Master Statistics and Probability Problems with Expert Help

Let $X_1,\dots <!--\; \dots \; -->,X_n$ be independent random variables with densities:
$f_{X_k}(x\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})=\{\begin{array}{ll}{e}^{k\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->x}& x\ge <!--\; \ge \; -->k\mathrm{\theta <!--\; \theta \; -->}\\ 0& \text{otherwise}\end{array}$
Find the pmf for $T=\underset{k}{min}\left(\frac{X_k}{k}\right)$.

Show that $\frac{1}{n}(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->\frac{1}{1-<!--\; -\; -->X_i}{)}^3$ is a sufficient statistic for $\mathrm{\beta <!--\; \beta \; -->}$ in a Beta $(\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->})$ density

Finite Math (Probability/Venn Diagram)
13) A survey revealed that 25% of people are entertained by reading books, 39% are entertained by watching TV, and 36% are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.
Is this problem stated correctly? How can 36% of the people be entertained by both books and TV, when only 25% of the people are entertained by reading books?
EDIT
Here are two other questions from the exam, that the instructor said followed the same logic as the question above.
14) Of the coffee makers sold in an appliance store, 5.0% have either a faulty switch or a defective cord, 1.6% have a faulty switch, and 0.2% have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
15) A survey of senior citizens at a doctor's office shows that 42% take blood pressure-lowering medication, 45% take cholesterol-lowering medication, and 13% take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.

Why is sample size important when analyzing data?

Choose the correct answer:
We use the t distribution to perform a hypothesis test about the population mean when:
a. The population from which the sample is drawn is approximately normal and the population standard deviation is known
b. The population from which the sample is drawn is not approximately normal and the population standard deviation is known.
c. The population from which the sample is drawn is approximately normal or sample size is 30 or larger and the population standard deviation is unknown.
d. The population from which the sample is drawn is not approximately normal and the population standard deviation is unknown.

Finding a population Function
I have been given the population of the USA from 1790 - 1980 (increasing in intervals of 10) and I am asked to solve this differential equation.
Using t as time in years, and P as size of population at any time t.
It shows $dP/dt=(b-<!--\; -\; -->d)P$.
I assume b and d are birth and death rates per 1000.
I have subbed in $B-<!--\; -\; -->d=13-<!--\; -\; -->8$.
I'm kinda puzzled I don't know what to do. I have made a table and graph on Excel with the data but I'm clueless. Any ideas guys and gals?

What is the exact value of $\sin <!--\; \; -->(45+30)$?

Getting P-value While Using Variance
Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean ${\mathrm{\mu <!--\; \mu \; -->}}_1$ and unknown variance ${\mathrm{\sigma <!--\; \sigma \; -->}}_1^2$, A second random sample from another normal population with unknown mean ${\mathrm{\mu <!--\; \mu \; -->}}_2$ and unknown variance ${\mathrm{\sigma <!--\; \sigma \; -->}}_2^2$ yields the measurements: 13, 7, 9, 11.
a) Test for evidence that ${\mathrm{\sigma <!--\; \sigma \; -->}}_1>1.0$. Complete the P-value for this test as accurately as possible. Draw a conclusion at $\mathrm{\alpha <!--\; \alpha \; -->}=0.05$.
Here's what I've done so far:
Step 1: Calculate ${\mathrm{\sigma <!--\; \sigma \; -->}}_1$
${\mathrm{\sigma <!--\; \sigma \; -->}}_1=\sqrt{\frac{(10-<!--\; -\; -->14{)}^2+(13-<!--\; -\; -->14{)}^2+(15-<!--\; -\; -->14{)}^2+(15-<!--\; -\; -->14{)}^2+(17-<!--\; -\; -->14{)}^2}{5}}=\sqrt{\frac{28}{5}}=2.366$
Step 2: Set up Hypothesis Test
$H_0:{\mathrm{\sigma <!--\; \sigma \; -->}}_1=2.366$
$H_a:{\mathrm{\sigma <!--\; \sigma \; -->}}_1>1.0$
How do I proceed from here? Thanks.
EDIT:
Also have this question, and would appreciate some insight.
b) Use the pivotal method(and a pivotal statistic with F distribution) to derive a 95% confidence interval for $\frac{{\mathrm{\sigma <!--\; \sigma \; -->}}_2}{{\mathrm{\sigma <!--\; \sigma \; -->}}_1}$. Work it out for these data. And test the null hypothesis that ${\mathrm{\sigma <!--\; \sigma \; -->}}_2={\mathrm{\sigma <!--\; \sigma \; -->}}_1$ at the 5% level of significance.

A survey is mailed to a random sample of residents in a city asking whether or not they think the current mayor is doing an acceptable job. What type of bias do you think would most likely be introduced in this type of situation? You can argue for more than one answer but please select the answer that would be the main source of bias. (This question is from your textbook) Group of answer choices
A. Response bias.
The wording of the survey may be confusing or provoke a certain response.
B. Undercoverage.
The entire population is not reached,
C. Nonresponse bias.
People who feel strongly about the mayor are more likely to respond.

We have data come from a normally distributed population with standard deviation 2.6, what sample size is needed to make sure that with 99% probability, the mean of the sample will be in error by at most 0.25?

How do you find the exact value of $\csc <!--\; \; -->(\frac{17\mathrm{\pi <!--\; \pi \; -->}}{6})$?

Solve clustering problem. After step $1$ i.e putting each data-point to a cluster now calculate the mean or average of all of the data-points in a specific cluster. let suppose We have some data-points like $A(2,3),B(4,5),C(6,7),D(8,9)$ in a cluster. How can we calculate their average?

Why is a sample statistic a random variable?

Degree of freedom and corrected standard deviation
It is often said that degree of freedom causes the need for standard deviation formula to be corrected. When explaining degree of freedom, it is often said that when one knows the mean of the formula, only $n-<!--\; -\; -->1$ data are actually needed, as the last data can be determined using mean and $n-<!--\; -\; -->1$ data. However, I see the same thing occuring in population - not just in sample. So what's going on here, and how is this justification really working?
For example, in simple linear regression model, variance of error terms are often sum of variance of each data divided by $n-<!--\; -\; -->2$. This is justified as said above. But if this justification is also true for population, not just sample, how is this really working?

Assume we have $Y_1,...,Y_n$ iid sample from the uniform distribution $U(0,\mathrm{\theta <!--\; \theta \; -->})$. Assume $T(Y_1,...,Y_n)=Mi{n}_{1<=i<=n}(Y_i)$

$f$ cts on $[a,b]$ and $f(x)\ne <!--\; \ne \; -->0$ for all $x\in <!--\; \in \; -->[a,b]$ implies that $f(x)$ is either always positive or negative on $[a,b]$.

According to frequentists, why can't probabilistic statements be made about population paramemters?
The confidence interval can be expressed in terms of samples (or repeated samples): "Were this procedure to be repeated on multiple samples, the calculated confidence interval (which would differ for each sample) would encompass the true population parameter 90% of the time."[1] Note that this does not refer to repeated measurement of the same sample, but repeated sampling.
And:
The confidence interval can be expressed in terms of a single sample: "There is a 90% probability that the calculated confidence interval from some future experiment encompasses the true value of the population parameter." Note this is a probability statement about the confidence interval, not the population parameter.
And:
A 95% confidence interval does not mean that for a given realised interval calculated from sample data there is a 95% probability the population parameter lies within the interval, nor that there is a 95% probability that the interval covers the population parameter.[11] Once an experiment is done and an interval calculated, this interval either covers the parameter value or it does not; it is no longer a matter of probability.

Calculate an average or typical time. If I would just use the median (or other "normal" types of calculating an average), for example $23:59$ and $00:01$ would yield $12:00$ when it should $00:00$. Is there a better method?

Binomial distribution sample vs. population mean
I'm a little confused at this question posed by my prof. He asked us to generate a binomial distribution in R and input whatever variables we wanted.
$x=rbinom(50,10,0.83)$
Then he asks us to compute the sample mean, sample variance, population mean and population variance of the distribution.
sample mean: mean(x)
sample var: var(x)
But I have no idea what he intends we do to get the population mean and variance. Don't you need a larger set of data to be the population and a smaller set to be the sample? I only (seem to) have one set here.

Estimating Gaussian parameters of a set of data points
I have a set of data points. When I draw a histogram of them, plotting their frequency of occurrence against them, I get a curve that looks like a normal curve. I am also able to perform test on the data set to know whether it follows a normal distribution or more precisely whether the population it comes from follows a normal probability distribution. I am using Shapiro Wilk test for it.
However, how can I know what the equation of that normal curve will be? Moreover, is there a way I can test whether other standard distributions fit the points more accurately, and estimate their parameters?