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Question 8A sample of size n-36 is drawn from a population with proportion p-0.58.a) find the mean of the sample proportion b) find the standard deviation of the sa...

Question

Question 8A sample of size n-36 is drawn from a population with proportion p-0.58.a) find the mean of the sample proportion b) find the standard deviation of the sample proportion c) Find the 'probability of the sample proportion being greater than 0.21 d) Find the probability of the sample proportion being less than 0.40Edit Viciy Insert Format Tools Table Paragraph 4 A~ 2y T~ | 12ptAe Peel Dro

Question 8 A sample of size n-36 is drawn from a population with proportion p-0.58. a) find the mean of the sample proportion b) find the standard deviation of the sample proportion c) Find the 'probability of the sample proportion being greater than 0.21 d) Find the probability of the sample proportion being less than 0.40 Edit Viciy Insert Format Tools Table Paragraph 4 A~ 2y T~ | 12pt Ae Peel Dro



Answers

Answer the following questions for the sampling distribution of the sample mean shown. (a) What is the value of $\mu_{\bar{x}} ?$ (b) What is the value of $\sigma_{\bar{x}} ?$ (c) If the sample size is $n=16,$ what is likely true about the shape of the population? (d) If the sample size is $n=16,$ what is the standard deviation of the population from which the sample was drawn?

Now let's look at what information we can gather from uh diagram that they give us about a distribution here. They give us the diagram of the distribution of the sample means, notice the notation down here saying that it's experts distribution. So it's the sampling distribution of the sample means, oh first, what they're asking us for is what is the mean of the distribution of the sample means? Well from our diagram we see the 12 is right underneath the highest point of the curve, And so 12 is the population mean of our um sampling distribution of the sample means next, it wants us to give them the standard deviation of the distribution of the sample means. So here, from our sampling distribution, we see that this 12.05 is underneath where the curvature changes at a transition point, And also the 11.95 is at a transition point. And those values that are one transition point away from the mean is one standard deviation away from the mean. So we can find our standard deviation of our Sampling distribution by taking the 12.05 and subtracting the mean of 12.00 And we get 0.05, So are stigmas of x bar is 0.05. Now, if n is equal to nine, what is likely true about this shape of the distribution of the population? The population that we grab the samples out of? Well, notice from this diagram, they gave us that diagram is the diagram of a normal shape and it's from the sampling distribution of the X bars. Well, the sampling distribution of expire will have a normal shape if either it was, the samples were taken from a normal distribution to start with. If the sample size is small, Um if we had large sample size, then it population could have a shape that might not be normal but notice they're saying and is equal to nine. Remember nine is considered a small sample size, it's not considered a large sample size until your sample sizes 30 or more. So since an equal nine sample size is small and the distribution of X bar is approximately normal, that means that it's likely true that the distribution of the population had a shape that was normal. Yeah. And then lastly, if an equal nine find sigma, the population standard deviation. Well remember that sigma's of X. Bar can be found by taking sigma divided by the square root of N. We don't know sigma but we know sigma Cybex far from earlier in the problem that 0.5 And we know n. n is nine. So we have 0.5 is equal to sigma divided by the square root of nine. Well the squared of nine is three, so when I saw this equation, I'll just multiply both sides by three to solve for sigma. And we get the sigma is equal to, well, three times 0.50 point 15 Yeah.

So in this question, were given a normally distributed population with men of 57.7 and a standard deviation, find the probability that a single randomly selected element X is less than 45. So this is a normally distributed population, so they can just convert that to our standard normal random variable, We have our population mean and standard definition. So that's probability oh z less than -1.052, which is .14 69 What? We were us to find the mean and standard deviation of the sample means that the mean of the sample mean is the population means Which is 57.7. The standard division is 12.1 over the square root of simple size, Which is 3.025. Find the probability that the mean of a sample of size 16 is less than 45, probability that the means of the sample is less than 45. So we take the sample means for for five minus 57 point 7/3 57.0.25 Which is equal to probably it easy, less than -4.19 that is zero.

In this exercise were given data for two populations. We have I mean for 7.9 for the fast population and the standard deviation of 5.4. Also given I mean of 7.1 for the second population and the standard division of 4.6. So we're supposed to I use this theater to find the mean and the standard deviation of all possible differences between the two sample means. So that's the first part of the problem. A And we are also given that the sample sizes will be three and six respectively. That means n one equals three and and to equals six. So first we get the mean of all possible differences between the two sample means. And that's given as follows, mu X r one minus X box too, which is obtained from the formula new one minus new too. So we use the values, we substitute the values into the equation and we obtain 7.9 minus 7.1 and that equals 0.8. Next you find the standard division of all possible differences between the two sample means and that is denoted fine sigma X one bar minus x two bucks, which is obtained by the formula square root of sigma one squared over n one plus sigma two squared of an end to now. When you substitute the values we have the following. You have five 0.4 squared over three. Loss 4.6 squared over three. When you work that out, you obtain the following, you get 9.72 plus 3.53 which equals square root off 13.25 which equals 3.64 So that's going to be the standard division of all possible differences between the two sample means. Next in party. The problem, we are checking what are the variable under consideration must be normally distributed on each of the two populations. For the answer that we have given in part A and the answer to that is no. This is because the formulas for the mean and standard deviation of all possible differences between the two sample means hold regardless of the distributions of the variable on the two populations. So the formulas in part A for uh for the mean and also for the standard deviations would hold even if the variable was not normally distributed. Uh in part, see the problem, we're going to be uh determining yeah. Whether we can conclude that the variable X one bar minus X two bar, which represents all possible differences between the two sample means is normally distributed. No. In this case we see no, because it has not been explicitly explicitly stated that the viable uh under consideration is normally distributed. And also the sample sizes are very small for us to use the central limit story, um to approximate that uh the variable is normally distributed.


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