be normalized for arbitrary n.
Question. Is is actually possible to normalize the variance of a sample data set without knowing the true mean and variance?

Doomsday Prediction
I have a calculus problem I can't seem to figure out. Any help would be appreciated!
Doomsday prediction. In 1960, three electrical engineers at the University of Illinois published a paper in Science titled "Doomsday." Based on world population growth data from 1000 AD to 1960 AD, the engineers found that population growth was faster than proportional to the population size. Using the data, they modeled the growth of the population as
$dP/dt=0.4873P^2$
where P is the population size in billions and t is centuries after 1000 AD.
1) Solve this differential equation.

Discrete version of continuous SIR model
I'm working with a SIR infection model, which is
$$\begin{array}{rcl}\frac{dS}{dt}& =& -<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}IS\\ \frac{dI}{dt}& =& \mathrm{\beta <!--\; \beta \; -->}IS-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}I\\ \frac{dR}{dt}& =& \mathrm{\gamma <!--\; \gamma \; -->}I\end{array}$$
in continuous time, where S, I, and R are the proportion of Susceptible, Infected, and Recovered, respectively.
However, since I am working with fixed-width discrete-time data, I think it would be more appropriate to modify the equations accordingly. I know this is incorrect (based on getting a negative values for $\stackrel{^<!--\; ^\; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}$ and $\stackrel{^<!--\; ^\; -->}{\mathrm{\beta <!--\; \beta \; -->}}$, neither of which should be negative):
$$\begin{array}{rcl}{S}_{t+1}& =& -<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}{I}_t{S}_t\\ I_{t+1}& =& \mathrm{\beta <!--\; \beta \; -->}{I}_t{S}_t-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}{I}_t\\ R_{t+1}& =& \mathrm{\gamma <!--\; \gamma \; -->}{I}_t\end{array}$$
Ultimately, I would like to estimate $\mathrm{\beta <!--\; \beta \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ by doing regression on
$$I_{t+1}=\mathrm{\beta <!--\; \beta \; -->}(I_t{S}_t)+\mathrm{\gamma <!--\; \gamma \; -->}(-<!--\; -\; -->I_t)+U_t\text{, where }{U}_t\text{~}N(0,{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$$
or whatever the discrete version is.

I have a question about t-statistics in first place I need to use t-statistics to estimate the population mean from a sample set, as you see I have a data set the mean and standard deviation are known for sample but not for the population.
How can I use t-statistics to determine the population mean I have X bar and S and n values for sample $n=45$how can I compute the degree of freedom and how can I determine the confidence level? thanks for your time huge regards