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Four-vear-olds in China average 45 unsupervised hours per day: Most of the unsupervised children live in rural areas, considered safe Suppose that the amount of uns...

Question

Four-vear-olds in China average 45 unsupervised hours per day: Most of the unsupervised children live in rural areas, considered safe Suppose that the amount of unsupervised time is normally distributed with standard deviation 15 A Chinese 4-year-old is randomly selected from rural area_ We are interested in the amount of time the child spends alone per day:In each appropriate box you are to enter either rational number in "plq" format or a decimal value accurate to the nearest 0.01(.2

Four-vear-olds in China average 45 unsupervised hours per day: Most of the unsupervised children live in rural areas, considered safe Suppose that the amount of unsupervised time is normally distributed with standard deviation 15 A Chinese 4-year-old is randomly selected from rural area_ We are interested in the amount of time the child spends alone per day: In each appropriate box you are to enter either rational number in "plq" format or a decimal value accurate to the nearest 0.01 (.25) In words, the random variable in which we are interested is defined by X = (pick onel b. (.25) Find the probability that the child spends less than one hour per day unsupervised: P(X < 1) = (.25) What is the probability that the child will spend over ten hours per day unsupervised_ P(X 10) (.25) The 70th percentile of the distribution of unsupervised time is given by 0.70



Answers

In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
a. In words, define the random variable X.
b. X ~ _____(_____,_____)
c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
d. What percent of the children spend over ten hours per day unsupervised?
e. Seventy percent of the children spend at least how long per day unsupervised?

So we're looking at people wearing seatbelts and not wearing seatbelts Children and finding out how many days they stay in the I. C. U. And so we're going to assume that those students with a seat belt have equal stay in the hospital as those who do not wear a seatbelt. And alternately that the seat belted Children have a mean stay less than those who were not seat belted. And so we're assuming that the difference between these two seat belt minus not seatbelt is zero and we're actually getting something that's negative and this will be our p. Value. So let's look at the data and let's get our test statistic and we have sample sizes to sample sizes. One sample size was 1 23 and the other was to 90. So that was the for the seat belt and this is for the non seatbelt. And so we're going to use degrees of freedom of 122 to be conservative. And they're relatively large sample sizes anyway. And so let's find our test statistic and we have our 0.83 minus are 1.39 And then we're going to divide that by the standard deviation of the seat belted group squared, divided by the sample size, and then the standard deviation of the non seat belted Children divided by the sample size of it. And when we do that calculation, we get the test statistic comes out to be negative 2.330 And so now we want to find, and that's so that's what this value is. We want to find the likelihood, if the difference is actually zero or the means are equal, how likely is this number or more extreme to come up? And so I'm going to use my uh my T c D E f to find this mighty CDF. And I'm going to use negative uh minority has negative one times 10 to the 99th hour in it. And then our upper value is going to be that negative 2.330 My degrees of freedom is again the 1 22. And I will paste and let that do the calculation. And I have a value of a p value of about 1%. And that is less than my 5% significance level. So I definitely have sufficient evidence to reject the null meaning that we believe that the Children that had seat belts on have a smaller uh stay in the uh in ICU than uh those that are non seatbelt. And again you write my sons for that. So now we want to find the confidence interval and the appropriate confidence intervals. Since this was a one tailed test and we had 5% in this tale, the confidence interval has 5% down here, 5% up there. We would need to look at a 90% confidence interval. And so my table does not have a T. V. Star value for 122 degrees of freedom. So I'm going to use my inverse T. Value my inverse teeth. And on the inverse T. I want the area below it to the point oh five. So I'm going to put in the 0.5. It will be finding basically this low limit which will be symmetrical. What the upper one and then our degrees of freedom are that 1 22? And we get that value to be Yeah. Uh And this was 1 22 comes out to be negative 1.657 And so now let's get that confidence center, let's look at the difference first. So we need this difference the 0.83 minus 1.39 And that difference comes out to be negative 0.56 plus or minus. And we have that T star value 1.657 because we have one down here and went up here and then we have this same uh same standard deviation we had here. So let's see if I get 1.77 squared and that is a 3.6 squared mm And our sample sizes are 1 23 and 2 90. Okay, And let's get that margin of air, Yeah. Mhm And 1.657 times square root of 1.77 squared divided by 1 23 plus 3.6 square divided by 2 90. And that margin of air comes out to be 0.398 and two. And it keeps going on but I'm just going to store that is X. And so let's get those two values. We have the negative 20.56 minus the X. Value. And that comes out to be negative 0.958 And then we can change that into an addition sign. And we find out that that is still negative and notice that that does not include zero. So we would again from this interval see that the difference does not appear to be zero as he appears to be negative. So it does appear as though um that the seat belts make a difference seatbelts beautiful. They're good that they appear to have a lesser chance of a lesser stay in the hospital in the ICU for Children.

Okay for this problem. We're going back to some school data of the kindergarten Children. So, uh, going to reinvestigate this problem about Children that are normally 39 inches on average and a standard deviation that is two inches average kindergarten heights. So And reports ABC Andy, we're terminating an individual child like, Okay, so I'm just gonna emphasize this enables one. So the big thing about this chapter, we're talking about doing a bunch of samples and find the sample mean and then we're gonna compare that, in contrast, contrast that in this case with just taking a single sample. So we're looking at the probability that the student height is between 38 and 40 inches. So, um, you could show that over here. So it's kind of close to the middle there. So 38 9 Is there some 34? We're kind of getting my sliver in the middle, So 38 40. So I'm gonna use the calculator since we have it up. We've done enough disease scores off the tables, and we just know this is a distribution. The biggest thing is using your showing a direction by using the scores were showing off on the white board, So it's looking normal. CDF We're talking about wanting to look its wars between 38 forties of the lower boundary is 38. Upper boundary is 40 and we know on average the mean the population Means says it's 39 and the variability is represented by the standard deviation of too. So really, we're just using all this information to get Z scores and disease where they get turned into probabilities. And we've done enough now in kind of see what we get. So the probability from one single sample of getting a single student that's that height is, uh, point 383 Now it's like a party. So now we're talking about Part B. We're talking about not taking Weinberger's. And what if we take thirties sample of 30 students and look at the average? We're gonna take those and average them? So what's the class mean going to be? If we do that, well, that's gonna get us closer to that truth values. It was probably a higher likelihood, but because of the end, we're going to see that it should be a higher number. I'm just gonna rewrite it for effect. Like to show em my student, show it and then show it symbolically. And then with the calculator, we've pretty much define that. We want to look between 30 and 40. We cannot use the calculator, which is just taking what you've gotten from your tables and we're looking at the cumulative cumulative distributive came out of density frequency of between 38 forties. That's gonna be the same. Same insane, insane. But here's the big key thing here. Instead of having, uh, the two inch plus or minus the center deviation of two inches, we're actually gonna take the two inches, and we're gonna divide that by the square root of the sample size. So in this case, they said, the sample size is 30. So are very building its lesson because we're dividing by the square of 30 and you can actually do that right under calculator. If you know how to click. Right button so squared of 30. Should that up, just a big decimal. But it's in the end. It's gonna have a probability. That's the big kick right there. It's word of 30. Your teacher want you to show that out you can show that on your page that puts us all the way up. 2.993 is almost 99% chance. If we take those numbers and averaging, that will hit that middle there. So point my 94 will even say so. That's if we took 30. Has sample of 30 an average debt that will get close to that middle number. It's definitely between 30 and 40. If individual kindergarten child is selected at random was a probability, it's taller than 40. So blue here is now. We're talking about a little more extreme, not in the middle. So makes sense. I've gotta mean in the middle. But now the problem ceases for a single pick. Just emphasize that a single pick n equals one. What's the probability that the height is greater than 40? Now we're talking about tall guess who are getting a single kid That's that tall. So let's see. Is it? That's 30. So 40 is kind of a boundary over here. I'm just going to use the blue to talk about this because it says probability is greater than 40 is really what we're talking about here. So we're come about the right side over there, So let's see what we see. Let's look at our distribution. Gets a single sample so normal CDF between Instead of saying between 30 and 40 were going to say that it's 40 is the lower end in the upper boundary is really infinity as far as your calculator is concerned. I was ready. You nines there. The meaning is still supposed to be 39 a standard deviation served up since we're talking about a single pick is, too. So no adjustment for sample size. So it says if you take a single pick, there's a likelihood that you'll get someone over 40 30% chance, so there's a decent probability. And if you look at the diagram, that seems reasonable. So it's 300.30 nine that you get a single student that's over the 40 inch, right? Look again. So party talks about Well, same thing. What's the probability of the class mean? Okay, so the probability about the will make this emphasis just act will say X bar X bar, the class mean is over 40 have actually much less likely because we know it's supposed to be between closer to 39 so getting the average to be that high should invest a lower number. But let's live the calculator. Do us in disease support translations for us. So we know we're going to look at a normal CDF Ah, from 42 positive Infinity 39. Supposed to be the mean And here's the kicker here, the class mean of sample of 30. So instead of just dividing by the standard deviation, we'll take the standard deviation adjusted for the sample size, which we've been told is 30. We need to affect our variability, so it's gonna be like, much less likely. So there we go. So probably I get an average. That is, that much higher, above which the average should be is pretty low, so we can see the probability of getting a class size mean of over 40 is only 400.0 three

Problem. 21. We have a report in 2010 that indicates that Americans between the ages, the ages of the Americans between eight and 18 years it's been an average of new equals 7.5 hours per day using some sort of training device, and in this report, the Standard Division was it will 2.5 hours per day. It's required to find the probability of selecting an individual who uses electronic devices more than nine hours a day. Then for about A it's required to get the probability of X is greater than nine hours birthday. We can get this probability using the school instead. The score equals X minus mu, divided by sigma we have X here equals nine minus mu, which is given 7.5 Divided by sigma equals 2.5. Then the score for nine hours per day equals 4.6. By entering the table, we can get the probability that zit is greater than zero and smaller than 2.6. The tables we have gives the probability starting from zero to the specific is it which is also in six gives in this area the probability from the table equals four point oh 793 then took it. The probability that is greater than 4.6 or greater than X equals nine. Then the probability of X is greater than nine equals half, which is half the area of the normal distribution buying us the shaded area. Sorry, this value is not opened or 73 equals point two 57 Then the probability is half minus 0.2257 which equals 4.2743 which equals 27.43%. Then the probability to select an individual who uses electronic devices more than nine hours per day equals 27.43%. For Part B. We want to get the portion of the same age from 8 to 18 year old who spent between eight and 12 hours. She means we need to get the probability for Ex is greater than eight hours and smaller than 12 hours. Birthday, of course. Then, to answer this, we get to the scores the first season score for it. We use X as 88 hours per day, eight minus 7.5, divided by 2.5 gives open to and for the two we get X equals 12, then 12 minus 7.5. Divided by 2.5, it equals 1.8. Then we get the probability for each Z score from the tables for from zero that is greater than zero, said one. And between 1.6 Sorry. Here it's open to from the tables. It equals four point oh 793 And for the second school between 1.8, it equals 4.4 641 This means we have the normal distribution here is open to and here is one point. The first area here is for that equal 0.2. It's the probability is this value and we have another venue from zero to 1.8. We have poll this area. Then to get the area to get only this area, we'll get the probability for X. Between eight and 12 hours per day will be the shaded area with blue, which is opened 4641 minus the shaded area with black Oh, point Oh 793 then equals Oh, 0.38 for it or we can write it in percentage 38.48% and this is the final answer of body and this is the final answer of 40 and the problem.


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