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Suppose that a person becomes unemployed. What is the probability that this person also was college educated?
I have this question that I am having a hard time understanding. And how I should go about regarding questions like this. Could someone explain how my thought process should be regarding questions like these or help me out?
Question:
Unemployment is affecting different groups differently. According to a American survey the unemployment on a national level was 13%, but only 6% for people with a college education. Suppose that 21% of the people in the workforce that is college educated. Now suppose that a person becomes unemployed. What is the probability that this person also was college educated?
The answer should be 9.7%, but I can't figure out how to get to this number.

"In a recent survey it was found that 50% of investors placed their money in Apple shares while 60% invested in Google. Of the people surveyed, 15% invested in neither of these.
Find the proportion of investors that had both Apple and Google shares."
I have tried calculating as two independent events:
$$Pr(A\cap B)=Pr(A)\cdot <!--\; \cdot \; -->Pr(B)=0.5\cdot <!--\; \cdot \; -->0.6=0.3,$$
which is incorrect.
According to the answers, the correct answer should be 0.25, but I'm not able to find this at all. Any help is greatly appreciated!

Q: In a survey of a group of 100 computing students, it was found that 32 knew Java, 33 knew C, 32 knew Python, 5 knew both Java and C, 7 knew both C and Python while none of the students knew both Java and Python. How many students knew none of the three languages?
Attempt:
First attempt was I guessed 3 because it only adds up to 97 so I assumed 3 students knew none of them.
Then I tried it out and got 27 for java, 21 for C and 25 for python, added those up and got 72. Minus 72 from 100 and got 28 as my final answer for students who know none.
My teacher has no marking scheme for this question, this is a practice q. Anyone know the solution?

Survey: prove 10% of population "Don't know"?
I have a simple question that I expect is a standard situation, but I can't seem to find the right answer (maybe because it's too simple).
I have a survey with the following answers:
(What is your current risk tolerance for your personal investments?)
Very Aggressive 1
Aggressive 11
Moderate 29
Conservative 7
Very Conservative 4
I don't know my risk tolerance 6
(total) 58
I want to be able to say something about the 6 "Don't know". I have tried to prove that 10% of the population "don't know", by using a chi-square test:
Null hypothesis: 10% of the population "don't know"
yes no
expected: 52.2 5.8
observed: 52 6
Chi-square = (52-52.2)^2/52.2 + (6-5.8)^2/5.8 = 0.0077
Chi-square for a two-tailed 1-df test is 3.84
Since 0.0077 < 3.84 I cannot reject the null hypothesis and accept that 10% of the population "don't know".

Question : In a road traffic survey the number of cars passing a marker at a road was counted. The streams of cars in the two directions were a priori modelled as independent Poisson processes of intensities 2 per minute and 3 per minute, respectively. It was decided to stop the counting once 400 cars had passed. Let $T_{400}$ be that (random) time point and compute, using appropriate approximations, a time α such that $P(T_{400}\le <!--\; \le \; -->\mathrm{\alpha <!--\; \alpha \; -->})=0.90$
Claimed Answer 85.1 minutes.
My Attempt : Since 400 cars passing with a rate of 5 cars/minute to pass in approximately $\frac{400}{5}=80$ min. However this is not correct.

A survey of 64 medical labs revealed that the mean price charged for a certain test was Rs. 120, with a standard deviation of 60. Test whether the data indicates that the mean price of this test is more than Rs. 100 at 5% level of significance.
I have solved this question but I don't know whether the answer is correct or not.
H0: mean = 120 (null hypothesis) H1: mean > 100 (alternative hypothesis)
we will use z test as the sample count is more than 30
$$\begin{gathered} z=|120-<!--\; -\; -->100|/60/\sqrt{64} \\ z=2.67 \end{gathered}$$
at 5% of significance, the critical value of z is 1.96. Since the z value we obtained is more than 1.96, so we reject the null hypothesis and therefore the mean price of the test is more than 100
Please tell whether the answer is correct or there is some mistake in this. Help is appreciated.

I have just started a course in statistics and have some general questions that have arisen trying to solve the following question:
A survey organisation wants to take a simple random sample in order to estimate the percentage of people who have seen a certain programme. The sample is to be as small as possible. The estimate is specified to be within 1 percentage point of the true value; i.e., the width of the interval centered on the sample proportion who watched the programme should be 1%. The population from which the sample is to be taken is very large. Past experience suggests the population percentage to be in the range 20% to 40%. What size sample should be taken?
I think I have to use this and solve for n
$$1.96\sqrt{\frac{\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->})}{n}}=.01$$
where $\mathrm{\pi <!--\; \pi \; -->}$ is the sample estimated proportion of people who watch the programme.
Now does this mean that I am 95% sure that I am within 1% accuracy? I also am not aware as to how I can find π though I have read I could use the population standard deviation instead and suspect I would have to use that as I am given some information- that the pop proportion is 20%-40%.
Finally in general what is being said here:
$$\mathrm{\pi <!--\; \pi \; -->}\pm <!--\; \pm \; -->1.96\sqrt{\frac{\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->})}{n}}=.01$$
My notes at the moment just say it contains the population mean 95% of the time.... why? I think if I had some graphical understanding of what was going on everything would be much simpler for me.

Whilst i used to be writing a literature survey on Moore-Penrose pseudoinverse (literatures like this one, and this one), I encountered with the following equality which was named as opposite order regulation:
$$(AB{)}^{+}=B^{\ast <!--\; \ast \; -->}(A^{\ast <!--\; \ast \; -->}AB{B}^{\ast <!--\; \ast \; -->}{)}^{+}{A}^{\ast <!--\; \ast \; -->}.$$
They claimed this equality, by noticing
$$AB=((A^{\ast <!--\; \ast \; -->}{)}^{+}{A}^{\ast <!--\; \ast \; -->}A)(B{B}^{\ast <!--\; \ast \; -->}(B^{\ast <!--\; \ast \; -->}{)}^{+})=(A^{\ast <!--\; \ast \; -->}{)}^{+}(A^{\ast <!--\; \ast \; -->}AB{B}^{\ast <!--\; \ast \; -->})(B^{\ast <!--\; \ast \; -->}{)}^{+},$$
and then applying a 'general reverse order law' $(PNQ{)}^{+}=Q^{+}{N}^{+}{P}^{+}$ here. In these papers, this law $(PNQ{)}^{+}=Q^{+}{N}^{+}{P}^{+}$ was widely used to derive equalities on $(AB{)}^{+}$.
My question here is that $(PNQ{)}^{+}=Q^{+}{N}^{+}{P}^{+}$ cannot be true for arbitrary matrix $P,N,Q$. Are there any extra requirements for it to hold, how is it proved, and how does it fit in the above case? Are there any references?
References on the first discovery of equality $(AB{)}^{+}=B^{\ast <!--\; \ast \; -->}(A^{\ast <!--\; \ast \; -->}AB{B}^{\ast <!--\; \ast \; -->}{)}^{+}{A}^{\ast <!--\; \ast \; -->}$ are also appreciated.

Given a vector x in the n=6-dimensional Euclidian space ${\mathbb{R}}^n$, do there exist n−1 continuous functions $f_1$ to $f_{n-<!--\; -\; -->1}$ such that the matrix
$$(x,f_1(x),\dots ,f_{n-<!--\; -\; -->1}(x))$$
is invertible for all x in ${\mathbb{R}}^n$.
I am aware of the results of
T. Wazewski, Sur les matrices dont les elements sont des fonctions continues. Composito Mathematica, tome 2 (1935), p. 63-68
B Eckmann, Mathematical survey lectures 1943-2004. Springer 2006.
There may be no topological obstruction for n=6 while there may be for n unusual. I know the answer for n=2 and four. it is based totally on diversifications with appropriate signs and symptoms in an effort to make $f_i(x)$ orthogonal to x. this doesn't paintings for n=6. as a result the question for n=6 or even large n=8, …

A problem of Set Theory The question is :
In a market survey,a manufacturer obtained the following data :
Did you use our brand? Percentage answering yes
1. April 59
2. May 62
3. June 62
4. April and May 35
5. May and June 33
6. April and June 31
7. April,May and June 22
Is this correct?
That is the question.Now I don't really understand how to determine yes or not.Please help me.

I had this question when I was shopping for some rates to refinance my house: how do know if I have found the lowest rates after searching through a bunch of offers with a reasonable amount of confidence, in a mathematical/statistical term. For example, if found "x" offers from different vendors, is the amount of "x" large enough to represent the bigger population i.e. all offers out there?
The reason I am asking is that I vaguely remember there are statistical sampling techniques from college days that survey companies used to represent the bigger populations. For instance, after surveying 2000 people, they can conclude something for the U.S. population. In other words, they "know" 2000 people can represent the U.S. population with some scientific backing.
Just curious, if there is a way to figure out the magical number "x", then I can go with the lowest rate after browsing through "x" vendors as well. The question is purely theoretical in this simple case: the lowest rate wins. Any leads or ideas are appreciated.

Figuring out probability of two random events both happening
So here's the problem:
The table below shows the distribution of education level attained by US residents based on data collected during the 2010 American Community Survey:
Highest level of education %
Less than 9th Grade 0.10
9th to 12th no diploma 0.09
High school grad - GED 0.25
Some college No degree 0.23
Associate's degree 0.08
Bachelor's degree --
Graduate or professional degree 0.09
Answer the following questions (give all answers to 2 decimal places):
a) Fill in the empty box for the proportion of US residents whose highest education level attained was a bachelors degree.
b) If two individuals are chosen at random from the population, what is the probability that both will have at least a bachelors degree?
c) If two individuals are chosen at random from the population, what is the probability that at least one will have some college or a college degree of some sort?
d) If two individuals are chosen at random from the population, what is the probability that exactly one will have some college or a college degree of some sort?
I managed to figure out a) which was 0.16. However, I tried a different methods to get b) but none of them worked. I answered .24, .25, and .01 but none of them were correct.
To answer b) I used the following formula:
P(A or B) = P(A) + P(B) - P(A & B) = (.16) + (.09) - (.16)(.09)
which gave me .24 but that was incorrect. What am I doing wrong?

How many different ways can a group of students be hired to work a survey?
The Statistics Survey Centre wishes to hire some undergraduates to help in a phone survey. On the call list are 6 second-year students, 3 third-year and 3 fourth-year students. Policy is to always hire at least one third year and one fourth year student. Other than that as many students as are needed can be hired without restriction.

Statement:
A manufacturer has been selling 1000 television sets a week at $480 each. A market survey indicates that for each $11 rebate offered to a buyer, the number of sets sold will increase by 110 per week.
Questions :
a) Find the function representing the revenue R(x), where x is the number of $11 rebates offered.
For this, I got (110x+1000)(480−11x). Which is marked correct
b) How large rebate should the company offer to a buyer, in order to maximize its revenue?
For this I got 17.27. Which was incorrect. I then tried 840999, the sum total revenue at optimized levels
c) If it costs the manufacturer $160 for each television set sold and there is a fixed cost of $80000, how should the manufacturer set the size of the rebate to maximize its profit?

How Can I calculate the standard deviation of a yes/no question?
Is it possible to calculate the standard deviation of a single yes/no question? For example, 100 people took the survey and 60 of them chose "yes", 40 of them chose "no". I'm thinking using binomial distribution sd formula: $$\sqrt{np(1-<!--\; -\; -->p)}$$ so it would be $\sqrt{100\ast <!--\; \ast \; -->(60/100)\ast <!--\; \ast \; -->(1-<!--\; -\; -->60/100)}$is this correct?

How do I 'reverse engineer' the standard deviation?
My problem is fairly concrete and direct.
My company loves to do major business decisions based on many reports available on the media. These reports relates how our products are fairing in comparison to the competitor's offerings.
The latest report had these scores (as percentages):
Input values (%) 73.5, 16.34, 1.2, 1.15, 0.97, 0.94, 0.9, 0.89, 0.81, 0.31
Our product in the 'long' tail of this list. I argued with them that besides spots #2 and #1 all the other following the tail are on a 'stand', since, probably, the standard deviation will be much bigger that the points that separates everyone in the tail.
So the question is: How may I calculate the standard deviation having only these percentual values available?

A recent survey of post-secondary education students in Canada revealed that 73% know what type of job they want, when they graduate. You are to randomly pick 32 post-secondary education students across the country, and ask each the following question: Have you selected a particular career path?
You have defined the random variable X to represent the number, out of 32 post-secondary students chosen, who responded YES
If you continued to select additional students, what is the probability that the 50th student selected will be the 39th student to respond YES?

Which summary statistic and hypothesis test should be used?
A survey is conducted on 398 people that are either of type A (198) or type B (200) and answer 1 question on a 5 point scale 1=strongly agree, 2= agree,...5=strongly disagree.
What is the correct summary statistic and what hypothesis test should be used in answering: if the 2 types of people have differing opinions regarding the question?
attempt
Summarize using a 2 way table of type vs response (1,2,...,5), with a vector $\stackrel{^<!--\; ^\; -->}{p}=\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\\ p_4\\ p_5\end{array}\right)$ as the summary statistic
This seems like a test of homogeneity is appropriate to answer the question but these people are drawn from the same population and cross-classified (right?) so then a test of independence (however a test of indep is unrelated to the question?)

Australia is having a national "postal survey" about same-sex marriage.
The options on the ballot paper are "Yes" and "No", but voting is not compulsory.
The survey is being conducted over 8 weeks during which eligible voters can return their response.
At the end of the third week, 60% of voters have returned their response.
Exit-polling indicates of those who have already voted, 60% voted "Yes" and 40% voted "No".
Question: All other things being equal (e.g. early/late responders no indication of which way they vote) what % of the remaining 40% who haven't returned their response would need to vote "No" in order to achieve parity with the "Yes" vote (here's the kicker...) taking into account that not all eligible voters will ultimately return their ballots?
Is this work-outable?

I recently have this question:
I have a bag of toys. 10% of the toys are balls. 10% of the toys are blue.
If I draw one toy at random, what're the odds I'll draw a blue ball?
One person provided an answer immediately and others suggested that more details were required before an answer could even be considered. But, there was a reason I asked this question the way that I did.
I was thinking about probabilities and I was coming up with a way to ask a more complicated question on math.stackexchange.com. I needed a basic example so I came up with the toys problem I posted here.
I wanted to run it by a friend of mine and I started by asking the above question the same way. When I thought of the problem, it seemed very clear to me that the question was "what is $\mathbb{P}(blue\cap <!--\; \cap \; -->ball)$." I thought the calculation was generally accepted to be
$$\mathbb{P}(blue\cap <!--\; \cap \; -->ball)=\mathbb{P}(blue)\cdot <!--\; \cdot \; -->\mathbb{P}(ball)$$
When I asked my friend, he said, "it's impossible to know without more information." I was baffled because I thought this is what one would call "a priori probability."
I remember taking statistics tests in high school with questions like "if you roll two dice, what're the odds of rolling a 7," "what is the probability of flipping a coin 3 times and getting three heads," or "if you discard one card from the top of the deck, what is the probability that the next card is an ace?"
Then, I met math.stackexchange.com and found that people tend to talk about "fair dice," "fair coins," and "standard decks." I always thought that was pedantic so I tested my theory with the question above and it appears you really need to specify that "the toys are randomly painted blue."
It's clear now that I don't know how to ask a question about probability.
Why do you need to specify that a coin is fair?
Why would a problem like this be "unsolvable?"
If this isn't an example of a priori probability, can you give one or explain why?
Why doesn't the Principle of Indifference allow you to assume that the toys were randomly painted blue?
Why is it that on math tests, you don't have to specify that the coin is fair or ideal but in real life you do?
Why doesn't anybody at the craps table ask, "are these dice fair?"
If this were a casino game that paid out 100 to 1, would you play?
This comment has continued being relevant so I'll put it in the post:
Here's a probability question I found online on a math education site: "A city survey found that 47% of teenagers have a part time job. The same survey found that 78% plan to attend college. If a teenager is chosen at random, what is the probability that the teenager has a part time job and plans to attend college?" If that was on your test, would you answer "none of the above" because you know the coincident rate between part time job holders and kids with college aspirations is probably not negligible or would you answer, "about 37%?"