Improve Your Understanding of College Level Statistics Problems

Interpretaton of confidence interval
I have read two books that explicitly state that the $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$% confidence interval should be interpreted as:
If you construct 100 such confidence intervals, $\mathrm{\alpha <!--\; \alpha \; -->}$ of them are expected to not contain the true population statistic and $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ of them are expected to contain the true population statistic.
and not as
There is a $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ probability that the true population statistic is contained in the confidence interval.
In my view, they both equivalent: If you make the first statement, you implicitly make the second statement. You are looking at any one arbitrary confidence interval, which in itself is a random variable, the generic confidence interval should, therefore, be subject to the second statement. Do things change when this random variable is actually realized?

Estimators and Confidence intervals
I was curious as to what the relationship between probabilistic values, estimators and confidence intervals are. I was wondering, if I have an estimator of some parameter $\mathrm{\lambda <!--\; \lambda \; -->}$, and a probability value that depended on $\mathrm{\lambda <!--\; \lambda \; -->}$ and the estimator, what would be the relationship between those two and a confidence interval for an arbitrary distribution?

Let f be a joint density function of a pair of continuous random variables X and Y. What properties does f possess?

For any positive real number $r,r^n$ converges.
False: Take any positive $r$, then as $n\to \infty r$ diverges.
If $x_n\ast <!--\; \ast \; -->y_n$ converges, then $x_n$ and $y_n$ both converge.
False: Suppose $x_n$ or $y_n$ dont converge. Then take $x_n=n^2$ and $y_n=\frac{1}{n}$. Then $x_n\ast <!--\; \ast \; -->y_n=n$, which diverges. Thus it does not converge.
True or False?

How do you find the exact value of the expression by using appropriate identities $\sin <!--\; \; -->(79)\cos <!--\; \; -->(49)-<!--\; -\; -->\cos <!--\; \; -->(79)\sin <!--\; \; -->(49)$?

Confidence Interval. Bernoulli Distribution
I am reviewing the construction of confidence intervals for a random sample with Bernoulli distribution. The book uses the statistics of the central limit theorem that distributes N(0,1) to estimate the interval :
$$Z_n=\frac{X_1+X_2+\cdots <!--\; \cdots \; -->+X_n-<!--\; -\; -->n\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}\sqrt{n}}$$
Why are the intervals constructed from these statistics symmetrical around the origin?
The book says: "Since it is desirable that the length of the interval be as small as possible and since the standard normal distribution is symmetrical around the origin, it turns out that the minimum length interval must also be symmetric around the origin", but I don't understand this.

A chemist weighed three separate test samples at 0.17 gram, 0.204 gram, and 2.3 grams. What is the total weight of the three samples?

Confidence interval multiplication
The question looks pretty simple but I can't get my hands on it:Say I have a probability which is the product of two other independent probabilities $p=p_1{p}_2$.
I have estimated each probability $p_1$ and $p_2$ and found some 95% confidence interval for each. How do I obtain a 95% confidence interval for p?
Taking the product of the bounds of the interval won't work as I would be taking 95% of a 95% confidence interval resulting in an approximately 90% which is not what I want.So conversely I would be taking 97.5% confidence interval for each and by multiplying the bounds I will obtain a 95% confidence interval, is that right?
I feel like something is going wrong. In my situation I deal with probabilities but it could be anything so this question can be generalised to any type of confidence intervals.
If my reasoning is correct, could someone convince me that's the correct way of doing so?

Why do statisticians use samples?

Difference of fixed points by subgroup action
If G is a finite group and X is a finite G-set, then we have a class equation which tells us
$$|X|=|X^G|+\sum <!--\; \sum \; -->|G/G_x|$$
where $X^G$ is the set of G-fixed points, $G_x$ is the stabilizer of $x\in <!--\; \in \; -->X$ and the sum is taken over representative elements of classes of non-singleton G-orbits.
My question, which arise from a proof in a paper which seems to use precisely this fact, is the following. If $H\le <!--\; \le \; -->G$ is a normal subgroup, are we able to write the cartinality of $X^H\setminus <!--\; \setminus \; -->X^G$ as sum of cardinalities of G/H orbits? If this is the case, over what that sum is taken?
I guess the result should descend by comparing the two class equations for H and G, but I'm struggling to write that down.

How do you find the exact value for $\sin <!--\; \; -->({\cos }^{-<!--\; -\; -->1}<!--\; \; -->(-<!--\; -\; -->\frac{1}{4}))$?

Which of the following statement is true?
(a) Risk behaviors are illegal for everyone.
(b) Risk behaviors can harm your health.
(c) Most teens engage in risk behaviors.
(d) There is no way to avoid risk behaviors.

The mass of Denise's rock sample is 684 grams. The mass of Pauline's rock sample is 29,510 centigrams. How much greater is Denise's sample than Pauline's sample?

If a 95% confidence interval for the mean was computed as (25,50), then if several more samples were taken with the same sample size, then 95% of them would have a sample mean between (25,50)
And I know this statement is false, but I want to know exactly why.

Confidence interval interpretation difficulty
When we are dealing with 95% confidence interval we mean that if we repeat process of collecting samples of same size and calculate 95% intervals for those samples then 95% of those intervals will contain the true population parameter.
Let the infinite number of intervals be represented by 100 for simplicity. Then 95 of these intervals will contain true population parameter.
Suppose we got an interval at the starting of the above process (L,U). Then if I ask what is the probability that this interval (L,U) contains the true population parameter then shouldn't it be $95/100=0.95$? (Because this interval (L,U) can be anyone of 100 and it would contain true population parameter of its one of those 95).
But this interpretation of confidence interval is considered incorrect. Can someone explain me why is this so?

Classify the following random variables as continuous or discrete: the quantity of fat in a sausage.

Confidence Intervals for an Exponential Distribution
$y_1$ is distributed $f_Y(y\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})=\mathrm{\theta <!--\; \theta \; -->}{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}y}{I}_{(0,\mathrm{\infty <!--\; \infty \; -->})}(y)$, where $\mathrm{\theta <!--\; \theta \; -->}>0$. Analyze the confidence interval for $\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}$ given by $[L(Y),U(Y)]=[Y,2Y]$
1: Determine the confidence coefficient of this interval.
Thoughts: I realized that my single data point is distributed
$\mathit{\text{exponential}}(\mathrm{\beta <!--\; \beta \; -->}=\frac{1}{\mathrm{\theta <!--\; \theta \; -->}})=\mathit{\text{gamma}}(\mathrm{\alpha <!--\; \alpha \; -->}=1,\mathrm{\beta <!--\; \beta \; -->}=\frac{1}{\mathrm{\theta <!--\; \theta \; -->}})$. So I did the following steps to get my confidence coefficient:
$$\begin{gathered} P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\in <!--\; \in \; -->[Y,2Y]) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->2Y) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}) \\ =\underset{\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}}{\overset{\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}}{\int <!--\; \int \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}y}\mathrm{d}y \\ =e^{-<!--\; -\; -->.5}-<!--\; -\; -->e^{-<!--\; -\; -->1} \\ =.23865 \end{gathered}$$
So I got .23685 as my confidence coefficient, and I'm fairly confident that that is the correct answer.
2: Determine the expected length of this interval.
Thoughts: So I'm not too sure what to do here, but this is what I have written down.
$$\begin{gathered} P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\in <!--\; \in \; -->[Y,2Y]) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->2Y) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}) \\ =P(e^{-<!--\; -\; -->1}\le <!--\; \le \; -->Y\le <!--\; \le \; -->e^{-<!--\; -\; -->.5})=.23865 \end{gathered}$$
Since my 'a' and 'b' values were $e^{-<!--\; -\; -->1}$ and $e^{-<!--\; -\; -->.5}$, respectively. Now I don't know where to go from there to find the expected length. Do I use the gamma probability table? Any help would be greatly appreciated.
3: Find a different confidence interval for $\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}$ with the exact same confidence coefficient, but the expected length is smaller than part 2's expected length. Find an interval where the characteristics, same confidence coefficient and expected length is smaller than part 2's, are satisfied.
Thoughts: I let $W=Y^{\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}}$, and I performed the transformation to get the distribution of W, and I ended with:
$$f_W(w\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})={\mathrm{\theta <!--\; \theta \; -->}}^2{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{w}^{\mathrm{\theta <!--\; \theta \; -->}}}{w}^{\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->1}$$
, which I think looks sorta like the gamma distribution, but then I don't know what to do from there. Any help would be appreciated.

So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one. 1)
$$\underset{x\to <!--\; \to \; -->0}{lim}{\left(\frac{e^4{(1-<!--\; -\; -->x^2)}^{\frac{1}{x^2}}}{{(1-<!--\; -\; -->2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$
I am having problems with this one to make it into one that can be used for L'Hospital's rule, but i cannot get it into $\frac{0}{0}$, there are others options as well, but i am stuck.
2) $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\frac{\arcsin <!--\; \; -->(\frac{1}{n})+\arcsin <!--\; \; -->(\frac{2}{n})+...+\arcsin <!--\; \; -->(\frac{n}{n})}{n})$
So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L'Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.

When is examining only one direction of a bijective map $\mathrm{\varphi <!--\; \varphi \; -->}:X\to <!--\; \to \; -->Y$ enough to categorize it as an isomorphism?
Let X,Y be sets with $\mathrm{\varphi <!--\; \varphi \; -->}:X\to <!--\; \to \; -->Y$ being bijective. If we consider X and Y various structures and ask what conditions do we have to impose on $\mathrm{\varphi <!--\; \varphi \; -->}$ for it to be an isomorphism, we can break these cases up into two groups:
Group 1:
- If X,Y are vector spaces, then $\mathrm{\varphi <!--\; \varphi \; -->}$ needs to be linear.
- If X,Y are groups, then $\mathrm{\varphi <!--\; \varphi \; -->}$ needs to respect the group operation.
- If X,Y are metric spaces, then $\mathrm{\varphi <!--\; \varphi \; -->}$ needs to be an isometry.
- etc.
Group 2:
- If X,Y are topological spaces, then $\mathrm{\varphi <!--\; \varphi \; -->}$ needs to be continuous and open.- If X,Y are smooth manifolds, then $\mathrm{\varphi <!--\; \varphi \; -->}$ needs to be a diffeomorphism.
- etc.
In case one, we only need to verify that $\mathrm{\varphi <!--\; \varphi \; -->}$ satisfies certain properties (that $\mathrm{\varphi <!--\; \varphi \; -->}$ is linear, respects group operation, isometry, etc.). Whereas in group 2, we need to verify that both $\mathrm{\varphi <!--\; \varphi \; -->}$ AND ${\mathrm{\varphi <!--\; \varphi \; -->}}^{-<!--\; -\; -->1}$ satisfy certain properties (continuous, smooth, etc). My question is, what is it that distinguishes the members of group 1 verses group 2? Why is it enough to check a certain condition in one direction only enough in certain cases?