5

Texting: A sample of 300 teenagers are selected at random: Let X be the number of teenagers who have sent text messages on their cell phones within the past 30 days...

Question

Texting: A sample of 300 teenagers are selected at random: Let X be the number of teenagers who have sent text messages on their cell phones within the past 30 days According to study by the Nielsen Company; the probability distribution of X is as follows:P(X)0.050.100.250.370.23Questions:a) What is the variable in this data? Be specific. (2pts) b) Compute the probability (in decimal form rounded to 2 decimal places) that:three teenagers sent text messagesthree or more of the teenagers sent text

Texting: A sample of 300 teenagers are selected at random: Let X be the number of teenagers who have sent text messages on their cell phones within the past 30 days According to study by the Nielsen Company; the probability distribution of X is as follows: P(X) 0.05 0.10 0.25 0.37 0.23 Questions: a) What is the variable in this data? Be specific. (2pts) b) Compute the probability (in decimal form rounded to 2 decimal places) that: three teenagers sent text messages three or more of the teenagers sent text messages fewer than two of the teenagers sent text messages anywhere from 2 to 4 teenagers sent text messages Compute P(z > 2)- d) Compute P (1 < X < 3) and explain what the result represents. Compute the mean and interpret the result: No need of showing your work f) Compute the standard deviation, Ox _ Round to 2 decimal places Work is not needed Extra Credit: g) Of the 300 teenagers surveyed how many teenagers have sent 2 text messages on their cell phones within the past 30 days. h) Of the 300 teenagers surveyed how many teenagers have sent anywhere from 1 to 3 text messages on their cell phones within the past 30 days.



Answers

Let $x$ be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that $x$ has a distribution that is approximately normal, with mean $\mu=7500$ and estimated standard deviation $\sigma=1750$ (see reference in Problem 15 ). A test result of $x<3500$ is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.
(a) What is the probability that, on a single test, $x$ is less than $3500 ?$ (b) Suppose a doctor uses the average $\bar{x}$ for two tests taken about a week apart. What can we say about the probability distribution of $\bar{x} ?$ What is the probability of $\bar{x}<3500 ?$ (c) Repeat part (b) for $n=3$ tests taken a week apart. (d) Interpretation Compare your answers to parts (a), (b), and (c). How did the probabilities change as $n$ increased? If a person had $\bar{x}<3500$ based on three tests, what conclusion would you draw as a doctor or a nurse?

So we're dealing with a situation where we're looking at the glucose level and uh so we know that the average is supposed to be 85 with a standard deviation of 25 it's approximately normal. And we know that if you have a value that is less than that, it's supposed to be excessive insulin that you're running high on insulin, you're using up and turning it. Um Use out the insulin excusing using up the glucose. So we want to find for this distribution if we only take one sample in a week, what's the likelihood of getting a sample if this is true of less than 40? And so we know we convert that to a Z value, 40 minus the 85 divided by the 25. And when we do that, 40 minus 85 uh huh. It's negative 45 on top. And then divided by 25. And let's see what it comes out to be. That comes out to be a value of negative a Z value of negative 1.8. And I've got a quick look at the table negative 1.8 below that comes out to be 0.3 0.359 So it's not real likely that if someone has a true mean of 85 that you're going to end up measuring less than 25 but notice that is a large standard deviation. Now they ask what happens if you take two samples during the week and you look at the mean of those two being less than 40 how likely is that to happen if again this is really your average. So let's convert it to a Z value. We have 40 minus 85. So again that negative 45 on top and then we'll have that 25 over the square root of two, So negative 45 divided by and we have 25 divided by the square root of two. And let me close that prophecy off. And that gives us a Z value. Yeah. Of negative 2.5. It would round 25 And so I'm gonna quick look that up in my table negative 2.55 And that gives me a 0.54 So that's not very likely to happen now. Let's go with a sample size of three. Yeah. Yeah. And we know I'm gonna cut to the chase. I'm gonna cut to the Z value and because we know and asking that same question so the numerator is going to be negative 25 and then we're gonna have 25 over the square root of three. And thank you. I'm sorry. That's negative. 45 over the score. About 25 square 23 Yeah I just got to change that one value. My calculator. I get that. That Z value comes out to be negative 3.12 Yeah. And negative 3.12 is only 0.3 zeros and a nine. And now we increase all the way up to a sample size of 55 times during the week. And again we're answering this question and we'd have a negative 45 in the top and we'll have the 25 over the square root of five in the bottom. Yeah. And I just have to go back and change that one number in my calculator. And that Z value comes out to be yeah, negative 4.2 And that is approximately zero. So if you did this reading five times during the week and you got a value that was less than 40. It is almost impossible for your actual mean to be 85. It's extremely unlikely. So we would think that you definitely have a problem with excessive insulin. Mhm.

Here we are told that 30% of all students want to buy a new text for a certain course, and we define that as the success. And so the 70% of students want a used copy, and we have a random selection of 25 purchasers. So we have an equals 25. We have probability of success. Is it what is 0.3? So if we define the random variable X as the number of successes that is the number of students or people who want to buy the new book, it follows a binomial distribution based on the size of 25 and a probability of success of 0.3. Now for part A were asked for the mean value and the standard deviation on the number who want the new copy. So that's the mean and standard deviation on the number of successes. So, for a binomial random variable, the mean is given by 10 times p just 25 times 0.3, which comes out his 7.5, and the standard deviation on X is given by and that comes out to you about 2.29 now for part B were asked, What is the probability that a number of successes will be more than two standard deviations away from the mean value? So I mean value of 7.5 and standard deviation is about 2.3. So basically, the question is asking, you know, what is the probability that that number of successes is? Well, I'll give the range first. So 7.5 is the mean closer minus two standard deviations of two times 2.29 So that's 5.21 or 9.79 So we're really looking for the probability of being more than two standard deviations above 7.5 or more than two standard deviations below 7.5. So we can say we're looking for the probability at the number of successes is less than or equal to now successes. It's a discreet distribution, so we can only have entered your numbers of successes. So if we have five or less successes, we're more than two standard deviations below 7.5. But we can also have at least 10 successes, which would put us more than two standard deviations above 7.5 this could be written as probability of at most five successes, plus one minus the probability of that most nine successes. So this is equal to zero point 193 plus one minus zero point 811 and that comes out to zero point 383 So now for Apart, See told that a bookstore has 15 new copies and 15 years copies and that 25 people are coming to purchase the text. So what is the probability that all of them we'll get the book that they want? So since there are 15 copies of the new text available, we want the number of Successes x, where the number of people looking to buy the new book to be at most 15. If it's any higher than 15 then somebody won't be satisfied. But we also want the people who are looking for the used book to be satisfied so out of 25 and we only have 15 used books as well. So out of the 25 people we want, at most 15 looking for the used book, which means that we want at least 10 people that's 25 minus 15. We want at least 10 people who wants the new book. So what we're really looking for is probability of the number of successes being between 10 and 15 inclusive, which is equal to the probability of at most 15 successes, minus the probability how that most nine successes. And so this comes out to approximately one minus zero point 811 gives us 0.189 and finally for party. We're told that new copies cost $100 used copies or $70 and we're told that the bookstore has 50 new and 50 used copies, so everybody will be satisfied. So what is the expected value of the total revenue from the sale of the next 25 copies of the book? So revenue will be given by a function of the number of successes, which is $100 times the number of successes. So for each new book sold, we get $100 revenue plus $70 times the number of used books sold, which will be the difference of 25 the number of successes. So this is our than your equation for revenue, and we can simplify this too. This So if we want to find the expected revenue, you take your expectation of both sides. And because of the linearity of expectation, this comes out to so the expected value of a constant is just that number. So this comes out to 17 50 plus 30 times the expected value expected value of X, which we already know from previously. That's 7.5, So this comes out to $1975. That's the expected revenue for the next 25 customers.

So we're looking at a distribution of our glucose and it's normal With a mean of 85 and a standard deviation that's pretty big. 25. And we want to know what's the likelihood, if you sample A person that has this distribution, what's the like likelihood that somebody will test under 40? And that means they have again, a high too much to producing too much insulin using up the sugar. And so we will need to convert that to a Z value and we know we'll take 40 minus uh the 40 minus the 85 which is negative 45. And then divide that by 25. And around that too -1.8 I guess don't need you around it. And we can look that value up And, you know, so that's about here for that individual distribution and they below negative 1.8, that would happen about 3.6% of the time If someone has an average glucose level of 85, that they would test and have. So even if you are having that average, you'll test that you're producing too much insulin about about 3.5% of the time. Now, what if we take an average of two readings? How likely is it that you're mean up to will be less than 40? Well, now we have that Z value being the 40 minus the 85. So still need that negative 45 on top divided by and we're going to take that standard deviation over the square root of and and let's see what that Z value comes out to be. And and so negative 45 divided by uh huh. Our 25 divided by the Square Root of two. Yeah. Yeah. And when we do we get that Z value is negative 2.5 around 55. And then the probability of being below negative 2.55 is only .0054. So as that's a rarity if you're mean is actually 85 it's not likely that you get to readings having an average less than 40. No they asked us to repeat that pretty sample size of three and let's find out what that Z value is. And so we'll just replace that with a three. So I can narrow back up. Just change that number two A three. And that gives me a z value of negative 3.1. It would round two And no below 3.12 isn't .0009. So it's even less likely with three. Well what if we use five readings and find the average? Uh huh. And again we'll just go back up and change that to a five. And that z value corresponds with a negative 4.2 And I don't have a table that even goes that walk that far. But I can use my normal C. D. F. And my normal CDF if I type in that upper as negative four point oh two. And it comes out to be a very small number .1234 Zeros to nine. So this is very unlikely to happen. So if you use five measurements and got the mean of five measurements to be less than 40. Yes. We would say there is a problem. We would think that actually your average is not 85. We would think it's actually lower than that because it's extremely unlikely For a distribution centered at 85 to produce the average of five readings being that load. So we would be suspicious that actually your glucose level was lower or you had a big insulin problem. Okay.

Okay, So we're giving a probability distribution which I have written at the top here, and we're trying to find the mean value and the standard deviation for this probability distribution. So in order to find the okay, in order to find the the mean of our probability distribution, we have to find the expected value of X. It is equal to the expected value of X. And the formula to find that is just the summation From X is equal to zero to whatever our last X value is. In this case it's for and we multiply our X value times our probability at that X value. And so now we're just gonna plug in some of our X values and probabilities into this equation and find our view of X. So this is equal to zero times 0.5 plus one, Times .15 Plus two Times .25 for us three Times .25 plus four times 0.3. And if we add this all together we'll see that this is equal to 2.6 now to find the standard deviation? Well, the formula for that is the square root of our expected value of x squared minus the expected value of x squared. And that can get a little tricky. But you just have to remember the first one is, the X squared is actually within this expected value, and the second one is, we're actually just taking are expected value of X, whatever that number is, and then squaring it. So find the expected value of X squared, We have to find the summation from X is equal 0-4 of X squared times are probability at that X value. So this is gonna cool zero squared times five plus one squared, which is just one times point oh 0.15 Plus two squared which is four Times .25, Yeah plus plus three squared which is nine times .25 Plus four squared which is 16 times 0.3. You know, if we have this all together we're going to get a value of 8.2. Now we have to do is plug this back into our equation and solve for our standard deviation. So we have the standard deviation is equal to 8.2 -2.6 squared. All square rooted. So if you plug this into our calculators, We'll see that this is equal to 1.2. Mhm. And now for part B we're trying to find again is a you are mean and a standard deviation Except this time we're trying to find the mean value and standard deviation of 150 students buying tickets with the probability that we have above. So in order to find the mean value, we can use this formula, which is the main value is our main value of our probability distribution Times our end value, which is in this case 150. So we get 2.6 Times 150, Which is equal to 390. Now to find the standard deviation, What we can do is again use a formula using the standard deviation we found in part A which is our standard deviation, found in part a Multiplied by the square root of our end value, which again in this case is 150, So we have 1.2 times the square root Of 150. And if we plugged that into a calculator We get 150, Sorry, the square root of 150 Time to 1.2, which is equal to around 14.7. Yeah. And now for part C, What we're trying to find is the probability that given that we have a maximum amount of 500 seats available, that 150 people buying tickets will all be accommodated for. So for every single one of the students to be accommodated for what we need is the Amount of tickets bought to be less than or equal to 500. So we're trying to find the probability that our tea Which is the amount of tickets bought by 150 students is less than or equal to 500. And so I'm going to do this is by using a Z. Score and transforming our probabilities so that we can use the standard normal distribution table to find our probabilities. So how I do this is we have Z is equal to t minus our U. of T value, all divided by our standard deviation that we found in part B. So if we do this, we have t, which is 500 minus UFT is 390 Divided by our standard deviation, which is 14.7. So if we plug this into a calculator, we get 110 divided by 14.7, which is equal to 7.4 eight. So now we're trying to find the probability that he is, sorry, that Z Is less than or equal to 7.48. And if you look at this already, we can see that some .48 is a massive value for a standard normal um distribution table. So This is going to be around .9999. Given that our standard deviation is 14.7 and we want to have a maximum number students of 500 and are mean number of students is 390, it kind of makes sense for the probability to be insanely high


Similar Solved Questions

5 answers
One eight-ounce glass of apple juice and one eight-ounce glass of orange juice contain total of 175.5 milligrams of vitamin C, Two eight-ounce glasses of apple juice and three eight-ounce glasses of orange juice contain total of 432.7 milligrams of vitamin C. How much vitamin C is in an eight-ounce glass of each type of juice? apple juice mg orange juice mg
One eight-ounce glass of apple juice and one eight-ounce glass of orange juice contain total of 175.5 milligrams of vitamin C, Two eight-ounce glasses of apple juice and three eight-ounce glasses of orange juice contain total of 432.7 milligrams of vitamin C. How much vitamin C is in an eight-ounce ...
5 answers
ExERCISE 2 Consider the following IVPy" +y = 3cos(wt) y(0) = 0 y (0) = 0Find the solution for w # 1.What happens astakes one values closer and closer to 1?What happens changing the IV to y(0) = 1, % (0) = 1?
ExERCISE 2 Consider the following IVP y" +y = 3cos(wt) y(0) = 0 y (0) = 0 Find the solution for w # 1. What happens as takes one values closer and closer to 1? What happens changing the IV to y(0) = 1, % (0) = 1?...
5 answers
Inktegrate Pomer scties QuestionWrite out the first four non-zero terms of the power series rep esentation for f(z) In(4 41) by integrating the power series for f' Express your answer J5 sum_Provide your answer bclow:
Inktegrate Pomer scties Question Write out the first four non-zero terms of the power series rep esentation for f(z) In(4 41) by integrating the power series for f' Express your answer J5 sum_ Provide your answer bclow:...
5 answers
(a) Let f : R- Rbe function defined by f () sin % Then f is continous at = [ Hinl. Use a trig: Identity below: For any 01, 02 € R; 01 + 02 01 02 sin 0 1 sin 02 = 2 cOS Sin
(a) Let f : R- Rbe function defined by f () sin % Then f is continous at = [ Hinl. Use a trig: Identity below: For any 01, 02 € R; 01 + 02 01 02 sin 0 1 sin 02 = 2 cOS Sin...
5 answers
Evaluate lim tan n+ 0 A graph of some of the first terms of the sequence follows5.54.43.32.21.9 12 15 18 21 24 27 30 33 36 39 42 45 48-l-2.23.3limit =Points possible:License
Evaluate lim tan n+ 0 A graph of some of the first terms of the sequence follows 5.5 4.4 3.3 2.2 1. 9 12 15 18 21 24 27 30 33 36 39 42 45 48 -l -2.2 3.3 limit = Points possible: License...
5 answers
8) points Given f(r)2x + 1.0hCompute 47 f(r)
8) points Given f(r) 2x + 1. 0h Compute 47 f(r)...
5 answers
Homework Exercises: any 2 2 U 20,28, x?1 ~ zez x2) 36 { 32 19 2 4 42 where f (x) 4 4 2 2 % f() 2 <intervol e7 1 2< down 1 dlo ( concave down fumop 0.5 E8 T) and Identify dn (oo ( 7=0 +
Homework Exercises: any 2 2 U 20,28, x?1 ~ zez x2) 36 { 32 19 2 4 42 where f (x) 4 4 2 2 % f() 2 <intervol e7 1 2< down 1 dlo ( concave down fumop 0.5 E8 T) and Identify dn (oo ( 7=0 +...
5 answers
A proton I5 ejected from @ very far d stance (where olectre porental energy 0) {owards an Alpha particle . The proton was complelely stopped at a distance T 3 X10 "m of the Alpha particlo; How much kinetic energy must the prolon have initially to get this close %0 the alpha particle? (charge of Alpha particle +20)
A proton I5 ejected from @ very far d stance (where olectre porental energy 0) {owards an Alpha particle . The proton was complelely stopped at a distance T 3 X10 "m of the Alpha particlo; How much kinetic energy must the prolon have initially to get this close %0 the alpha particle? (charge of...
5 answers
Let R denote the region bounded by the curves y =x? y =6-x,and y = 0. Find the volume of the solid that results from rotating the region around cach of the following: The x-axis:The y-axis:The line xThe line y = -
Let R denote the region bounded by the curves y =x? y =6-x,and y = 0. Find the volume of the solid that results from rotating the region around cach of the following: The x-axis: The y-axis: The line x The line y = -...
5 answers
Amund Cun ral AtonFormulzLoxis StruclureTotal (aloms Ioni D;lingElectron group alancemenl WnitlatHoecuar DrarinqUrd *Molecue Vector fot Poian Polar dyto) BonosBond ang eValence Electronebonded Fildms0af9Motecular Shape NameCHChShapo:HCN (C i5 Iniu cuntine alomiShapo:IShal,
Amund Cun ral Aton Formulz Loxis Struclure Total (aloms Ioni D;ling Electron group alancemenl Wnitlat Hoecuar Drarinq Urd * Molecue Vector fot Poian Polar dyto) Bonos Bond ang e Valence Electrone bonded Fildms 0af9 Motecular Shape Name CHCh Shapo: HCN (C i5 Iniu cuntine alomi Shapo: IShal,...
5 answers
Find the domain and range of the function, and b) sketch a comprehensive graph of the function clearly indicating any intercepts or asymptotes.$$f(x)= rac{1}{x}-4$$
Find the domain and range of the function, and b) sketch a comprehensive graph of the function clearly indicating any intercepts or asymptotes. $$f(x)=\frac{1}{x}-4$$...
4 answers
Outsourcing of Jobs The cumwlative numter of jods approxima edurcec Jyereese JY U.S Jased mwnatonaTrom 2005through 2009N(:)O.0s(t + 1.1)2.2 0.7: + 0.3 (0 < t { 4)where NG:) mesumec whol numcecmilliona How fast was the number cf U.S.jobsthatwere outsourced changing 2006 (= 1)? (Round Your anskiers to the nearestjob: Per Year
Outsourcing of Jobs The cumwlative numter of jods approxima ed urcec Jyereese JY U.S Jased mwnatona Trom 2005 through 2009 N(:) O.0s(t + 1.1)2.2 0.7: + 0.3 (0 < t { 4) where NG:) mesumec whol numcec milliona How fast was the number cf U.S.jobsthatwere outsourced changing 2006 (= 1)? (Round Your a...
5 answers
A Newfoundland & Labrador license plate consists of of three letters followed by three digits (ex ABC-123). If there are no restrictions on the letters or digits, how many different NL license could be created?01) 29,312,0002) 8,468,0003) 2,600,0004) 17,576,0005) 2,340
A Newfoundland & Labrador license plate consists of of three letters followed by three digits (ex ABC-123). If there are no restrictions on the letters or digits, how many different NL license could be created? 01) 29,312,000 2) 8,468,000 3) 2,600,000 4) 17,576,000 5) 2,340...
5 answers
Which response includes all of the following substances that have AG"; = 0,and no other substances?HCI(g) 2 Fe(s) HCI(aa) Fzlg)3) 1 € 2b) 2 & 4c) 1,2 & 4d) 1,2,3 & 4
Which response includes all of the following substances that have AG"; = 0,and no other substances? HCI(g) 2 Fe(s) HCI(aa) Fzlg) 3) 1 € 2 b) 2 & 4 c) 1,2 & 4 d) 1,2,3 & 4...
5 answers
Let f be the function defined Iby f (z)612 9r 4for 0 < $ < 3 Which of the following - f is decreasing = statements Is true? on the interval (0, 1) because f' (r) Oon the interval (0, 1). 'Is increasing on Ihe interval (0, 1) because /" (z) Oon the interval (0,1) decreasing " on Ihe interval (0, 2) because [" (2) on Ihe interval (0,2). f is decreasing on the interval (1,3) because f' (r) Oon Ihe Inlerval (1,3)
Let f be the function defined Iby f (z) 612 9r 4for 0 < $ < 3 Which of the following - f is decreasing = statements Is true? on the interval (0, 1) because f' (r) Oon the interval (0, 1). 'Is increasing on Ihe interval (0, 1) because /" (z) Oon the interval (0,1) decreasing &quo...
5 answers
MqBr[)eq:2)NICI LlClt PBr, 41 2 Eq t-BuOk_ heat 5) | c4 HBr; S0= 6) Nach 7) PCC3)
MqBr [)eq: 2)NICI LlClt PBr, 41 2 Eq t-BuOk_ heat 5) | c4 HBr; S0= 6) Nach 7) PCC 3)...

-- 0.021043--