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Problem [ (30 points)The population mean for the average weigh of 12 year old children is [ 2 pounds and the population standard deviation is 40. A random sample of...

Question

Problem [ (30 points)The population mean for the average weigh of 12 year old children is [ 2 pounds and the population standard deviation is 40. A random sample of 50 drawn from population Find the probability that the sample taken will have the average weight of the children will be between |40 and /14. What is the probability that the variance for the sample will be between 120 and 200?

Problem [ (30 points) The population mean for the average weigh of 12 year old children is [ 2 pounds and the population standard deviation is 40. A random sample of 50 drawn from population Find the probability that the sample taken will have the average weight of the children will be between |40 and /14. What is the probability that the variance for the sample will be between 120 and 200?



Answers

A normally distributed population has mean 1,214 and standard deviation 122 . a. Find the probability that a single randomly selected element $X$ of the population is between 1,100 and 1,300 b. Find the mean and standard deviation of $x-$ for samples of size $25 .$ c. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300

All right. So this population we're looking at has a meat of 200 standard deviation of 50 at a sample size of 100 has taken to find a point estimate for the population. Mean now we're supposed to find the probability that the sample mean is within Lester minus five or plus or minus 10 of the population. Mean All right, let's find our stare deviation for our sampling distribution. Ah, because it doesn't give us a population size. We're going to assume it's infinite. So this equals population standard deviation over the square root of, and it's gonna be 50 over the square root of 100. So that's 50 over 10 were just five. All right? What? We could z score and figure out the probabilities and then do some subtraction. We don't need to do that, because keep in mind, this is following a normal distribution by central limit. Yeah, Central limit. Durham. So thus we've established in a previous section that the probability that this is the mean this is one standard deviation. This is two standard deviation. This right here, 68.5% of all data points and I didn't space this well enough. Hold on. We're gonna move us up for the sake of this point. This year is 95% of all date of all the data. So right here, this is one standard deviation. So we could say that part is 68.5% probability in part B. It's two standard deviations, so that's 95% probability, and there you have it.

All right, So we're given some information about a population, namely, that it's meat is 200 standard deviation is 50. We also have a simple random sample of size 100 and we're supposed to find the probability that our point estimate for mean will be within five of our population mean and then within 10. So let's start by finding our standard deviation of our sampling distribution. So that's gonna be using this formula. We can assume that, um, excuse me, an infinite we can use the formula for an infinite size population in virtue of the fact that it doesn't really give us a population size. So, uh, that's gonna be 50 over the square root of, ah 100 just 50 over 10 which is five. And what we could Z score plus five and negative five plus seven negative 10 against this new standard deviation and our point estimate for a sample mean we don't really need to do that, because if five is our standard deviation, the then the if we were to draw this out, since it's a roughly normal just region by the central limit, their, um Here we go. Ah, we know that about 68% of the data points fall between the first standard deviation in 95% fall between the second standard deviation. So therefore, this is 68% and this is 95% since one standard deviation, two standard deviations, and there you have it.

All right, So this question asks us about a sampling distribution for a sample proportion where the point estimate is point for and we have a sample size of 200. So the first thing we should do is start establishing our sampling distribution. So remember that for proportions, the mean of our distribution is just equal to our sample proportion. But then our standard error is the square root of our sample proportion times one minus the sample proportion all over the sample size. So in this case, that's square root of point for times 0.6 over 200. And this works out to be 0.0 three 464 So now we know they're sampling. Distribution is normal with a mean of 0.4 and a standard error of 0.0 surgery 464 So now we can compute some probabilities. So part, eh? Once the probability that we're within 0.3 of the mean, which is the same thing is asking the probability that were between 0.37 and 0.43 which we get by adding and subtracting 0.3 from the mean. And this is an area problem with a normal curve. So we use normal CDF with a lower bound of 0.37 an upper bound of 0.43 a mean of 0.40 and a standard error of 0.34 64 And this plugging it into your calculator works out to be point 613 five. Then Part B asks us the probability that we're within 0.5 so higher this time. Ah, higher tolerance, which converting this it's the same thing is asking probability between 0.35 in 0.45 which again could be written as an area problem. So normal CDF are lower bound is 0.35 Our upper bound is 0.45 Armenia's point for, and our standard air is still 0.3464 and that works out to be point 8511

This question wants us to find the mean variance and standard deviation of a binomial distribution. We're told that the parameters of this binomial distribution are that and the number of trials is equal to 50 and p. The probability of success is equal to 0.4. Let's start with the mean which we call Mu the Greek letter you view we know for a binomial distribution is equal toe end times p the number of trials times the number of successes. In this case, that's gonna be 50 times 0.4. And when we do that multiplication, we find that Mu is equal to 20. Now the variance is looks similar, but in this time is gonna be n times p times Cute. Now we don't have a Q listed up here as one of our parameters, so we need to figure what what that is. Q. Is the probability of failure, and it's always equal to one minus p so we can do this. Quick. Calculation one minus 0.4 is, of course, 0.6, so Q is 0.6. That's the probability that an individual trial fails, and when we multiply all of these together we'll have Sigma squared is equal to 50 times 0.4 times 0.6, which is equal to 12. And that is our variance. We have a variance of 12 and then we want to find our standard deviation under standard deviation is just the square root of our variants. So it's gonna be the square root of 12 which we can plug into a calculator and find is about equal to 3.46 So these air your final answers. We have a mean of 20 variance of 12 and a standard deviation of about 3.46


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