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The average welght of American men over the age of 20 is [95 pounds and the standard deviation is 33.6 pounds: The distribution ol men"> weights is skewed r...

Question

The average welght of American men over the age of 20 is [95 pounds and the standard deviation is 33.6 pounds: The distribution ol men"> weights is skewed right due to & few very heavy men. Let F represent the mcan weight of 64 fndomly sclected American men_What Is thc standard error Ior the sampling distribution ofthe sample mean weightsituation?0792t 3 PotntsTable; Keypat choane40 pound;6 17 pounda0.59 poundsSinkte ths populinod

The average welght of American men over the age of 20 is [95 pounds and the standard deviation is 33.6 pounds: The distribution ol men"> weights is skewed right due to & few very heavy men. Let F represent the mcan weight of 64 fndomly sclected American men_ What Is thc standard error Ior the sampling distribution ofthe sample mean weight situation? 0792t 3 Potnts Table; Keypat choane 40 pound; 6 17 pounda 0.59 pounds Sinkte ths populinod



Answers

The heights of American men follow a normal distribution with mean 70 in. and standard deviation 3 in. Suppose that the weight distribution $(1 b s)$ for men that are $x$ inches tall also has a normal distribution, but with mean $4 x-104$ and standard deviation. $3 x-17 .$ Let $Y$ denote the
weight of a randomly selected American man. Find the (unconditional) mean and standard
deviation of $Y .$

Yeah, given their in a variable X, which represents height, and why would represents wait for an American male. And we're told that X and y have you, by variant normal distribution. But the mean and standard deviation of the heights are 73 inches in the mean and standard deviation of the weights are £170.20 pounds. That the correlation coefficient is ro equals 0.9 in part A, whereas to determine the distribution of why, given that X is equal to 68. So the weight distribution for a 5 ft eight inch American male well, we have that the conditional variable. Why, given X equals X is normal since why necks are by variant with a mean, which is the mean of why you to plus the correlation coefficient rho times the standard deviation of y sigma too times X minus sigma or new one over sigma one and variance which again, by a formula from this section which is one minus the correlation coefficient, Rho Square times standard deviation or the variance of why, um so in part A. We want to just substitute into these expressions and therefore it follows that d mean is going to be equal to We were given that the average weight is 170 so 170 plus were given that rose 0.9 were given that the standard deviation of why is 20 times and then we're given that we're looking at 68 inch persons of 6 88 minus the means 70 over the standard deviation which were given is three. This is approximately 158 because we're looking at weight. This is pounds, that's the mean and the variance is going to be one minus were given. This is 10.9 squared times be variants of why we're giving the standard deviation of why is 20. So this is times 20 squared, and this works out to be 76 and therefore it follows that the standard deviation of why is the square root of this, which is calculated to be approximately 8.72 pounds, therefore follows that the weight distribution of American males with the height of 5 ft eight inches well, this is going to be normal with a mean of £158 and stare a deviation of £8.72 next in part B, whereas to determine the distribution of why, given the X is equal to 70. In other words, the weight distribution for a 5 ft 10 inch American males and rest to find the ways this distribution is similar to that party and how it differs. So once again, now we're going to exchange 68 in the above calculation with X equals 70. So you're going to repeat A With X equals 70. We get that the mean in this case you can find this on your own is £170. Now the standard deviation is now still about £8.72. This isn't going to change because X wasn't involved in the equation for the variance. So it follows that the weight distribution of 5 ft 10 inch tall American males is a normal distribution, with parameters 170 and 8.72 Now, comparing it with the one from Part A, you see that the distributions in A and B are both normal. That's one thing, and they also have the same standard deviation, which is about 8.72 But the average weight differs by height. Finally, in part C, we're asked to calculate the probability that the random variable Y is less than 180 given that X is equal to 72. In other words, probability at a 6 ft tall American male weighs less than £180. Well, again, we're going to use the same formulas from part A except well. Now use X equals 70 instead, and we get the mean is now equal to 182 pounds. And the standard deviation? Well, this is still going to be approximately £8.72. So we have that the variable. Why, given X equals 70. Sorry, given X equals 72. My mistake is normally distributed with a mean of 182 and the standard deviation of 8.7 to approximately. Therefore, the probability that why is less than 1 80 given that X equals 72 is equal to five of 180 minus the mean, which is 1 82 over the staring deviation, which is 8.72 This is approximately five of negative 50.23 and using a calculator, computer or table, we can look up the value of this, and it's approximately 0.4090 So about 41%.

I have to verify a few things in this problem by calculation by hand. This is how I calculate the mean simply add them all up and divide by. The number off observations is 1 60.25 This is the standard deviation. I simply apply the formula substitute the values and the value that I get is 18.49 Standard error is standard deviation upon route off sample size, which is 4.6 to these other value that I'm going to need in order to calculate the P statistic. And the T statistic is simply substituting in the values over here. This will be 1 60.25 and this turns out to be 0.5

In question 41. We have the distribution of heights of adult American Man is approximately normal with mean of 69 inches and a standard deviation of 2.5 inches. Draw a normal curve on which the mean and standard deviation are correctly labeled. Now, in all normal curves, we have the mean in the center and then on the right side we have three standard deviations. And on the left side we have three standard deviations and they're spaced out in this case by 2.5. So writing what we know, we know that this is a normal distribution with a mean of 69 a standard deviation of 2.5 inches. We're gonna start with our normal curve. We're gonna label the center, and then we're gonna start spacing this out by 2.5. The 2.5 is represent our standard deviations. So 69 is in the center. That's our mean, and then we have 71.5 is one standard deviation away. 74 is two standard deviations and 76.53 standard deviations away on the left. You can also see that with 66.5, 64 61.5. Lastly, we'll finish this up with labeling man's height, and this is all in inches.

Okay for this problem we're looking at, um, some weight values, overweight men. So we're looking at a random sample of 60 and we're going to find confidence intervals. So it's, you know, it's really no over here, so we know that we have, ah, sample of 60 came. And we know that the number of men so from our sample are X bar next cream you, the sample mean equals two £30. So these men's on average were £30 based on the 60 men, and the standard deviation of the sample equals to £4.2. And what we want to know is we want to know what the point estimate ISS A. What's the point Estimate? Well, the point estimate means the population are best prediction for the population is the same thing as the sample mean, so we know our best estimate is £30. Based on the data we have, that's pretty part B asks us to find you 95% confidence interval of this. So maybe a conference in every will see, I what? We're gonna use our technology for this. We're gonna use the calculator we see up over here So when you stop and tests and we're going to get into a paged work here going to get into a ah zi interval so and we have not dated. But we have statistics. Silveria, look at our statistics here, so use what we kind of written down already. So we have been told already that the, uh, 4.2 the standard deviation that was given to us. So plus or minus 4.2 of the typical deviation from the mean and the mean value off what we were told was that £30 overweight. But these men were, and this is based on a sample size of 60 and the confidence interval of the confidence level will question be asked us to find wants us to find the 95% confidence level. So by doing this, the car this is going to tell us how wide oven interval we need to be to be confident that we're going to capture the Truman mean weight of these men. So the confidence interval for this means that the weights are between 28.9 for 95% confident that the true mean is between 28.9 and £31.1 to be 95% certain. Guess there's 95% confidence interval and then Part C asked. Justifiable. What if we want to be 99% sure 99% confident that were captured, the true meaning? Well, a good thing to know is that when you up your certainty, it's going to make it a little whiter in terms of, uh, the spread of what our prediction is. So we're gonna do the same thing. Test it is the interval and our data's deliver. Next, we just put it in 74.2 30. But we use me to change this to 8.99 when the 99% certainty certain along for 1% error in our prediction and so we can see that the for the 99% confidence interval the range of values. The true prediction is between 28.6 and £31.4 and party is the best question. It says which one is whiter? I've kind of mentioned it before, but we'll answer it out. So, uh, the 99% confidence terrible is wider because we want to have a more certainty. So look at the question here was that you have always larger and why, Uh, that was 9% confidence. Intervals is whiter because we want more certain TV. Want more certainty about possible values. So our spread of our prediction is whiter way confidence that holds for means of two different confidence levels.


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