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EnteredAnswer PreviowRerui 0.026204770.02620477irattoiThe answer above NOT correct:point) poll of 90 students found that 209 wore Iavor of ralsing tullon bulld this...

Question

EnteredAnswer PreviowRerui 0.026204770.02620477irattoiThe answer above NOT correct:point) poll of 90 students found that 209 wore Iavor of ralsing tullon bulld this Poll = now toatball stadium. The standard dovlatlon - Is 596. What would be the standard doviation the sample size wore increased from 90 t0 2337 Answer: 0.02620477

Entered Answer Previow Rerui 0.02620477 0.02620477 irattoi The answer above NOT correct: point) poll of 90 students found that 209 wore Iavor of ralsing tullon bulld this Poll = now toatball stadium. The standard dovlatlon - Is 596. What would be the standard doviation the sample size wore increased from 90 t0 2337 Answer: 0.02620477



Answers

Answer the questions about the specified normal distribution. In a survey of men in the United States (ages $20-29)$ the mean height was 69.4 inches with a standard deviation of 2.9 inches.
(a) What height represents the 90 th percentile? (b) What height represents the first quartile?

All right. The information in this problem is normally distributed. So therefore we'll start by drawing a normal bell shaped curve. And this survey is about women and their heights and the mean height of women ages 20-29 was found to be 64,2", with a standard deviation Of 2.9". So for part A we want to calculate the Height that represents the 95th%ile. So what we're saying is 95 of what heights are below an unknown value. So in order to find that we're going to start by finding the Z score associated with it. Keeping in mind that AZ of zero represents average, So 64.2 would transition into a Z. of zero. We are trying to figure out the Z score that's going to have 95 of the heights to its left, or less than it. In order to do. So we can use the inverse norm, function on a graphing calculator. And when you use inverse norm, you must provide the area in the left tail, followed by the mean and the standard deviation. So for our situation, we have 95 of the curve in the left tail. The average. Since we are finding a Z score, the average of the Z scores, or the standard normal curve is always zero And the standard deviation of the standard normal curve is always one. So we're going to bring in our calculator and to access inverse norm, you would hit the second button and the bears button and if you look above the abbreviation vares, you're going to see the abbreviation for distributions. So we're going to access the distributions menu of our calculator, Which is # three in this menu, we'll put the area in the left tail, followed by the mean of the standard normal curve and the standard deviation of the standard normal curve. In doing so, we are going to get a Z score of approximately 1.645. So the Z score right here Is 1645. Now you have a formula that the Z score can be found by using x minus mu divided by sigma. So if I were to use some algebraic manipulation and calculate my cross products, I would have x minus mu equivalent to Z time sigma. And if I were then to add mu to both sides so that I can isolate X, I would have a new formula X equals mu plus Z times sigma. So to find our height of women that represents the 95th percentile, I will use that formula mu plus Z times sigma. So the average height of women ages 20-29 was 64 to The Z score associated with this boundary line for the 95th%ile was 1.645 And the given standard deviation was 2.9. So x will be 68 9705 inches. So that means that 95 of women ages 20-29 will be shorter than 68.9705 for part B. We want to calculate the height that represents the first quartile. So again I'm gonna draw my curve and my average will be placed in the center which again corresponds to a Z score of zero and this time I want the first portal. So the first quartile is the same as The 25th%ile. Yeah, so I'm trying to find the height Which represents the 25th%ile. So I'm going to calculate my Z. By using inverse norm. Again I'm going to place the area that's in the left tail, followed by the mean of the standard normal curve and the standard deviation of the standard normal curve. So I'm bringing in my calculator. 2nd, there's inverse norm area in the left tail, followed by the mean, and the standard deviation of the standard normal curve. And you're going to get approximately a -67. So the z right here is negative .67. So for me to calculate the raw data, I'm going to use the formula we derived mu plus Z times sigma. The mean height of women in that age Bracket was 64 to The Z score we just acquired was a -67 and the standard deviation 2.9. So for the 25th%ile I will get a height of approximately 62.257". So just to recap, The 95th%ile is represented by the height of 68 9705 inches, and the first quartile Is represented by a height of 62 257".

Mhm. Okay. This question were given the end 725 and X musical 275. It's so cute simply X summer. And Just to 75-85, Let's get this year 35 mm he said of moving to value. Mhm. For the 95 confidences Is 1.96. Get everything. Okay. Okay. And the 95% confidence interval about the population proportion of adults follow college football and then be given in this general over the people. That's what is? I don't have to discouraged. He persecute from the peak. Okay. Right. Which would give us Europe 3 5 Less than -1.96 spirit. You have been 35 times 655 over. And which is given to the 75? Yeah, That's 0.317. What's your .383? It's a buster miners.

Hey, everybody, my name is Colin and let's go ahead and jump right into this problem that deals with the students in Mr Shanks class and were asked to calculate a 90% confidence interval for the slope of population aggression. Mind so to do that, let's first just go ahead and jump right in. Let's go ahead and calculate our Alfa value that we're going to need for this problem and I'll actually do you a favor. Go ahead, draw better Alfa than that. And to do that, we're gonna as we as you'll recall, we're gonna take one and we're going to subtract our confidence level over 100 which for this case is going to be one minus 90 over 100 since we're asked for a 90% confidence interval and that's just going to equal 0.1. And from this, we can calculate are critical probability or RP Star, which is just going to be one minus Alfa over to or one minus point 1/2. This is 0.95 and then the last thing we need to go ahead and calculate that critical value or that T value is our degrees of freedom. Now you were called The degrees of freedom is just the number of trials with number of constraints minus two. And for this case, our end is equal to the number of students in his class that air sampled, which are 18 students. So are, and in this case is 18. Which means there are degrees of freedom is 18 minus two or 16. So from this, we're going to be looking for a critical value or a T star with a point one alfa value and 16 degrees of freedom. And will you You can go. You can go look that up in the table quite easily. Teoh do that. What we're gonna do is we're gonna look at the table when you're gonna find the column that matches our Alfa value of 0.1 and are degrees of freedom of 16. And when you do that, you see it. The T value there is 1.746 and we can I use that to go ahead and calculate our confidence interval. With this, we're going to calculate our margin of error. That's kind of the next step that we need And that margin of error is just going to be that t value that we just calculated are that critical value. And we're gonna multiply that by the standard error for the slope of this regression line or 1.746 times. That's standard error. Standard deviation column, arm span, row 0.8 091 And would you do that? We get a margin of error, a 0.14127 which is all that we need to calculate that confidence interval. So we're 90% confidence interval in this case is just going to be that slope that we have for this particular regression line. 0.84 zero for two plus or minus our margin of error 0.141 to 7. And when you do that, you will get a confidence interval off 0.6992 2.9817 And that right there is our 90% confidence that our which means that we can say for this particular aggression analysis, we are 90% confident that the slope off the regression line will fall within these parameters

The following is a solution # six. And this asks 785 people whether they follow college football. And of those 785 people, 275 of them said that they followed college football. So that was the prompt. And we're asked to find a 95% confidence interval in this scenario. So if you wanted to you could find the P. Hat. You just take the X. Over the end but you really don't need to. Especially if you're gonna use your calculus which I'm going to show you here in a second. But to 75 out of 7 85 um That's gonna be our P. Hat. And uh this is a proportion uh they in fact it's a one proportion Z interval. Okay? Because it doesn't ask anything about mean or anything like that. It just gives you a proportion. So that's what we're gonna do. So you can use the formula if you so wish it just may take you a little longer. I'm gonna go to the T. I. T. Four and if you go to stat and air over two tests and you go all the way down to the a option where it says one prop Z ent we're doing a one proportions E interval. And uh the X value remember was 2 75. That was the number of favorable the people that said that they follow college football out of a total possible 785 people that were randomly selected. And we want to be 95% confidence of 950.95 will be the sea level. And then whenever we calculate that this top band here that's our confidence interval. Now you can write the P. Hat if you want. It's about 35% but this top band is actually the answer. So I'll go in round I liked around three decimal places, so .317 or 31.7 And .384 or 38.4%. Now it doesn't actually say to do this. But if you wanted to interpret this, you would just say we can be Um, confident that the true population proportion of people who follow college football is between 31.7 and 38.4%.


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