# The problem reads: Suppose P(X_1)=.75 and P(Y_2|X_1)=.40. What is the joint probability of X_1 and Y_2? This is how I answered it. P(X_1 and Y_2) =P(X_1)\times P(Y_1|X_1)=.75\times .40=0.3. What I don't understand is how do you get the P(Y_1|X_1)? I am totally new to Statistices and I need to understand each part of the process in order to get the whole concept. Can anyone help me to understand why the P and X exist and what they represent?

Question
The problem reads: Suppose $$\displaystyle{P}{\left({X}_{{1}}\right)}={.75}$$ and $$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}={.40}$$. What is the joint probability of $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$?
This is how I answered it. P($$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$) $$\displaystyle={P}{\left({X}_{{1}}\right)}\times{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}={.75}\times{.40}={0.3}.$$
What I don't understand is how do you get the $$\displaystyle{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}$$? I am totally new to Statistices and I need to understand each part of the process in order to get the whole concept. Can anyone help me to understand why the P and X exist and what they represent?

2021-01-11
$$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$ both represent the occurrence of someevent, e.g. the number that comes up after rolling a fair die or if a person is a smoker or a non-smoker. $$\displaystyle{P}{\left({X}_{{1}}\right)}$$ represents the probability that the event $$\displaystyle{X}_{{1}}$$ occurs. $$\displaystyle{P}{\left(\frac{{Y}_{{2}}}{{X}_{{1}}}\right)}$$ represents the probability that the event $$\displaystyle{Y}_{{2}}$$ occurs given that the event $$\displaystyle{X}_{{1}}$$ has occurred. In other words, we wantto know what chance $$\displaystyle{Y}_{{2}}$$ will occur knowing that $$\displaystyle{X}_{{1}}$$ has already happened.
The joint probability of $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$, denoted by $$\displaystyle{P}{\left({X}_{{1}}\cap{Y}_{{2}}\right)}$$ or $$\displaystyle{P}{\left({Y}_{{2}}\cap{X}_{{1}}\right)}$$
$$\displaystyle{P}{\left({X}_{{1}}\cap{Y}_{{2}}\right)}={P}{\left({Y}_{{2}}\cap{X}_{{1}}\right)}$$
by the commutative property, representsthe probability that both events $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$ occur.
Use the following identity to compute
$$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}={\frac{{{P}{\left({X}_{{1}}\cap{Y}_{{2}}\right)}}}{{{P}{\left({X}_{{1}}\right)}}}}$$
The formula for $$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}$$ can be derived in the following manner: If the event $$\displaystyle{X}_{{1}}$$ occurs, then in order for the event $$\displaystyle{Y}_{{2}}$$ to occur, then they both have to occur. This explains the numerator. Since we know that the event $$\displaystyle{X}_{{1}}$$ has occurred and we are only interested in a small part of $$\displaystyle{X}_{{1}}$$ (case where $$\displaystyle{Y}_{{2}}$$ occurs when $$\displaystyle{X}_{{1}}$$ has occured), then $$\displaystyle{X}_{{1}}$$ represents the set of all possibleoutcomes. This explains the denominator.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
Two stationary point charges +3 nC and + 2nC are separated bya distance of 50cm. An electron is released from rest at a pointmidway between the two charges and moves along the line connectingthe two charges. What is the speed of the electron when it is 10cmfrom +3nC charge?
Besides the hints I'd like to ask you to give me numericalsolution so I can verify my answer later on. It would be nice ifyou could write it out, but a numerical anser would be fine alongwith the hint how to get there.
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when $$Z=1.3$$ and $$H=0.05$$;
Assume that you do not have vales of the area beyond $$z=1.2$$ in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table $$[for\ z=1.3\ and\ H=0.05]$$.
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$
Twelve people join hands for a circle dance. In how manydifferent ways can they do this?
part (b)
Suppose 6 of these people are men and the other six are womenin how many ways can they join hands for a circle dance, assumingthey alternate in gender around the circle?
How do you solve this problem? I don' t even know whereto begin.
A Ferrari with a mass of 1400 kg approaches a freeway underpassthat is 10 m across. At what speed must the car be moving, inorder for it to have a wavelength such that it might somehow"diffract" after passing through this "single slit"? How dothese conditions compare to normal freeway speeds of 30m/s?
$$\begin{array}{cc}\hline & \text{Afraid to walk at night?} \\ \hline & \text{Yes} & \text{No} & \text{Total} \\ \hline \text{Male} & 173 & 598 & 771 \\ \hline \text{Female} & 393 & 540 & 933 \\ \hline \text{Total} & 566 & 1138 & 1704 \\ \hline \text{Source:}2014\ GSS \end{array}$$
If the chi-square $$(\chi 2)$$ test statistic $$=73.7$$ what is the p-value you would report? Use Table C.Remember to calculate the df first.
Group of answer choices
$$P>0.250$$
$$P=0.01$$
$$P<0.001$$
$$P<0.002$$

$$\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}$$