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?

If I draw one toy at random, what're the odds I'll draw a blue ball?

listgrein6u
2022-09-17
Answered

If I draw one toy at random, what're the odds I'll draw a blue ball?

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asked 2022-09-10

First off, this is a vague question about a survey which is, I guess, meant to be vague. So bear with me

In Morel's "Motivic Homotopy Theory" survey he mentioned the following fact in motivating the Tate circles. One notes that ${\mathbb{P}}^{1}$ is equivalent to ${S}^{1}\wedge {\mathbb{G}}_{m}$ while, if we follow our topological intuition, ${\mathbb{P}}^{1}$ ought to be ${S}^{2}$ which is equivalent ${S}^{1}\wedge {S}^{1}$. So we need to keep track of this difference between the "usual" topological situation and count the number of ${\mathbb{G}}_{m}$'s.

Now Morel justified the first equivalence by saying that ${\mathbb{A}}^{1}$ is invertible in the the unstable homotopy category. With just this information, is there an "intuitive", or maybe "geometric" (i.e. without going to the details of the motivic unstable category) way to see how this fact leads to the equivalence above?

In Morel's "Motivic Homotopy Theory" survey he mentioned the following fact in motivating the Tate circles. One notes that ${\mathbb{P}}^{1}$ is equivalent to ${S}^{1}\wedge {\mathbb{G}}_{m}$ while, if we follow our topological intuition, ${\mathbb{P}}^{1}$ ought to be ${S}^{2}$ which is equivalent ${S}^{1}\wedge {S}^{1}$. So we need to keep track of this difference between the "usual" topological situation and count the number of ${\mathbb{G}}_{m}$'s.

Now Morel justified the first equivalence by saying that ${\mathbb{A}}^{1}$ is invertible in the the unstable homotopy category. With just this information, is there an "intuitive", or maybe "geometric" (i.e. without going to the details of the motivic unstable category) way to see how this fact leads to the equivalence above?

asked 2022-09-20

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?

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?

asked 2022-09-12

Hi I am stuck on this probability question and don't know what to do.

Q. A survey is carried out on a computer network. The probability that a log on to the network is successful is 0.92. Find the probability that exactly five out of nine users that attempt to log on will do so successfully.

Q. A survey is carried out on a computer network. The probability that a log on to the network is successful is 0.92. Find the probability that exactly five out of nine users that attempt to log on will do so successfully.

asked 2022-09-14

A survey of students revealed that 30% of them have a part-time job. If students are chosen at random what is the probability of the following events: That less than 2 have part time work? More than 3 are working part-time? None have a part-time job?

This is a question from a mock exam paper as I prepare for exams.

I guess I could start by imagining a number of 1000 college students surveyed with 300 having part-time jobs.

so for part (i) -> (30/100)(70/10)(70/10)(70/10)(70/10)=7%

That ain't right. Can someone shed some light on this?

This is a question from a mock exam paper as I prepare for exams.

I guess I could start by imagining a number of 1000 college students surveyed with 300 having part-time jobs.

so for part (i) -> (30/100)(70/10)(70/10)(70/10)(70/10)=7%

That ain't right. Can someone shed some light on this?

asked 2022-09-06

Survey of the tax increase

In a survey of 1,000 people, 420 are opposed to the tax increase.

a) Construct a 95% confidence interval for the proportion of those people opposed to the tax increase.

b) Interpret the CI in terms of the question.

c) Is the estimate in part (a) valid? Explain.

My work:

I was able to solve a) and get the answer of [0.4205,0.4195].

For b) I am confused as to how to interpret the CI in terms of the question... could it be that its the percentage range of the possible number of people opposed to the tax increase?

For c) I think that it is valid because the number of people opposed to the tax increase is 420, which lies in the range of the confidence interval.

EDIT:

To solve for a) I did $$p\pm 1.96x[\sqrt{\frac{0.42x0\times 0.58}{1000}}$$, which equalled 0.42 $$\pm $$ 0.0005...

I now see that I forgot to apply the sqrt before multiplying by 1.96. The correct answer should then be: 0.42 $$\pm $$ 1.96x0.0156 = 0.42 $$\pm $$ 0.0306, which results in the range of [0.4506, 0.3894]

In a survey of 1,000 people, 420 are opposed to the tax increase.

a) Construct a 95% confidence interval for the proportion of those people opposed to the tax increase.

b) Interpret the CI in terms of the question.

c) Is the estimate in part (a) valid? Explain.

My work:

I was able to solve a) and get the answer of [0.4205,0.4195].

For b) I am confused as to how to interpret the CI in terms of the question... could it be that its the percentage range of the possible number of people opposed to the tax increase?

For c) I think that it is valid because the number of people opposed to the tax increase is 420, which lies in the range of the confidence interval.

EDIT:

To solve for a) I did $$p\pm 1.96x[\sqrt{\frac{0.42x0\times 0.58}{1000}}$$, which equalled 0.42 $$\pm $$ 0.0005...

I now see that I forgot to apply the sqrt before multiplying by 1.96. The correct answer should then be: 0.42 $$\pm $$ 1.96x0.0156 = 0.42 $$\pm $$ 0.0306, which results in the range of [0.4506, 0.3894]

asked 2022-09-07

How is this the right answer?

In a survey of customer satisfaction, participants are asked to give a score of 1,2,3 or 4 to each of the 6 questions. If participants are instructed not to give the same numerical score to more than 4 questions, how many responses are possible?? I know the answer is 4,020 from the solutions, however I don't understand why. The solution says:

$$4\ast 3\ast {\textstyle (}\genfrac{}{}{0ex}{}{6}{5}{\textstyle )}=72$$

Hence, the number of possible responses is:

$$4,096-4-72=4,020$$

Why and how is this the answer? I don't understand where the −4 comes from.

In a survey of customer satisfaction, participants are asked to give a score of 1,2,3 or 4 to each of the 6 questions. If participants are instructed not to give the same numerical score to more than 4 questions, how many responses are possible?? I know the answer is 4,020 from the solutions, however I don't understand why. The solution says:

$$4\ast 3\ast {\textstyle (}\genfrac{}{}{0ex}{}{6}{5}{\textstyle )}=72$$

Hence, the number of possible responses is:

$$4,096-4-72=4,020$$

Why and how is this the answer? I don't understand where the −4 comes from.

asked 2022-09-04

Sorry if my title sounds vague and inaccurate. I can't think of a better way to put it lol. Anyway, I stumbled upon this problem today while preparing for Oxford tsa:

A survey of households in a town showed that (allowing for sampling errors) between 75% and 85% owned a dishwasher, between 35% and 40% owned a tumble dryer and less than 5% owned neither.

How many people own both a tumble dryer and a dishwasher?

In which the answer is: "between 10% and 30%"

I spent lots of time trying to figure out how to solve it but to no avail. The inclusion of the ranges in a typical Venn diagram question totally throws me off. Can anyone help me with this?

A survey of households in a town showed that (allowing for sampling errors) between 75% and 85% owned a dishwasher, between 35% and 40% owned a tumble dryer and less than 5% owned neither.

How many people own both a tumble dryer and a dishwasher?

In which the answer is: "between 10% and 30%"

I spent lots of time trying to figure out how to solve it but to no avail. The inclusion of the ranges in a typical Venn diagram question totally throws me off. Can anyone help me with this?