Improve Your Understanding of College Level Statistics Problems

An airline offers a certain flight once per day that usually contains about 250 passengers. The flight offers seats in first class (most expensive), business class, and economy class (least expensive). The airline wants to survey 500 passengers of this flight about their overall satisfaction. The passengers will be selected using a cluster random sample where each flight is a cluster. Why might the airline choose a cluster random sample instead of a simple random sample in this setting?

Confidence interval for difference confidence level
Lets say I measure quantity $\mathrm{\mu <!--\; \mu \; -->}$ and compute confidence interval for it $[\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_1,\mathrm{\mu <!--\; \mu \; -->}+{\mathrm{\delta <!--\; \delta \; -->}}_2]$ with a 68% confidence limit.
What if I want a confidence interval for 95% CL, can I simply scale errors like this? Let $x=\frac{0.95}{0.68}$
$[\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->x{\mathrm{\delta <!--\; \delta \; -->}}_1,\mathrm{\mu <!--\; \mu \; -->}+x{\mathrm{\delta <!--\; \delta \; -->}}_2]$

How we interpret confidence intervals.
For the formula $\stackrel{¯<!--\; ¯\; -->}{X}\pm <!--\; \pm \; -->z_{\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}}\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}$, we substitute ${\stackrel{¯<!--\; ¯\; -->}{x}}_{obs}$ for $\stackrel{¯<!--\; ¯\; -->}{X}$, but isn't that observed sample mean ${\stackrel{¯<!--\; ¯\; -->}{x}}_{obs}$ only for one particular sample?
Also why is it incorrect to interpret a confidence interval as the probability that the actual $\mathrm{\mu <!--\; \mu \; -->}$ lies in a interval (a,b) is $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})100\mathrm{\%<!--\; \%\; -->}$

Let f be a joint density function of a pair of continuous random variables X and Y. What are the expected values of X and Y?

How do binomial and hypergeometric random variables differ? In what respects are they similar?

Advantages and disadvantages of making observation of behavior as it occurs in natural situations?

Suppose we have $X_1,\dots <!--\; \dots \; -->,X_n$ i.i.d, $X_i\sim <!--\; \sim \; -->\text{Exp}(1,\mathrm{\mu <!--\; \mu \; -->})$ (pdf is $f_{\mathrm{\mu <!--\; \mu \; -->}}(x)=e^{-<!--\; -\; -->(x-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})}$ for $x\ge <!--\; \ge \; -->\mathrm{\mu <!--\; \mu \; -->}$ and $0$ for $x<\mathrm{\mu <!--\; \mu \; -->}$). Is there any one dimensional (i.e. $T:{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$) sufficient statistic for parameter $\mathrm{\mu <!--\; \mu \; -->}$? Obvious two dimensional sufficient statistic is $T(x_1,\dots <!--\; \dots \; -->,x_n)=(\sum <!--\; \sum \; -->x_i,min{x}_i)$

Give examples of various ecological levels and explain how a personal network will assist the process of bereavement within the various levels.

Let X and Y be independent random variables each having the uniform density on {0,1,...,N}. Find $$P(X\ge <!--\; \ge \; -->Y)$$

Homomorphisms between different types of groups
I've been self-studying some Algebra, and recently did a set of exercises proving both that, for a surjective homomorphism $\mathrm{\varphi <!--\; \varphi \; -->}$ between two groups G and H, that if:
(i) G is cyclic, then H is cyclic, and;
(ii) G is abelian, then H is abelian.
Proving these two was simple enough, but now I'm thinking about removing the surjectivity constraint on $\mathrm{\varphi <!--\; \varphi \; -->}$. It does not seem to me that these statements necessarily hold, in either case, but I am having some difficulty coming up with solid counterexamples to either.
I was considering using the Klein group as my H group for part (i), and the dihedral group of order 3 ($D_3$) for part (ii), but I was getting stuck here, as I'm not sure precisely to what I should compare it in order to construct . Any hints/suggestions as to where to look would be greatly appreciated.

Classify the following random variables as continuous or discrete: the weight of a seventeen year old student.

T-Intervals and % Confidence Interval
A random sample of six 2009 sports cars is taken and their "in the city" miles per gallon is recorded. The results are as follows: 23, 19, 24, 17, 16, 22. Assuming the population distribution is normal, calculate the 99% confidence interval for μ, the population mean "in city" mpg for 2009 sports cars.

When using the F distribution to test variances from two populations, should the random variables from each population be independent or dependent?

True and False?
If $A$, $B$, and $C$ are three sets, then the only way that $A\cup <!--\; \cup \; -->C$ can equal $B\cup <!--\; \cup \; -->C$ is $A=B$.

Confidence Intervals (A level)
Firstly let me apologise for asking this question on here - I see this as getting a sledgehammer to crack a nut, but I have nobody else that I can ask for advice on this topic.
I'm an A level Mathematics student so please forgive me for lack of/poor notation that you may normally come to expect/be familiar with.
I am looking at the specification for my exam that is coming up and these are three sections:
A) Construct symmetric confidence intervals for the mean of a normal distribution with known variance
B) Construct symmetric confidence intervals from large samples, for the mean of a normal distribution with unknown variance.
C) Construct symmetric confidence intervals from small samples, for the mean of a normal distribution with unknown variance using the t -distribution.

How do you use a power series to find the exact value of the sum of the series $1+\sqrt{2}+\frac{2}{2!}+\frac{(\sqrt{2}{)}^3}{3!}+\frac{4}{4!}+$...?

Determine the correlation coefficient of the random variables X and Y if var(X) = 4, var(Y ) = 2, and var(X + 2Y ) = 15.

Confidence interval for Uniform($\mathrm{\theta <!--\; \theta \; -->}$, $\mathrm{\theta <!--\; \theta \; -->}+a$)
Consider a random sample $X_1,\dots <!--\; \dots \; -->,X_n$ from a Uniform($\mathrm{\theta <!--\; \theta \; -->}$, $\mathrm{\theta <!--\; \theta \; -->}+a$) distribution, where $\mathrm{\theta <!--\; \theta \; -->}$ is unknown and a is known. We wish to determine a confidence interval for $\mathrm{\theta <!--\; \theta \; -->}$.
The reader may verify the following details: The statistics $Y={\text{min}}_i{X}_i$ and $Z={\text{max}}_i{X}_i$ are jointly sufficient for $\mathrm{\theta <!--\; \theta \; -->}$. For $\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->c_1\le <!--\; \le \; -->c_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a$, $P\{c_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->c_2\}=[(c_2-<!--\; -\; -->c_1)/a{]}^n$. For $0<\mathrm{\gamma <!--\; \gamma \; -->}<1$, set $d_1=(1-<!--\; -\; -->\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}})/2$ and $d_2=(1+\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}})/2$. Then $\mathrm{\gamma <!--\; \gamma \; -->}=P\{\mathrm{\theta <!--\; \theta \; -->}+a{d}_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a{d}_2\}=P\{Z-<!--\; -\; -->a{d}_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->Y-<!--\; -\; -->a{d}_1\}$. Thus, $\mathrm{\gamma <!--\; \gamma \; -->}=P\{\mathrm{\theta <!--\; \theta \; -->}+a{d}_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a{d}_2\}=P\{Z-<!--\; -\; -->a{d}_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->Y-<!--\; -\; -->a{d}_1\}$ is a γ confidence interval for $\mathrm{\theta <!--\; \theta \; -->}$.
Now here's the difficulty: If we observe $Z-<!--\; -\; -->Y>a\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}}$, then $Z-<!--\; -\; -->a{d}_2>Y-<!--\; -\; -->a{d}_1$, so our formula yields a nonsensical answer. Have I made an error in my calculations? Or is this one of those problems with confidence intervals?

Is the growth on an exponential function faster than on a power function? Is this explanation valid?
I'm trying to find simple mathematical proof that lengthy passwords are more secure than complex (with multiple symbols) ones. I don't want to get into the concept of entropy - it is not useful to me atm. I've found this explanation at Security Stack, which it seemed pretty reasonable to me. My question is: is this explanation valid and, if so, is there a limit to it's validity?. My doubt comes from testing it out considering a password with n digits, but restrict to a group of 26 characters (the alphabet) vs a password with fixed 10 digits, with n possible characters:
$$f(n)={26}^n\text{vs}f(n)=n^{10}.$$
For these two functions it just doesn't work out as the answer in Security Stack explained. So, how does this work?
I guess the title may be a bit misleading, because this question markes as duplicate does not answer my question! I get that exponential growth is faster than polynomial growth, but why doesn't it apply to these two functions I'm comparing? I mean, what is wrong in my line of thought?