For the following situation find the mean and standard...

For the following situation, find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 505, 638, 549, 611, and 586. Use a sample size of 2. The standard deviation of the population is 46.65 (Round to two decimal places as needed.) Identify all samples of size 2 with the correct accompanying means below. A. 549, 638, x = 593.5 B. 505, 586, x = 545.5 C. 586, 505, x = 545.5 D. 586, 549, x = 567.5 E. 638, 549, x = 593.5 F. 638, 611, x = 624.5 G. 505, 611, x=558 H. 549, 549, x=549 I. 638, 505, x = 571.5 J. 586, 638, x=612 K. 586, 611, x = 598.5 L. 611, 549, x = 580 M. 586, 586, x = 586 N. 638, 586, x=612 O. 611, 611, x = 611 P. 549, 505, x = 527 Q. 611, 505, x = 558 R. 611, 586, x = 598.5 S. 505, 505, x = 505 T. 505, 549, x=527 U. 549, 611, x = 580 V. 611, 638, x=624.5 W. 505, 638, x=571.5 X. 638, 638, x = 638 Y. 549, 586, x=567.5 Z. 586, 625, x = 605.5 [. 611, 638, x = 557.5

Transcript

00:01 So in this question we have five original uh samples here so we have a 503 3 636 542 604 and 576 so we have our original sample five and we want to list all the in this case we want to find what is the mean of this sample mean distribution and what is the standard deviation for the sampling distribution, which is the distribution of the sample means.

00:36 So to do this, they are considering that we should consider now a sample of two from this.

00:42 And to list all the samples, you're going to find all of them from a to z, except in this case, letter p.

00:53 P is not one of them.

00:55 So we have a to z, which means 26 values, but you remove p, which will give us 25 possible values for the sample of 2 from this original sample, which allows repetition.

01:08 So we could have 503 and 636 or 636 and 503.

01:17 There will be two different samples, which is the same as, for example, getting 503 twice.

01:24 It is also possible.

01:25 So we have a total of 25 possible samples of size 2.

01:30 So you need to select all of these, a to z, except p.

01:36 P and if we consider that value, i'm gonna list here some of them.

01:42 For example, we could have, i'm gonna put x1, x2, x bar and the probability.

01:52 We could get, like i said, 503 and 503, which means that the sample mean will be also 503.

01:59 And because we have 25 values, this means that this specific sample has a probability of 1 divided by 25 because we have 25 possible samples of size 2.

02:11 So we could do the same for for example 503 and 636 which will give us the sample of 569 with the same probability and we could go until we get to the last possible value which is again getting 576 and 576, which will give us me equals to 576 with probability 1 divided by 25.

02:42 This one here, actually for this one here, which is like 569, we have two possible here, and it is 0 .5, two possible ways of getting the value, which is when we get 503 and then 636, or let me say that we have here in the middle, we got 636 first, and then we got 503.

03:14 This will also give us the same value for the mean, okay? which means, which i'm going to put here, the same probability, 1 divided by 25.

03:25 But then when we list all these 25 values, so let me put three dots here, when we list only the values for the mean, you're going to see here that some of them are repeated.

03:39 So when they are repeated, we put here how many times they repeat...

Question

Please give Ace some feedback