Master Statistics and Probability Problems with Expert Help

What is the shape of a chi-squared distribution?

In the example experiment on eye-witness memory of a car accident, if the question they used ("Was there shattered glass?") had been too easy and all participants got it right, this would have compromised the:
A. internal validity of the dependent variable
B. construct validity of the independent variable
C. internal validity of the independent variable
D. construct validity of the dependent variable

What is the difference between $\frac{1}{N}\sum <!--\; \sum \; -->\frac{y_i}{x_i}$ and $\sum <!--\; \sum \; -->\frac{\stackrel{¯<!--\; ¯\; -->}{y}}{\stackrel{¯<!--\; ¯\; -->}{x}}$
I have a data set and I was looking at the population ratio and trying to estimate it using different methods. I was expecting eq1 and eq2 to be different but was very surprised that it was by a factor of almost 3x. I was just wondering why it was so different, is it partly due to n cancelling on eq2. Any explanation would be appreciated.

Why is the statistic $t=r\sqrt{\frac{n-<!--\; -\; -->2}{1-<!--\; -\; -->r^2}}$$\approx t(n-2)$?

Determining sample size of a set of boolean data where the probability is not 50%
I'll lay out the problem as a simplified puzzle of what I am attempting to calculate. I imagine some of this may seem fairly straightforward to many but I'm starting to get a bit lost in my head while trying to think through the problem.
Let's say I roll a 1000-sided die until it lands on the number 1. Let's say it took me 700 rolls to get there. I want to prove that the first 699 rolls were not number 1 and obviously the only way to deterministically do this is to include the first 699 failures as part of the result to show they were in fact "not 1".
However, that's a lot of data I would need to prove this. I would have to include all 700 rolls, which is a lot. Therefore, I want to probabilistically demonstrate the fact that I rolled 699 "not 1s" prior to rolling a 1. To do this, I decide I will randomly sample my "not 1" rolls to reduce the set to a statistically significant, yet more wieldy number. It will be good enough to demonstrate that I very probably did not roll a 1 prior to roll 700.
Here are my current assumptions about the state of this problem:
- My initial experiment of rolling until success is one of geometric distribution.
- However my goal for this problem is to demonstrate to a third party that I am not lying, therefore the skeptical third party is not concerned with geometric distribution but would view this simply as a binomial distribution problem.
A lot of sample size calculators exist on the web. They are all based around binomial distribution from what I can tell. So here's the formula I am considering:
$$n=\frac{N\times <!--\; \times \; -->X}{X+N–1}$$
$$X=\frac{{Z_{\mathrm{\alpha <!--\; \alpha \; -->}/2}}^2­\times <!--\; \times \; -->p\times <!--\; \times \; -->(1-<!--\; -\; -->p)}{{\mathsf{M}\mathsf{O}\mathsf{E}}^2}$$
n is sample size
N is population size
Z is critical value ($\mathrm{\alpha <!--\; \alpha \; -->}$ is $1-<!--\; -\; -->\mathsf{c}\mathsf{o}\mathsf{n}\mathsf{f}\mathsf{i}\mathsf{d}\mathsf{e}\mathsf{n}\mathsf{c}\mathsf{e}\mathsf{l}\mathsf{e}\mathsf{v}\mathsf{e}\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{b}\mathsf{a}\mathsf{b}\mathsf{i}\mathsf{l}\mathsf{i}\mathsf{t}\mathsf{y}$)
p is sample proportion
MOE is margin of error
As an aside, the website where I got this formula says it implements "finite population correction", is this desirable for my requirements?
Here is the math executed on my above numbers. I will use $Z_{a/2}=2.58$ for $\mathrm{\alpha <!--\; \alpha \; -->}=0.01$, $p=0.001$ and $\mathsf{M}\mathsf{O}\mathsf{E}=0.005$. As stated above, $N=699$ on account of there being 699 failure cases that I would like to sample with a certain level of confidence.
Based on my understanding, what this math will do is recommend a sample size that will show, with 99% confidence, that the sample result is within 0.5 percentage points of reality.
Doing the math, $X=265.989744$ and $n=192.8722086653\approx <!--\; \approx \; -->193$, implying that I can have a sample size of 193 to fulfill this confidence level and interval.
My main question is whether my assumption about $p=\frac{1}{1000}$ is valid. If it's not, and I use the conservative $p=0.5$, then my sample size shoots up to $\approx <!--\; \approx \; -->692$. So I would like to know if my assumptions about what sample proportion actually is are correct.
More broadly, am I on the right track at all with this? From my attempt at demonstrating this probabilistically to my current thought process, is any of this accurate at all?

Grassmannians are a pretty useful subject in numerous fields of mathematics (and physics). In fact, it was the first non-trivial higher-dimensional example that was given in an introductory projective geometry course during my education.
Later I learned you can use them to define universal bundles and that they are playing a role in higher-dimensional geometry and topology. Though I have never came across a book or a survey article on the geometry and topology of those beasts. The field is a little wide, so let me specify what I am interested in:
Topology and Geometry of Grassmannians $G_k({\mathbb{R}}^n)$ or $G_k({\mathbb{C}}^n)$
Connections with bundle and obstruction theory.
Differential Topology of $G_k({\mathbb{R}}^n)$ or $G_k({\mathbb{C}}^n)$ (for instance, are there exotic Grassmannians).
Homotopy Theory of $G_k({\mathbb{R}}^n)$ or $G_k({\mathbb{C}}^n)$.
Algebraic Geometry of $G_k(V)$, where V is a n-dimensional vectorspace over a (possible characteristic $\ne <!--\; \ne \; -->0$ field F)

Which of the following statements is true about the t-distribution?
a) For large sample sizes, the t-distribution has the same properties as the normal curve.
b) For small sample sizes, the t-distribution has the same properties as the normal curve.
c) Like the Normal distribution, the t-distribution is symmetric for small n.
d) Since population standard deviation is usually unknown, the standard error uses the sample standard deviation to estimate population standard deviation.

How do you find the exact value of ${\sin }^{-<!--\; -\; -->1}<!--\; \; -->(\sin <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}))$?

Let $Xi$ be a random variable distributed as $N(i,i^2),i=1,2,3$. As-sume that the random variables $X_1,X_2$ and $X_3$ are independent. Using only the three random variables $X_1,X_2$ and $X_3$ give an example of a statistic that has a $t$ distribution with two degrees of freedom.

Finding the Expected Value of $T=$$\sum <!--\; \sum \; -->X_i^2$

Bayesian Statistics - Basic question about prior
I try to get an understanding of bayesian statistics. My intuition tells me that in the expression for the posterior
$$p(\mathrm{\vartheta <!--\; \vartheta \; -->}|x)=\frac{p(x|\mathrm{\vartheta <!--\; \vartheta \; -->})p(\mathrm{\vartheta <!--\; \vartheta \; -->})}{{\int <!--\; \int \; -->}_{\mathrm{\Theta <!--\; \Theta \; -->}}p(x|\mathrm{\theta <!--\; \theta \; -->})p(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}}$$
the term $p(\mathrm{\vartheta <!--\; \vartheta \; -->})$ is the marginal distribution of the likelihood-function $p(\mathrm{\vartheta <!--\; \vartheta \; -->},x)$. It is obtained by
$$p(\mathrm{\vartheta <!--\; \vartheta \; -->})={\int <!--\; \int \; -->}_{X}p(\mathrm{\vartheta <!--\; \vartheta \; -->}|x)p_X(x)dx$$
where $p_X(x)$ should be the marginal distribution of the Observable data. Does that make sense?
To this point it makes sense with this example: Offering somebody a car insurance without knowing the person's style of driving (determined by $\mathrm{\vartheta <!--\; \vartheta \; -->}\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}$) to feed some statistical model, we still can make use of the nation's car-crash statistics as our prior, which is a pdf on $\mathrm{\Theta <!--\; \Theta \; -->}$. That would be the marginal distribution of the "driving styles" across the population.
Maybe I am just oversimplifying here, because my resources did not mention this.

Let $\{Xi\}\sim <!--\; \sim \; -->N(i\theta ,1)$ for $i=1,....,n$ be an independent, but not identically distributed sample. Check that $T=\underset{i}{\sum <!--\; \sum \; -->}{X}_i$ it is a sufficient statistic for $\theta$.

When conducting a hypothesis test for the population mean, you will not know the population standard deviation, so you will have to use the sample standard deviation instead. How will this affect the process?

How do you find the exact value for $\cos <!--\; \; -->240$?

Derive an expression for the $p$-value using a test with test statistic $T=\sqrt{n}({\stackrel{¯<!--\; ¯\; -->}{X}}_n-<!--\; -\; -->{\mathrm{\theta <!--\; \theta \; -->}}_0)/\mathrm{\sigma <!--\; \sigma \; -->}$

Derive the Cramer von Mises test statistic
$$n{C}_n=\frac{1}{12n}+\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{(U_{(i)}-<!--\; -\; -->\frac{2i-<!--\; -\; -->1}{2n})}^2$$
where $U_{(i)}=F_0(X_{(i)})$ the order statistics

How do you use a power series to find the exact value of the sum of the series $1-<!--\; -\; -->\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^2}{2!}+\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^4}{4!}-<!--\; -\; -->\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^6}{6!}+$...?

Solve PDE using method of characteristics with non-local boundary conditions.
Given the population model by the following linear first order PDE in u(a,t) with constants b and $\mathrm{\mu <!--\; \mu \; -->}$:
$$u_a+u_t=-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tua,t>0$$
$$u(a,0)=u_0(a)a\ge 0$$
$$u(0,t)=F(t)=b{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$
We can split the integral in two with our non-local boundary data:
$$F(t)=b{\int <!--\; \int \; -->}_0^{t}u(a,t)\mathrm{d}a+b{\int <!--\; \int \; -->}_t^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$
Choosing the characteristic coordinates $(\mathrm{\xi <!--\; \xi \; -->},\mathrm{\tau <!--\; \tau \; -->})$ and re-arranging the expression to form the normal to the solution surface we have the following equation with initial conditions:
$$(u_a,u_t,-<!--\; -\; -->1)\bullet <!--\; \bullet \; -->(1,1,-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tu)=0$$
$$x(0)=\mathrm{\xi <!--\; \xi \; -->},t(0)=0,u(0)=u_0(\mathrm{\xi <!--\; \xi \; -->})$$
Characteristic equations:
$$\frac{\mathrm{d}a}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=1,\frac{\mathrm{d}t}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=1,\frac{\mathrm{d}u}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tu$$
Solving each of these ODE's in $\mathrm{\tau <!--\; \tau \; -->}$ gives the following:
$$(1)\int <!--\; \int \; -->\mathrm{d}a=\int <!--\; \int \; -->\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}(2)\int <!--\; \int \; -->\mathrm{d}t=\int <!--\; \int \; -->\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}(3)\int <!--\; \int \; -->\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}tu\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$a=\mathrm{\tau <!--\; \tau \; -->}+F(\mathrm{\xi <!--\; \xi \; -->})t=\mathrm{\tau <!--\; \tau \; -->}+F(\mathrm{\xi <!--\; \xi \; -->})$$
$$\therefore <!--\; \therefore \; -->a=\mathrm{\tau <!--\; \tau \; -->}+\mathrm{\xi <!--\; \xi \; -->}\therefore <!--\; \therefore \; -->t=\mathrm{\tau <!--\; \tau \; -->}$$
$$\int <!--\; \int \; -->\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}\mathrm{\tau <!--\; \tau \; -->}u\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$\int <!--\; \int \; -->\frac{1}{u}\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}\mathrm{\tau <!--\; \tau \; -->}\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$\ln <!--\; \; -->u=-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2+F(\mathrm{\xi <!--\; \xi \; -->})$$
$$u=G(\mathrm{\xi <!--\; \xi \; -->})e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2}$$
$$\therefore <!--\; \therefore \; -->u=u_0(\mathrm{\xi <!--\; \xi \; -->})e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2}$$
Substituting back the original coordinates we can re-write this expression with a coordinate change:
$$\mathrm{\xi <!--\; \xi \; -->}=a-<!--\; -\; -->t\mathrm{\tau <!--\; \tau \; -->}=t$$
$$\therefore <!--\; \therefore \; -->u(a,t)=u_0(a-<!--\; -\; -->t)e^{-<!--\; -\; -->\frac{1}{2}{t}^2}$$
Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution?
$$u(0,t)=u_0(-<!--\; -\; -->t)e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{t}^2}=b{\int <!--\; \int \; -->}_0^{t}u(a,t)\mathrm{d}a+b{\int <!--\; \int \; -->}_t^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$

Using a "population" consisting of probabilities to predict accuracy of sample
Since I'm not sure if the title explains my question well enough I've come up with an example myself:
Let's say I live in a country where every citizen goes to work everyday and every citizen has the choice to go by bus or by train (every citizen makes this choice everyday again - there are almost no citizens who always go by train and never by bus, and vice-versa).
I've done a lot of sampling and I have data on one million citizens about their behaviour in the past 1000 days. So, I calculate the "probability" per citizen of going by train on a single day. I can also calculate the average of those calculated probabilities of all citizens, let's say the average probability of a citizen going by train is 0.27. I figured that most citizens have tendencies around this number (most citizens have an individual probability between 0.22 and 0.32 of going by train for example).
Now, I started sampling an unknown person (but known to be living in the same country) and after asking him 10 consecutive days whether he went by train or by bus, I know that this person went to his work by train 4 times, and by bus 6 times.
My final question: how can I use my (accurate) data on one million citizens to approximate this person's probability of going by train?
I know that if I do the calculation the other way around, so, calculate the probability of this event occurring given the fact that I know this person's REAL probability is 0.4 this results in: ${0.4}^4\cdot <!--\; \cdot \; -->{0.6}^6\cdot <!--\; \cdot \; -->10C4=\sim <!--\; \sim \; -->25\mathrm{\%<!--\; \%\; -->}$. I could calculate this probability for all possible probabilities between 0.00 and 1.00 (so, $0\mathrm{\%<!--\; \%\; -->}-<!--\; -\; -->100\mathrm{\%<!--\; \%\; -->}$ without any numbers in between) and sum them all, which sums to about 910%. I could set this to 100% (dividing by 9.1) and set all other percentages accordingly (dividing everything by 9.1 - so, our 25% becomes ~2.75%) and come up with a weighted sum: $2.75\mathrm{\%<!--\; \%\; -->}\cdot <!--\; \cdot \; -->0.4+X\mathrm{\%<!--\; \%\; -->}\cdot <!--\; \cdot \; -->0.41$ etc., but this must be wrong since I'm not taking my accurate samples of the population into account.

Let $X1,...,Xn$ be independent and identically distributed with density $P_{\theta }(x)=$
$$\{\begin{array}{ll}2x/{\theta }^2& \text{for }0\le <!--\; \le \; -->x<\theta \\ 0& \text{else}\end{array}$$
Demonstrate that $m_n=max(x_1,...,x_n)$ is a sufficient statistic .