The population mean annual salary for environmental

Here is a link to a Word document with the solution for the problem:

Probability

Please feel free to ask if you have any questions about this solution.

If you would still like to receive a phone call, please let me know.

Thanks,

Ryan

Are you asking about the link?

I'm not sure what you are specifically referring to.

If you are asking about the website, then, yes, in general the site is available 24 hours a day. Of course, it depends on whether there is anybody online who works in the category that your question is place in. If there are no available experts, then there would be some delay in getting an answer.

If you are asking about the link that I posted, then the answer is that the link will be available for about a week.

Is there anything else that I can help you with?

This site can be reached at www.justanswer.com.

If you are having trouble accessing your account, you should contact a Customer Service representative to get assistance. There should be a link on the page that you can use to contact them, or you can call them at(###) ###-####

Did you click on the link that I posted? The one labeled "Probability"?

If needed, I can post a new link. Or I can repost the solution as text here.

If you would prefer to use email, please let me know and we can make arrangements to do that.

Here is the solution in text:

Calculate the z value:

z = (x-bar - mu) / (sigma / √n) = (62,500 - 66,000) / (6400 / √40) = -3.4587

Then, since the question asks for the probability that the average is "less than 62,500", we want the area to the left of z = -3.4587:

P(x-bar < 62,500) = P(z < -3.4587) = 0.0003

You can get this value from tables, from a statistical calculator, or by entering the following in a cell in Excel:

=normsdist(-3.4587)

Solution: The probability is 0.0003 (rounded to four decimal places).

Great!

Is there anything else I can help you with?

The site doesn't charge on a "per day" basis like that. Also, please be forewarned that if you are offered some kind of a "subscription" for the site, that their policies do not allow subscriptions to be used in the homework categories.

The best approach for your situation is to post a set of questions, and adjust the price to a level that you are comfortable with for that amount of work involved. If one of the experts is willing to take on the assignment for the offered price, then they will contact you.

How many more questions do you need help with tonight?

The interpretation of the result from the previous question is the first choice:

Only 0.03​% of samples of 40 specialists will have a mean salary less than
​$62,500. This is an extremely unusual event.

For the problem about the gasoline prices, do you just need the final answer, or do you need the intermediate steps as well?

Ok.

The answer the gasoline price problem is:

Probability = 0.8106

If you have additional problems that you want help with, can you please include a problem number? That will make it easier to refer to specific problems, and to match up the answers with the correct problem.

Thanks,

Ryan

Yes.

D. About 81​% of samples of 33 gas stations that week will have a mean price between
$ 2.699 and ​$2.727.

Ok.

What they want you to do here is compare the probability of a single individual from the population having a height that is less than 65.1 inches, with the probability of a sample of 27 individuals having an mean (average) height that is less than 65.1 inches.

For the single individual, calculate the z-value as:

z = (x - mu) / sigma = (65.1 - 64.4) / 2.7 = 0.2593

The corresponding probability is:

P(x < 65.1) = P(z < 0.2593) = 0.6023

For the sample of 27 individuals selected from this population, calculate the z-value as:

z = (x-bar - mu) / (sigma / √n) = (65.1 - 64.4) / (2.7 / √27) = 1.3472

The corresponding probability is then:

P(x-bar < 65.1) = P(z < 1.3472) = 0.9110

The probability is much higher in the second case, so you are far more likely to select a sample of 27 individuals with a mean height that is less than 65.1 inches than you are to select a single individual who is less than 65.1 inches tall.

Did you enter the 0.6023 answer first?

Ok, the answer for the first part:

What is the probability of randomly selecting 1 woman with a height less than 65.1 ​inches?

is 0.6023

D. The percent of the population who fall under the​ "yes" category probably lies between​ 40% and​ 50%.

The z-core that corresponds to the upper 15th percent is 1.04 (rounded to two decimal places).

I'm not sure what to say about that. The only thing that I can think of is that they have some kind of wonky definition of "Upper P15". I interpreted it to mean the top 15%. Normally though, this would be referred to as "P85" instead of "Upper P15".

If we ignore the "Upper" part, and just compute P15, then the answer is -1.04. (Note the negative sign.)

zc = 1.372

round off as directed

Do you have more questions to work on tonight?

I see. If you are able to get back in, please feel free to post the next problem here.

So you can't get to this site on your computer?

Ok, good.

E = 0.5922

The 90% confidence interval is (616.5722,(###) ###-####

The 95% confidence interval is (613.5783,(###) ###-####

Please let me know if the above intervals do not match the answer form that you are expected to use. Sometimes the intervals are expressed as "mean ± margin of error".

For a 90% confidence interval, use z = 1.645.

The margin of error is then:

E = z * s / √n = (1.645)(16.5) / √100 = 2.7143

Then, the limits of the confidence interval are:

Lower limit = mean - E = 134 - 2.71 = 131.29

Upper limit = mean + E = 143 + 2.71 = 136.71

The new confidence interval is thus (131.29, 136.71)

The original confidence interval (with n = 50) is wider than the new confidence interval. This is because the larger sample size narrows the confidence interval.

A. The n = 50 confidence interval is wider because a smaller sample is​ taken, giving less information about the population.

Mean = 3.504

New standard deviation = 0.040

Sample size = 57

Margin of error = z * s / √n = (1.645)(0.040) / √57 = 0.0087

Lower limit = mean - E = 3.504 - 0.0087 = 3.495

Upper limit = mean + E = 3.504 + 0.0087 = 3.513

The new confidence interval is (3.495, 3.513).

The original confidence interval is wider.

A. The s = 0.070 confidence interval is wider because of the increased variability within the sample.

For the 99% confidence level:

Required sample size = (z * s / E)^2 = (2.576 * 11.7 / 0.44)^2 = 4693 (when rounded up to the next whole number).

I don't know. I don't have any access to the details of your account. A customer service representative can help you with that inquiry.

I'm afraid that I can't answer your question about billing because:

1) I am not authorized by the website to discuss those types of issues, and

2) I don't have any information about your account, what you paid, the terms of the agreement, etc

This one:

A doctor wants to estimate the HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99% confidence assuming
σ=11.7?
Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size​ required?

A​ 99% confidence level requires ____ subjects.

was already answered. The answer was 4693.

For this one:

A beverage company uses a machine to fill​ one-liter bottles with water​ (see figure). Assume that the population of volumes is normally distributed.

​(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a
90​% confidence interval for the population mean. Assume the population standard deviation is
55 milliliters.
​(b) Repeat part​ (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample​ size? Explain.

a) n = (z * s / E)^2 = (1.645 * 55 / 1)^2 = 8186 (rounded up to the next whole number)

b) n = (z * s / E)^2 = (1.645 * 55 / 22)^2 = 17 (rounded up to the next whole number)

The smaller error tolerance requires the larger sample.

And for this one:

A machine cuts plastic into sheets that are 30 feet ​(360 ​inches) long. Assume that the population of lengths is normally distributed. Complete parts​ (a) and​ (b).

​(a) The company wants to estimate the mean length the machine is cutting the plastic within
0.125 inch. Determine the minimum sample size required to construct a 90​%
confidence interval for the population mean. Assume the population standard deviation is
0.25 inch.

n = (z * s / E) = (1.645 * 0.25 / 0.125)^2 = 11 (rounded up to the next whole number)

Yes

Part of the trouble there is how one should interpret "44 points".

Taking 44 as a whole number, the required sample size would be:

n = (2.576 * 11.7 / 44)^2 = 1 (when rounded up to the next highest whole number)

This seemed unreasonable, which is why I went with the previous answer, but try it and see.

I don't know. It seems to be working now though.

No. The only thing I received recently was the bit about the sample size answer being wrong. I didn't get anything else.

A machine cuts plastic into sheets that are 30 feet ​(360 ​inches) long. Assume that the population of lengths is normally distributed. Complete parts​ (a) and​ (b).

​(a) The company wants to estimate the mean length the machine is cutting the plastic within 0.125 inch. Determine the minimum sample size required to construct a 90​% confidence interval for the population mean. Assume the population standard deviation is 0.25 inch.

n = 11 ​(Round up to the nearest whole number as​ needed.)
​

(b) Repeat part​ (a) using an error tolerance of 0.0625 inch.

n = (z * s / E) = (1.645 * 0.25 / 0.0625)^2 = 44 (rounded up to the next whole number)

B. The tolerance E =0.0625 inch requires a larger sample size. As error size​ decreases, a larger sample must be taken to ensure the desired accuracy.

a) n = (1.645 * 0.25 / 0.05)^2 = 68

b) n = (1.645 * 0.40 / 0.05)^2 = 174

A population standard deviation of

0.40

in. requires a larger sample size. Due to the increased variability in the​ population, a

larger

sample size is needed to ensure the desired accuracy.

0.40 for the first

large for the second

(19.28, 20.64)

C. With 90​% ​confidence, it can be said that the confidence interval contains the true mean closing stock price.

tc = 2.797

tc = 1.729

E = 5.5

For this you need to first calculate the number of degrees of freedom:

df = n - 1 = 14 - 1 = 13

Then, get the critical t value from tables or a statistical calculator. The critical value with c = 0.99 and 13 degrees of freedom is tc = 3.012.

The margin of error is then E = tc * s / √n = 3.012 * 3.8 / √14 = 3.1

Is the value of c 0.99 or 0.9999?

Does it give you any clues as to why? I can't find any error there.

Ok, no worries.

For this one

df = n - 1 = 16 - 1 = 15

The critical t-value with c = 0.80 and 15 degrees of freedom is tc = 1.341

E = tc * s / √n = 1.341 * 2.5 / √16 = 0.8381 (round off as directed)

(a) (9.6, 18.0)

(b) (10.6, 17.00)

The interval using the t-distribution is wider.

(107.0, 119.0)

The sample mean is the midpoint of the interval:

mean = (13.3 + 22.9) / 2 = 18.1

The margin of error is half of the width of the interval:

margin of error = (22.9 - 13.3) / 2 = 4.8

For the "incorrect" one, which order did you enter the answers in? I think I gave them to you in the reverse order.

Yes. The site seems to be having a bit of an issue (or maybe it's just on my end). I attempted to post an answer, but it seems to have gotten lost somewhere. I'll post it again.

Confidence interval: (50.83, 79.17)

The margin of error is 14.17

D. Use a normal distribution because n ≥ 30.

(1.48, 1.50)

A. Use a​ t-distribution because n < ​30, the miles per gallon are normally distributed and σ is unknown.

(19.5, 22.3)

A doctor wants to estimate the HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size​ required?
A​ 99% confidence level requires
46924692 subjects.
​(Round up to the nearest whole number as​ needed

For the 90% confidence interval, the required sample size is:

n = (1.645 * 11.7 / 0.44)^2 = 1914

The decrease in confidence leads to a decrease in the required sample size.

p = 299 / 1281 = 0.233

q = 1 - p = 1 - 0.233 = 0.767

p = 118 / 808 = 0.146

q = 1 - p = 1 - 0.146 = 0.854

The margin of error is (0.779 - 0.753)/2 = 0.013

The sample proportion is (0.753 + 0.779)/2 = 0.776

I'm sorry. That was either a typo or a brain hiccup. It should be 0.766 instead.

(0.702, 0.727)

a) 51

b) 56

Oops, I missed correcting the "11%" in your post to "1%". Most of the numbers are showing up doubled on my end, as in "3434%" instead of "34%".

The corrected answers are:

a) 6073

b) 6766

b) required sample size for an error tolerance of 22 milliliters = 17 bottles

E = 0.751

Margin of error = z * s / √n

For a c = 0.95, the value of z is 1.96.

Then E = 1.96 * 2.3 / √36 = 0.751

ok

Required sample size = 24

A doctor wants to estimate the HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 4 points with 99% confidence assuming σ=11.7? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size​ required?
A​ 99% confidence level requires 57 subjects.
​(Round up to the nearest whole number as​ needed.)
A 90% confidence level requires
nothing subjects.
​(Round up to the nearest whole number as​ needed)

Required sample size = 24

Hi,

I'm sorry about the delay in replying. I am away from my desk at the moment. I will send the solutions for these as soon as I can.

Ryan

Probably in three or four hours or so.

For this one:

Use the confidence interval to find the estimated margin of error. Then find the sample mean.

A biologist reports a confidence interval of left parenthesis (2.1,2.9) when estimating the mean height​ (in centimeters) of a sample of seedlings.

The estimated margin of error is (2.9 - 2.1)/2 = 0.40

The sample mean is (2.1 + 2.9)/2 = 2.50

For this one:

Use the confidence interval to find the margin of error and the sample mean.

(0.426,0.550)

The margin of error is (0.550 - 0.426)/2 = 0.062

The sample mean is (0.426 + 0.550)/2 = 0.488

Those were the first two that you posted this morning.

Would it be easier for you if we just started fresh and have you just post the ones that you currently need help with?

Ok, ready when you are.

To determine the critical value, either look up the value closest to 0.86 in a table of the normal (z) distribution, or use a statistical calculator.

If you have Excel, you can enter "=normsinv(x)" into a cell to get the z-value that has an area to the left of z equal to x.

Since the confidence interval must include 86% of the distribution (c = 0.86), each tail will include (1 - 0.86)/2 = 0.07. This must be added to the value 0.86 to get the total area to the left of the desired value of z. 0.86 + 0.07 = 0.093

Thus, enter "=normsinv(0.93)" into Excel, without the quotation marks.

zc = 1.48

Oops, typo. 0.86 + 0.07 = 0.93, not 0.093 as written in the previous post.

The confidence interval is (54.88, 58.52), so it appears that choice B is what they are looking for.

The z-value for a 98% confidence interval is z = 2.326.

The margin of error for the confidence interval is then E = (z * s / √n) = (2.326 * 0.3 / √48) = 0.1007

Lower limit = mean - E = 4.9 - 0.1007 = 4.80

Upper limit = mean + E = 4.9 + 0.1007 = 5.00

The 98% confidence interval is then (4.80, 5.00)

Same process as the last one. Here, the z-value is 1.96.

The resulting confidence interval is (14.7, 18.5).

Margin of error = (4.5 - 2.7)/2 = 0.90

Sample mean = (2.7 + 4.5)/2 = 3.60

Margin of error = (0.550 - 0.426)/2 = 0.062

Sample mean = (0.426 + 0.550)/2 = 0.488

For a 90% confidence level, use z = 1.645.

Then, n = (z * s / E)^2 = (1.645 * 8.7 / 2)^2 = 52 (rounded up to next whole number)

For c = 0.98, use z = 2.326.

n = (z * s / E)^2 = (2.326 * 8.1 / 2)^2 = 88.7 -> 89

Estimate margin of error = (2.9 - 2.1)/2 = 0.40

Sample mean = (2.1 + 2.9)/2 = 2.50

n = 60, x-bar = 119.00, s = 15.30, z = 1.645 for a 90% confidence interval

Margin of error = E = (z * s / √n) = (1.645 * 15.30 / √60) = 3.2492

Lower limit = x-bar - E = 119.00 - 3.2492 = 115.75

Upper limit = x-bar + E = 119.00 + 3.2492 = 122.25

The 90% confidence interval is (115.75, 122.25).

For the 95% confidence interval, do the same thing, except with z = 1.96.

Margin of error = E = (z * s / √n) = (1.96 * 15.30 / √60) = 3.8714

Lower limit = x-bar - E = 119.00 - 3.8714 = 115.13

Upper limit = x-bar + E = 119.00 + 3.8714 = 122.87

The 90% confidence interval is (115.13, 122.87).

A. With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%.

90% interval: (0.7662, 0.8083)

95% interval: (0.7621, 0.8124)

99% interval: (0.7542, 0.8203)

Round off values as needed.

Does it say how many digits to round off to?

How do they want you to round this?

Ok...

90% interval = (0.763, 0.805)

95% interval = (0.759, 0.809)

99% interval = (0.751, 0.817)

between 0.763 and 0.805

yes

95% interval = (0.759, 0.809)

A. between 0.759 and 0.809

B. between 0.751 and 0.817

B. The​ 99% confidence interval

Is the standard deviation 5, or 55?

Also, in part b, is the tolerance 2 milliliters, or 22 milliliters?

a) n = (1.645 * 5 / 1)^2 = 68

b) n = (1.645 * 5 / 2)^2 = 17

I don't know what else to say there. This problem is a repeat, and the issue with the previous answers was due to whether the values were 5 and 2 or 55 and 22. This doubling of digits is a bit of a pain to deal with.

The calculation itself is pretty simple. Are you getting some kind of an error message?

The error tolerance of 1 milliliter requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain sufficient information from the population the desired accuracy.

yes, that's correct

a) z = (x - mu) / sigma = (550 - 507) / 120 = 0.3583

P(x < 550) = P(z < 0.3583) = 0.6399 = 0.6399 * 100% = 63.99%

b) z = (x - mu) / sigma = (525 - 507) / 120 = 0.15

P(x > 525) = P(z > 0.15) = 0.4404

Expected number greater than 525 is 0.4404 * 1000 = ≈ 440.4

How many more do you have? We are well beyond an hour at this point.

Yes

Try 440.38 instead. That's the only thing I can think of.

Ok, I don't know what else we can do with that last one that would be different from what I have already done.

z = (x-bar - mu) / (sigma / √n) = (23.3 - 23) / (1.3 / √62) = 1.8171

P(x-bar < 23.3) = P(z < 1.8171) = 0.9654

This would not be considered unusual.

The sample mean would not be considered unusual because it has a probability that is greater than 5%.

z = (212 - 211) / (3.4 / √65) = 2.3713

P(x-bar > 212) = P(z > 2.3713) = 0.0089

The sample mean would be considered unusual because it does not lie
within the range of a usual​ event, namely within 2 standard deviations

of the mean.

It seems that this question was already answered, since the values for the mean and standard deviation of the distribution of means are already computed, and one of the choices about the graph is marked as "correct".

What do you need me to do here?

The mean of the distribution of sample means is $220 (the same as the sample mean)

The standard deviation of the distribution of sample means is 57 / √29 = 10.58

There are four choices (A, B, C, and D), but only three descriptions.

I would select whichever choice goes with the following:

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 188.2 to 251.8 in increments of 10.6 and is centered on 220. All values are approximate.

which is either Choice C or Choice D, depending on whether the descriptions come at the beginning of each choice, or at the end.

Select the one that has values from 188.2 to 251.8 running along the bottom edge, with the value 220 in the middle along the bottom.

Can you take a screenshot of the graphs and post that?

I'd lean toward Choice C then.

The values -2, 0, and 2 would most likely be z-values though. Unless the axis of the graph is marked as something like "z-value", then this choice is not likely to be correct.

A typical layout though would label the mean in the center, and the values that are 2 standard deviations above and below the mean. With a mean of 220, and a standard deviation of the mean of 10.58, the values that are 2 standard deviations away from the mean are:

220 - 2(10.58) = 198.84

220 + 2(10.58) = 241.16

These values match the values given in Choice C.

It depends on what kind of computer you have.

If it is a Macintosh, press the "Command" key, the "Shift" key, and the number 3, all at the same time.

If it is a Windows machine, there is likely a "Print Screen" button on the keyboard. Depending on the make and model, you might need to press a function key, or the "Alt" key at the same time to take the screenshot. Doing so will save the screen image to the clipboard, and you can then paste it into a document.

a) z = (x - mu) / sigma = (550 - 510) / 114 = 0.3509

P(x < 550) = P(z < 0.3509) = 0.6372

b) z = (525 - 510) / 114 = 0.1316

P(x > 525) = P(z > 0.1316) = 0.44765

n = 1000 * 0.44765 = 447.65

a) z = (675 - 505) / 112 = 1.5179

P(x < 675) = P(z < 1.5179) = 0.9355 = 93.55%

b) z = (550 - 505) / 112 = 0.4018

P(x > 550) = P(z > 0.4018) = 0.343915

n = 1000 * 0.343915 ≈ 343.92

n = (2.576 * 11.7 / 3)^2 = 101

n = (2.576 * 13.7 / 4)^2 = 78

You're welcome.

Ok. You will need to start a new question for that. If you wish to request me specifically, you can do so by simply adding "For Ryan" at the beginning of your new question.

Thanks.

Hi,

I'm sorry that I couldn't reply sooner. I have been having some computer issues today.

You will need to open a new question page for these problems. Please be sure to add "For Ryan" at the beginning of the new question page so that you are sure to have your question directed to me.

Also, please put ALL of the problems that you need help with into a single document and post that as an attachment. I need to know what I am agreeing to before I make a commitment.

Thanks,

Ryan

There should be a link to a list of your questions, and there should be a link or button there that you can click on to start a new question.

Once you are able to start a new question, just type in "For Ryan" at the beginning of the text.

If you need further assistance with getting a new question page opened, you should call Customer Service at(###) ###-#### I am limited in what I can offer because I have only a vague idea of how the site looks on your end.

I'm sorry, I don't know what else to tell you as I don't have any suggestions other than contacting Customer Service. They will be able to walk you through the process. Not only will they have your current account information (in case there is some issue with billing or something like that), they will also know what the screen looks like on your and and should be able to direct you on which page to go to and which buttons to click.

I'm sorry, I guess I don't know what you mean by that.

I don't mind combining questions today with questions on Thursday, as long as we can agree on a reasonable price for a certain number of questions. What happened last time is that there was no definite end to the number of questions, and the end result is that I was committed for several hours with no additional compensation. Particularly on the second day when "an hour" got stretched out to more like four hours. Even though there are gaps between the questions, I'm still tied to my computer, and my ability to get anything else done is very limited. I'm not here to get rich, but I'm also not here to work for free.

I'm ok with how that played out last time, but I don't want to do that again.

If you just type "For Ryan" at the beginning, that should take precedence over any assignment that the system doles out. If someone else replies to you, you are free to tell them that you want to work with me.

I'm sorry, I don't know if you were waiting to hear back from me, or if something else came up. I'm back, and ready when you are.

No problem.

I don't currently have anything scheduled for Thursday, so that should be fine. What time did you have in mind? (Include the time zone too please.)

I'm on the west coast of the US. Pacific Daylight Time.

3:30pm Pacific Time it is. See you here then.

Yes

ok, what is it?

I have some appointments tomorrow. What time did you have in mind?

(Sorry about the delayed response. This page is so long now that it takes forever to load. Yet another reason why we need to get a new page started.)

Can you be more specific about what "3 homework" means? Do you mean three separate problem sets?

I can't tell you how to open a new page, because I don't know what the site looks like on your end. Customer Service can help you with this. They will know which links you should use, and will be able to direct you to the correct page.

You can call them at(###) ###-####

Which question? Did you open a new one? If so, I haven't seen it. It is possible that it ended up in a different category for some reason, in which case I wouldn't be able to see it.

No, we can't just keep on adding to this page. Most importantly, there is no mechanism for me to get paid if we do that, since this "page" has already been paid for. Secondly, the page is so long at this point that it is very slow to load.

You may have made a payment to the site, but, as I understand it, that only deposits funds to your account on the site. Unfortunately, the process is a little more complicated than that. I only get paid when you rate the service that you have received. Since this page has already received a rating, there is no provision for you to add another rating, which means that there is no way on this page for you to “pay me again”. This is why a new page needs to be opened.

Please keep in mind that I am not an employee of the site. I am a user of it, just like you. I simply don’t have either the information or the access to assist you with account issues or other site issues. My role on the site is limited to answering math questions. You really need to contact Customer Service for assistance.

I never received any notification of the new page being opened, and you did not add "For Ryan" at the beginning, which left another expert free to answer your question. Since you have accepted the offer that was presented to you, a rating has been entered on the page, so even if the other expert were to step aside, there would be no way for me to get paid. At least not without you having to contact Customer Service again.

Ok.

I see your new question, but it got posted in the "General" category for some reason and I cannot post a response to it at this time. I have made a request to a moderator to move it to the proper category. I am confident that it will be moved, but I don't know how quickly it will happen.