Improve Your Understanding of College Level Statistics Problems

Why can we use empirical standard deviation when computing mean confidence intervals
I'm reviewing some basic statistics, and I'm asking myself questions on things I used to take for granted when I first saw them years ago. I'm going to state things as I understand them, so there might be a mistake in the following.
Consider a random variable X following a distribution of mean $\mathrm{\mu <!--\; \mu \; -->}$ and standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}$. We measure n samples of X, and observe an empirical mean $\stackrel{¯<!--\; ¯\; -->}{x}$ and empirical std s.
Because we're observing samples, $\stackrel{¯<!--\; ¯\; -->}{x}$ and s are random variables themselves. We should therefore not use $\stackrel{¯<!--\; ¯\; -->}{x}$ and $\mathrm{\mu <!--\; \mu \; -->}$ interchangeably, and the same goes for s and $\mathrm{\sigma <!--\; \sigma \; -->}$. Instead, people compute confidence interval on μ based on $\stackrel{¯<!--\; ¯\; -->}{x}$. By the central limit theorem, if n is large enough, we can say that $\stackrel{¯<!--\; ¯\; -->}{x}\sim <!--\; \sim \; -->N(\mathrm{\mu <!--\; \mu \; -->},\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}})$.
When computing confidence intervals, we usually use ${\mathrm{\sigma <!--\; \sigma \; -->}}_{\stackrel{¯<!--\; ¯\; -->}{x}}=\frac{s}{\sqrt{n}}$. If all of this is correct, my question is the following: since both s and $\stackrel{¯<!--\; ¯\; -->}{x}$ are random variables, why does it seems to be ok to consider $s=\mathrm{\sigma <!--\; \sigma \; -->}$ when computing confidence intervals for $\mathrm{\mu <!--\; \mu \; -->}$?

A researcher is interested in the cumulative grade point average (CGPA) of students who enrolled in a Statistics course this semester. There are three groups, namely A, B and C which consist of 30 students for group A, 40 students for Group B and 30 students for group C. A sample of 50 students are randomly selected in this study.
i) State the population and sampling frame.
ii) State the variable of interest.
iii) Give the appropriate sampling technique to be used for this study.
iv) Calculate the number of students that should be selected from each group.
v) Suggest the data collection method for this study. Give ONE advantage and ONE disadvantage.

a) What is the variance of the sum of n independent random variables? b) What is the variance of the number of successes when n independent Bernoulli trials, each with probability p of success, are carried out?

How do you solve $\sin <!--\; \; -->(3x)=-<!--\; -\; -->1$ and find all exact general solutions?

How large can the outer automorphism group be?
You can make the outer automorphism group very large by taking G be to be a vector space (say over a finite field) so that if $|G|=q^n$, then $Out(G)=G{L}_n({\mathbb{F}}_q)$ of size exponential in n.
So I have two questions related to this:
1) Is there a "family of groups" with a faster growing outer automorphism group?
2) What does the outer automorphism group of a "typical" group look like?
I am leaving both questions fairly vague since I am not sure exactly what kind of an answer I am looking for. The goal is to simply understand how much bigger the outer automorphism group is (compared to the group) in general.

Suppose that the random variables $$Y_1$$ and $$Y_2$$ have means $${\mathrm{\mu <!--\; \mu \; -->}}_1$$ and $${\mathrm{\mu <!--\; \mu \; -->}}_2$$ and variances $${\mathrm{\sigma <!--\; \sigma \; -->}}_1^2$$ and $${\mathrm{\sigma <!--\; \sigma \; -->}}_2^2$$, respectively. Use the basic definition of the covariance of two random variables to establish that $$Cov(Y_1,Y_2)=Cov(Y_2,Y_1)$$.

Calculating the confidence interval for population variance given the confidence interval for population mean
I am given a sample size of 9 from a normal population, as well as the 95% confidence interval for population mean. I want to calculate the 95% population variance.
I see that the confidence interval for an unknown population mean is $\stackrel{¯<!--\; ¯\; -->}{Y}\pm <!--\; \pm \; -->t_{\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}}\frac{S}{\sqrt{n}}$. Then since the midpoint of the confidence interval for population mean is the sample mean, I can plug that into $\stackrel{¯<!--\; ¯\; -->}{Y}$, 9 into n, and 0.025 into $\mathrm{\alpha <!--\; \alpha \; -->}$. From there, I can solve for S. How can I find the confidence interval for population variance from there?

X and Y are random variables with E [X]=E[Y]=0 and Var[X]=1, Var[Y]=4 and correlation coefficient p=1/2. Find Var[X+Y].

Classify each of the following random variables as discrete or continuous. X = age (in years) of the oldest person that a randomly selected adult has ever met.

How do you find the exact value of $\sin <!--\; \; -->(\frac{7\mathrm{\pi <!--\; \pi \; -->}}{6}-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{3})$?

How do you find the exact value of $\sec <!--\; \; -->120$?

Calculate confidence level from a given confidence interval
I'm fairly comfortable calculating the confidence interval. But now I'm seeing a problem where I'm giving a confidence interval CI(27.6621,30.3379) and I'm requested to calculate the confidence level. I'm also given the sample size $n=85$ and a standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}=7.5$. I can't find a formula for that, I feel like it's something simple I'm not seeing. Any help would be appreciated.

Give an example of another continuous random variable (in addition to temperature) and another discrete random variable that would influence the enjoyment of the event. Give the sample space for those random variables.

Confidence Interval problem
Which of the following will result in a wider confidence interval?
Larger sample size
Higher level of confidence
Larger population standard deviation
Larger population mean
Larger sample mean
So after running some numbers, I believe 2,3, and 4 are correct because they appeared to work however apparently that's incorrect. Could anyone please help me?

Nonisomorph groups of order 2002
While searching for non-isomorph subgroups of order 2002 I just encountered something, which I want to understand. Obviously I looked for abelian subgroups first and found $2002=2^2\ast <!--\; \ast \; -->503$ so we have the groups
$$\mathbb{Z}/2^2\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}/503\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}/2\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}/503\mathbb{Z}$$
Now I want to understand why those two are not isomorph. I know that for two groups $\mathbb{Z}/n\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}/m\mathbb{Z}\cong <!--\; \cong \; -->\mathbb{Z}/(nm)\mathbb{Z}$ it has to hold that $gcd(n,m)=1$. But I don't understand how we can compare Groups written as two products with groups written as three products as above, how does that work? And I think that goes in the same direction: How is it then at the same time that
$$\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong <!--\; \cong \; -->\mathbb{Z}/2012\mathbb{Z}≆<!--\; ≆\; -->\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong <!--\; \cong \; -->\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/1006Z$$
because $gcd(4,2012)\ne <!--\; \ne \; -->1,gcd(2,2)\ne <!--\; \ne \; -->1,gcd(503,1006)\ne <!--\; \ne \; -->1$. I don't understand the difference to the first comparison.

Two random variables, X and Y, have standard deviations 2.4 and 3.6, respectively. Which one is more likely to take a value close to its mean? Explain your answer.

How do discrete and continuous random variables differ?

What is the difference between discrete and continuous random variables ?

95% Confidence Interval for $\mathrm{\lambda <!--\; \lambda \; -->}$
Consider a random sample $X_1,X_2,..,X_n$ from a variable with density function
$$f_X(x)=2\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\pi <!--\; \pi \; -->}x{e}^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\pi <!--\; \pi \; -->}{x}^2}x>0$$
A useful estimator of $\mathrm{\lambda <!--\; \lambda \; -->}$ is
$$\stackrel{^<!--\; ^\; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{n}{\mathrm{\pi <!--\; \pi \; -->}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_1^2}$$
Derive a 95% confidence interval for $\mathrm{\lambda <!--\; \lambda \; -->}$ (a hint is provided suggesting to consider the distribution of $\frac{\mathrm{\lambda <!--\; \lambda \; -->}}{\stackrel{^<!--\; ^\; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}}).$).

the situation in the question :so we have a number of scatter plots with each showing an estimated regression line (based on a valid model) and associated individual 95% confidence intervals (CI) for the regression function at each x-value, as well as the observed data. A professor asks 'I don't understand how 95% of the observations fall outside the 95% CI as depicted in the figures'. Briefly explain how is is entirely possible that 95% of the observations fall outside the 95% CI as depicted in the figures.(We weren't given actual figures)
Anyway I thought that it may have been due to the fact that a lot of outliers affected the regression line calculated, and so a confidence interval formed from a bad regression line would be bad - resulting in 95% of observations falling outside the 95% CI.