Test for HIV. There's a false positive rate of 0.025 and a false negative rate of 0.08. Let's say a particular patient has a probability of testing positive for HIV of 0.005. The patient gets tested and it's positive. What are the chances that the patient actually has HIV?

The scatter plot below shows the average cost of a designer jacket in a sample of years between 2000 and 2015. The least squares regression line modeling this data is given by $\hat{y}=-4815+3.765x.$ A scatterplot has a horizontal axis labeled Year from 2005 to 2015 in increments of 5 and a vertical axis labeled Price ($) from 2660 to 2780 in increments of 20. The following points are plotted: $(2003,2736),(2004,2715),(2007,2675),(2009,2719),(2013,270)$. All coordinates are approximate. Interpret the y-intercept of the least squares regression line. Is it feasible? Select the correct answer below: The y-intercept is −4815, which is not feasible because a product cannot have a negative cost. The y-intercept is 3.765, which is not feasible because an expensive product such as a designer jacket cannot have such a low cost. The y-intercept is −4815, which is feasible because it is the value from the regression equation. The y-intercept is 3.765 which is feasible because a product must have a positive cost.

Confidence interval for a density function parameter
I am trying to solve this problem: Given ${\displaystyle X_1,\dots ,X_n}$ a random sample of a population of random variables with p.d.f.
${\displaystyle f(x,\theta )=e^{-(x-\theta )}I{\left(x\right)}_{x\ge \theta }}$, find a confidence interval of level ${\displaystyle 1-\alpha \text{for}\theta }$.SNKFollowing a hint, I first looked into the distribution of ${\displaystyle Y=\text{min}\left(X_i\right)-\theta }$, which I found to be Exponential(n).
Taking into account that the indicator function in the pdf limits to values of x greater than ${\displaystyle \theta }$, and using the exponential's cdf. and some bounds, I got that the confidence interval should be ${\displaystyle I_{1-\alpha }(X_1,\dots ,X_n)=[0,\frac{\ln \left(\alpha \right)}{n}]}$.
Would that be correct? I am not sure I am using the hint correctly. If it is wrong, what else could I try?

Assume that the true value, that is the population mean, the average height of french women in france is 167.5 cm. Students in a class of 160 students each took a random sample of french women of size 100, measured the height of the women and calculated a 95% confidence interval (confidence interval) for the average height.How many students can be expected to calculate a confidence interval that does not include the number 167.5?
15 8 5 20 1

The expectation of the number of people getting their correct hats in the hat check problem, using linearity of expectations.
n persons receive n hats in a random fashion.
Formula used: By linearity of expectation function,
If $X_1,X_2,...,X_n$ are random variables, then
$E(X_1,+X_2+,...+X_n)=E(X_1)+E(X_2)+...+E(X_n)$ (2)

Express the confidence interval ${\displaystyle 172.3<\mu <229.1}$ in the form of ${\displaystyle \bar{x}\pm ME.}$

if 1:1000 of people is sick. the probability to be false positive is 0.07. if a person is sick there is not chance the test for the disease is wrong. If someone random is got a positive result, what are the chances he's actually sick?
I got 1.5% and wanted to check because I feel it should be more since the diagnose is never wrong. I took 1/1000 and divided it by 0.07+1/1000 and multiplied by 100 to get the percent.

You are conducting a test of independence for the daim that there is an association between morning beverage and age.
$\begin{array}{|ccccc|}\hline & \text{ Juice }& \text{ Coffee }& \text{ Pop }& \text{ Tea }\\ \text{ less that 25 }& 19& 18& 5& 32\\ \text{ 25 or older }& 27& 30& 2& 43\\ \hline\end{array}$
What is the chi-square test-statistic and the corresponding p-value for this data?
${\displaystyle {\chi }^2=?}$, [three decimal accuracy]
p-value=?, [three decimal accuracy]