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At a large block party, there are 40 men and 30 women. You wantto ask opinions of how to improve the next party. You choose atrandom four of the men and separately ...

Question

At a large block party, there are 40 men and 30 women. You wantto ask opinions of how to improve the next party. You choose atrandom four of the men and separately choose at random three of thewomen to interview.a) What is the probability that any of the 40 men is selected tobe in your sample. Any of the 30 women?b) Why is your sample of the seven men and women not an SRS ofpeople at the party?

At a large block party, there are 40 men and 30 women. You want to ask opinions of how to improve the next party. You choose at random four of the men and separately choose at random three of the women to interview. a) What is the probability that any of the 40 men is selected to be in your sample. Any of the 30 women? b) Why is your sample of the seven men and women not an SRS of people at the party?



Answers

For a group of 40 people, what is the probability that exactly three people in the group will celebrate their birthdays on a Wednesday this year?
a. Find the probability by using the binomial probability formula.
b. Find the probability by designing and running a simulation.
c. Compare your results for parts (a) and (b). Explain any discrepancy.

Question, 24 says at a party there are 30 students over the age of 21 and 20 students under the age of 21. You choose three of those over 21 2 of those under 21 separately to interview about their attitudes towards alcohol. You've given every student in the party the same chance of being interviewed. What is that chance? It's one out of 10. So you took three out of 30 and then two out of one. So it was a one in 10 chains. The second part question asked, Why is the sample not? And that's our US because it is stratified. You're getting the same amount of our same chance of based off of each group. But is this a stratified? So it's not a random sample.

Okay, so let's first right out and organize all the information that we're given. So we have four of the faculty members that are female. Nine of them are male and for under 40 or right you of 40 2/4 of the female and 3/9 of the males. So now for Port A, we know that we're gonna have a denominator of 13. 13 is the total number of people. And what's the probability that a randomly selected person is a female or under 40? So we already know that there's four total females, and that's including either under 40 or above 40. And then we to also add the number of males that are under 40. And that's three in this equal 7/13. And that is the probability of having either a female or a male under 30 that's under 40. This the same exact thing. Okay, so for Part B could use a similar logic, so we know regardless, the denominators 13. There's 13 total people, so the probability that we have a male or someone over 30 is we know at least nine because there's nine total males and then two out of four of the females are over 40. So that means there's an 11 out of 13 chance of having either a male or a female over 40. This is these are your two answers.

All right, So this question asks us about a sampling distribution for a sample proportion where the point estimate is point for and we have a sample size of 200. So the first thing we should do is start establishing our sampling distribution. So remember that for proportions, the mean of our distribution is just equal to our sample proportion. But then our standard error is the square root of our sample proportion times one minus the sample proportion all over the sample size. So in this case, that's square root of point for times 0.6 over 200. And this works out to be 0.0 three 464 So now we know they're sampling. Distribution is normal with a mean of 0.4 and a standard error of 0.0 surgery 464 So now we can compute some probabilities. So part, eh? Once the probability that we're within 0.3 of the mean, which is the same thing is asking the probability that were between 0.37 and 0.43 which we get by adding and subtracting 0.3 from the mean. And this is an area problem with a normal curve. So we use normal CDF with a lower bound of 0.37 an upper bound of 0.43 a mean of 0.40 and a standard error of 0.34 64 And this plugging it into your calculator works out to be point 613 five. Then Part B asks us the probability that we're within 0.5 so higher this time. Ah, higher tolerance, which converting this it's the same thing is asking probability between 0.35 in 0.45 which again could be written as an area problem. So normal CDF are lower bound is 0.35 Our upper bound is 0.45 Armenia's point for, and our standard air is still 0.3464 and that works out to be point 8511

So in the question, it was stated. The difficulty of mathematics department at some college is composed of four females and nine mils. So a number of females close to four and number of males equals to nine. So of the four females, too, are under age 40. So under age 40 is it close to and up? 40 is close to two. And among the male of the males, three are under age 40. So under age under 43 and, uh, for should be thinks so in question A uh So now so in question, you just asked that I have to find out the probability of randomly selected faculty member a female or underage for people. So the probability of mhm female are our nerves age 40 under 40 years now and the total is nine plus four equals two. 13 No, among them, females are four, so they are nervous for 24 days calculated here plus three. Another 40 this tree. So finally the answer is seven by 13 and in question, be in question be I have to find out the probability of mail or over age 40 So the probability of male are over age for the equals two, the total period is starting of the male R nine. So this is nine, uh, upper age for these calculated here. So we have just helped with this too, which is 2 13. So the answer is 11 by 13. So they said the two probabilities.


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