Statistics Homework ;_;

I apologise for putting this in GC instead of Help, but I've been studying for my statistics exam for three hours now and I think there's one thing that's holding me back from getting an A on the exam.

When calculating the percentile rank I'm just not getting how to get the n_b variable. In the problems in the book we're the full distribution table, but on the exam we'll just have the mean, the standard deviation, and the total number of scores.

Anyone know how to derive n_b with the mean, standard deviation, and total cases?

Just for referrence, here's the formula for finding the percentile rank: P = (n_w[X-L] + in_b)¸ Ni

P = percentile rank
X = percentile point
L = lower real limit of X
n_w = number of scores within the score value
n_b = number of scores below X
N = total number of scores
i = 1

In this example, X = 65, L = 64.5, I'm assuming n_w = 1, N = 30, the standard deviation is 8, and the mean is 50.

So, so far I have P = (1[65 - 64.5] + 1n_b) ¸ 30

I hate math soooo much. ;_;

Wow. I used to be the best at math but I completely shelved the second I graduated highschool. I don't know.

*head explodes*

If edczxcvbnm sees this thread (if he is feeling up to it/not lazy), he might be able to figure this out for you.

You can use a t-test to find the percentage of the bell curve that is below a certain point by converting X to a t-score. Convert from percentage to number of scores by multiplying by N. I'm not 100% sure, but if I encountered such a problem on a test, that's how I would go about doing it.

edit: For reference...

t = (x - u)/s

where u - mean, s - stdev. Then use a calculator function or something (or do the integral by hand) to get the percent area to the left, then multiply by N. If I understand correctly, you should get nb.

OOC:	How do you do superscripts and subscripts?

Ah, I see... I don't know about everyone else, but in this class that's the formula for the z score, which is another part of the problem. So take 1.875 times 30 is 56.25. I don't know what to do next, but I think this is the right track...

OOC:	<*sup> <*sub> tags

Huh, as far as I know, the only difference between a z-score and a t-score is that z's use population mean/stdev's, while t's use sample mean/stdev's... it may still be valid.

all looks greek to me
xD get it? teehee!
anyways. gave up maths after failing A-level.
currently doing an essay on the social impacts of living on the island of Hawaii....yeay volcanoes!
*hugs volcano* xD :love:

Quote:

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Originally Posted by -N-

Huh, as far as I know, the only difference between a z-score and a t-score is that z's use population mean/stdev's, while t's use sample mean/stdev's... it may still be valid.

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Yeah, it's still valid. I'm still trying to get the n_b from that though... argh, I hate this so much. I have to leave to take this test in forty minutes.

Good luck, Shlup. I did great on Statistics myself, but it was a few years ago. That's kind of Dutch to me now because I never actually got to use any of it in real life. Wish I could help you though.

Math is for the weak like shlups :p Take intro to sleep or something like that not math :p

I understand most of this(t-scores, z-scores, standard deviations, etc.) except for this n-sub-b variable thing you're wondering about. xD

I haven't taken Statistics since AP Stat in 12th grade though, so I probably couldn't help anyway.

It's not like I'm taking this class of my own free will. It's required. :weep:

Oh well, I'll get at least partial credit... SCREW MATH GODDAMMIT!

Oh boo hoo. I feel so sorry for you.

Maybe if you'd taken AP Stat in high school, you could've passed out of it. :p

But if I take it now instead of high school then I wouldn't've gotten to ditch high school as much... It was worth it!

And I just figured it out anyway. z = 1.88, so the proportion of scores below z is 0.4699, that times N is 14, so n_b is 14. Woo!

The percentile point of 65 is 48. Does that sound right? I hope so. ;_;

EDIT: I'm off to class. ;_;

Someone can close this thread or move it to Help or leave it open or whatever. My fate is sealed!

Uh... z sounds right, but the proportion doesn't. For z = 1.88, p = .96995. Think about it - z/t scores represent the # of stdev's out from the mean, and the 68-95-99 rule states p for |z|<2 is ~95%, so it clearly couldn't be just .46995. What you calculated was the proportion from the mean up to 1.88 stdev's out, but you have to add 0.5 for the bottom half of the bell curve (from the mean to -infinity stdev's out).

I hope you figured it out by now though. Good luck.