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N'ez 2'_ 1'hn '= 6'ATTD~tuentt FltenltnA UCSD break theFroulem [10 pint-| Suppose depending On thcir irint Giluation' living camnus do...

Question

N'ez 2'_ 1'hn '= 6'ATTD~tuentt FltenltnA UCSD break theFroulem [10 pint-| Suppose depending On thcir irint Giluation' living camnus dora Those livin sndutu} JJo nrtmcnI?Those livingomCcatinn *Eat Jrar using 4 Markor We wi-h model the number of studlents each living dorins will conduct surizy which indicates that those" students living in chning Suppose apurtment with anal with probability Kill move off cumpu: star the dorms apartmenta Ilot with probability that

n'ez 2'_ 1' hn '= 6' ATTD ~tuentt FltenltnA UCSD break the Froulem [10 pint-| Suppose depending On thcir irint Giluation ' living camnus dora Those livin sndutu} JJo nrtmcnI? Those living omC catinn *Eat Jrar using 4 Markor We wi-h model the number of studlents each living dorins will conduct surizy which indicates that those" students living in chning Suppose apurtment with anal with probability Kill move off cumpu: star the dorms apartmenta Ilot with probability that those living in off campus Wll mOIe Dutoll and mot will stay in An apartment with probability dort with probability home Will moie dorn with probability and finally that those living Oma probability ? and will stay living at home Will mOTe apartmnent wich probability curent enrollment of 30,000 undergrads stays constant . with prolbility Suppose the Apcct 3q 01 cach living situation? students would the long term, how Iany ; ` `



Answers

The U.S. Census Bureau provides data on the number of young adults, ages $18-24,$ who
are living in their parents' home. Let
$$\begin{aligned} M &=\text { the event a male young adult is living in his parents' home } \\ F &=\text { the event a female young adult is living in her parents' home } \end{aligned}$$
If we randomly select a male young adult and a female young adult, the Census Bureau
data enable us to conclude $P(M)=.56$ and $P(F)=.42$ (The World Almanac, $2006 ) .$ The
probability that both are living in their parents' home is 24 .
$\begin{array}{l}{\text { a. What is the probability at least one of the two young adults selected is living in his or }} \\ {\text { her parents' home? }}\end{array}$
$\begin{array}{l}{\text { b. What is the probability both young adults selected are living on their own (neither is }} \\ {\text { living in their parents' home)? }}\end{array}$

For this problem. We're told that 2% of two million high school students would take the S A T every year, requires special accommodations and then were asked to consider a random sample of 25 students who have recently taken the test. And we're asking probabilities so weakened we can define X as the number of successes being the number of students who required special accommodation and because we're sampling without replacement from the population. The samples are not strictly independent, but because 25 is far less than 5% of the total population, X can be pretty reliably estimated as a binomial random variable based on 25 size of 25 and the probability of success of 0.2 part A were asked, What is the probability that exactly one required special accommodation? What is the probability that we have exactly one success? So that's the probability mass function for the binomial, and this comes out to approximately zero point 308 now for B were asked, what is the probability that at least one received special accommodation? So what is the probability that we have at least one success we can rewrite this as one minus the probability of getting exactly zero successes, and this comes out to approximately zero point 397 or C. We're looking for the probability of at least two successes, so this could be rewritten as one minus probability. That X is less than or equal to one, and this equals one minus 0.911 which equals 0.89 for a party, you were asked, what is the probability that the number of successes is within two standard deviations of the mean number of successes or the expected number of successes? So to calculate this, we must calculate the mean number of successes where this distribution and the standard deviation So for a binomial random variable, the mean or expected value for the number of successes is he going to end terms? P, which comes out to you 1/2 and the standard deviation of a binomial random variable is given by the square root of N times. P Times Q, which comes out to 0.7, so within two standard deviations of the expected value. We're within this range, so the mean value is 0.5 and so plus or minus two times 0.7, which is the standard deviation, and that gives us a range from minus 0.9 to 1.9. So that means we're looking for the probability of a certain number of successes. So we for outcomes we can only have integer numbers of students, and that can't be negative. So this would be replaced by zero Can't be any lesson zero, and the number two would be more than two standard deviations greater than the expected number. So therefore, we must look for outcomes that are less than or equal to one. So we're looking for the probability that the number of successes is between zero and one, which is the same as saying we're looking for the probability of at most one success, which comes out to 0.911 and finally for party. It's explained to us that students who receive special accommodation are allowed 4.5 hours for the S A T, and students who do not are allowed three hours to rate the S A T. So we're asked, what do we expect? The average time allowed for the 25 students to be, so we can begin by writing a function that defines the average time spent by the 25 students. So for a student who is given special accommodation to get 4.5 hours, so we have 4.5 hours times number of students who get the special accommodation and then the students who don't get three hours and the number of them will be 25 minus the number of successes. And to make it an average, we must divide by the number of students, divide by 25. So this function defines the average number of hours that the students are given for the S A. T out of a sample of 25 and this can be simplified as 75 plus 1.5 x over 25. So for the question, we're asked, what is the number of hours allocated to the students on average, that we expect. So what is their expected number for the average number of hours allocated to the students? So what is asking for us? The expected value of the function h of X. So we can say that is equal to the expected value of 75 plus 1.5 x over 25. And because of the linearity of expectation, we can factory with the 1/25 and rewrite the rest of it like this. Now we have 1/25 so the expected value of a constant is just that constant. So we have 75 plus 1.5 times the expected value for X. And we've already calculated that previously, as 0.5. So this comes out to 3.3 So are we expect that the average number of hours allocated to the students to the 25 students writing S A T will be 3.3 hours.

Question number six. The probability that they're all full or fewer UN rented units is more revel int to the landlord because if there are four or fuel on rented units, which is equivalent equivalent within with a minimum off 16 rented, you're honest, then the land, the land rule lord, can meet all monthly expenses, So the incivility full all fewer.

This is problems using, um, this joint relative frequency in the to a table is basing location versus whether or not you're staying in your home state. So we have yes and no to staying in your home state. And then we have nothing. We have three different states. So Nebraska, North Carolina, and other Nebraska, North Carolina and other states. So it's just filling in as needed. Some for part A. He wants us to calculate that was probability that, uh, plans to stay in his home stay if they're from Alaska. So we're looking for a probability of yes, given Nebraska. Okay, so we need our probability of getting of Nebraska. Yes, which is 0.0 for four. Okay. And then the probability of not leaving is I'm not looking for anything. 0.0 point for Okay. Said it. Find that probability. We're going to 0.0 for four, divided by the sum to get our total. Nebraska 0.44 plus 0.4. Okay, So put that calculator when 04 ford, about about 40.0 for four. Last point for there's a 9.9% or the probability is 0.99 Have them staying. Give another in Nebraska. OK, Part B says most probably that randomly selected student who does not plan to stay at home and we're gonna look at all that knows, um, and North Carolina. So we need the North Korean and other for no. So North Carolina know is 0.193 and other for No 0.256 case. We're gonna find the probability of North Carolina and given no yes, not capital. And I need to make that a lower case in for now. Case of over the North Carolina, 0.193 then divided by the sum of all of our nos. 0.4 plus 0.193 plus 0.256 Okay, let's put that in the calculator. Was there a 0.193 But about 0.4 waas, 0.193 plus 0.256 It's the probability is gonna be around 23% or 0.227 Okay, Now, part C that over here it wants us to know whether planning to stay in their home state and living in Nebraska are independent events. So basically, we see that compare the probability of Nebraska, North Carolina and other. So we need those data points to so 0.51 from North Carolina. And for other, we're gonna have 0.56 Case was found The probability of saying yes given North Carolina, that's going to 0.5 wine divided by the sum of those two. And so that probably is staying at your from North Carolina is gonna be around 21% 0.209 to which is also is already way different than Nebraska's WAAS. The probability of saying yes if you're from other stay again. We're just going to the probability of 0.56 divided by the sum for other. So I got 0.56 to about about 0.56 plus 0.256 So that probably is gonna agree around 18% or 0.179 So all these probabilities are different from each other. Um, the problem north China and other closer and probably of Nebraska is is definitely off. So I would make the claim that being Nebraska and claiming yes are gonna be dependent for They're not independent events. This is called not independent. And that's because that its probability of 0.99 is is a lot different from the 18% in the 21%. Okay, thank you very much.

So for number one, we're looking at male and female students with long hair or with short hair. And we're just riding out the symbols for each of these problems and a through J you're gonna let f represent a female on film. A male s is gonna represent short hair, and l is gonna represent you want to go through these, eh? Probability that a student does not have long hair is just written probability parentheses. And because we're focusing on not having long hair, we're gonna right along here with an apostrophe. This is gonna indicate the compliment of long hair, which is, in essence, the same thing as the probability of short hair. Be the probability that a student is male or has short hair probability print the seas male. We're just gonna represent with the Capitol or s, which stands for short hair. You see the probability that a student is a female and has long hair? We're gonna do probability female. We can write the word and ill for long Here, Indy. The probability that a student is male given that student here, this looks a little bit differently in this case on D, the probability that a student his mail given we're gonna represent that what they line down the middle, that they have long hair. So what we're doing here is actually changing our interest from the entire population to Justus with here in e the probability that a student has long here given that the student is a mill, you're gonna flip those letters around long here. Given that we have a male, Yeah, all the female students probability that a student has short here, so it doesn't have the word given in this one. However, there indicating that that's who were interested in this female students. So of all female students, the probability that a student have short hair. So we're looking at the probability that a short hair out of all the females g of all the students with long hair. So that's what we're given the probability that a student is female. So we're looking at female given long here h the probability that a student this female or has long hair just gonna separate those with or but probability that I randomly slept. The student is a male student with short hair, So a male student with short hair indicates that he's a male and he has short hair. The probability that a student is female, it's just purely probability of.


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