Post by: Yendall on October 20, 2012, 07:59:45 pm
(http://i292.photobucket.com/albums/mm37/yendall_2008/asdwas_zps1771ecb2.jpg)
Question 3
Using the 68–95–99.7% rule, the standard deviation for temperature is closest to
A. 1 °C
B. 2 °C
C. 3 °C
D. 4 °C
E. 6 °C
I came across a pretty cool equation to find the standard deviation:
From inspection I estimated the range to be:
Using the formula:
So i thought the standard deviation was 3 degrees. The answer is 2.
What is the proper way to find the Standard Deviation from inspection?
Post by: Daenerys Targaryen on October 20, 2012, 08:08:12 pm
If you divide by six you get 2.25, which is the closest to two. Not sure if this is the proper way, but i have come across questions and interperted it like this and got it correct.
Post by: Stick on October 20, 2012, 08:10:05 pm
Data:
(http://i292.photobucket.com/albums/mm37/yendall_2008/asdwas_zps1771ecb2.jpg)
Question 3
Using the 68–95–99.7% rule, the standard deviation for temperature is closest to
A. 1 °C
B. 2 °C
C. 3 °C
D. 4 °C
E. 6 °C
I came across a pretty cool equation to find the standard deviation:
From inspection I estimated the range to be:
Using the formula:
So i thought the standard deviation was 3 degrees. The answer is 2.
What is the proper way to find the Standard Deviation from inspection?
You've got the formula wrong. It's actually
Using that:
Which gives you your correct answer. :)
I only discovered this formula a few days ago myself. It should be stated more explicitly in the study design in my opinion. :S
Post by: Daenerys Targaryen on October 20, 2012, 08:12:37 pm
Post by: Stick on October 20, 2012, 08:16:03 pm
Post by: Yendall on October 20, 2012, 08:30:15 pm
I'll remember that for the exam, cheers guys.
Post by: Stick on October 20, 2012, 08:40:56 pm
I'm just assumingaccounts for 67% of data
accounts for 95% of data
accounts for 99.7% of data
I'll remember that for the exam, cheers guys.
No, that's not true either. I think you've just got your logic a bit mixed up about it all. If you want to find any percentage of the data, you're going to need to find out what one standard deviation is first and then multiply it (not just change the denominator like you've done).
Post by: Daenerys Targaryen on October 20, 2012, 08:51:07 pm
Post by: StumbleBum on October 20, 2012, 09:02:22 pm
I came across a pretty cool equation to find the standard deviation:
I believe that equation relates solely to small sets of data, that's what the essentials book states anyway...
Post by: panda31 on October 20, 2012, 09:16:06 pm
Could you explain it?
Thanks
Post by: StumbleBum on October 20, 2012, 09:44:03 pm
Hey Stick, what do you mean by finding the standard deviation then multiplying it? what do you multiply it with - and do you find the standard deviation of the whole of the data first?I believe he's referring to the 68-95-99.7 rule. Where you find what the standard deviation of a data set is, along with the mean. Then you multiply the standard deviation by 1 and add and subtract from either side of the mean to get approximately 68% of the data. For 95% of the data you add and subtract two multiplied by the standard deviation from the mean; and by 3 for 97.7% of the data.
Could you explain it?
Thanks
Although that's just what I assumed from what he said, could be talking about something else..?
Post by: Daenerys Targaryen on October 20, 2012, 09:50:25 pm
Find the SD, then use that and the mean how you usually would.
Post by: Yendall on October 20, 2012, 10:38:52 pm
No, that's not true either. I think you've just got your logic a bit mixed up about it all. If you want to find any percentage of the data, you're going to need to find out what one standard deviation is first and then multiply it (not just change the denominator like you've done).What I mean is that you can't use the first two formulas because we will always have to deal with 99.7% of data. We don't get smaller data sets from VCAA. We won't get a distribution with only 68% of the data shown (most likely), and it wouldn't make sense to get that and not 99.7%. But say you did get a range of data that was only 68% of a bigger set of data, the range and standard deviation of that particular 68% will work with the first formula, right?
I would always use the last formula because you have to account for -3SD and +3SD in all cases
Post by: StumbleBum on October 20, 2012, 11:19:45 pm
But say you did get a range of data that was only 67% of a bigger set of data, the range and standard deviation of that particular 67% will work with the first formula, right?
Why do you keep saying 67% and not 68%?
But yes that is correct, 68% of the data would be distributed about 2 SD's (assuming this is about the mean also) so the range of that 68% divided by two would give you the value for the SD.
Post by: Yendall on October 20, 2012, 11:21:40 pm
Post by: StumbleBum on October 20, 2012, 11:32:33 pm
Sorry I meant 68%, I'm fairly delirious.Haha, gets a bit like that after the monotony of Further exams.
Post by: Daenerys Targaryen on October 21, 2012, 11:37:24 am
But chances are they're gonna give a full set rather than a fraction of it.
Post by: StumbleBum on October 21, 2012, 11:51:17 am
Yeah, if you're given only 68% of the data then you divide by 2, 95% divide by 4, and so on.Mmm don't think throughout the whole year, including all the practise exams, i've ever seen a question where they haven't used a full set... Would be a nice little differentiating question in a VCAA exam though.
But chances are they're gonna give a full set rather than a fraction of it.
Post by: Daenerys Targaryen on October 21, 2012, 12:01:33 pm
Post by: Stick on October 27, 2012, 09:20:19 am
If the question states the data is normally distributed, you need to divide the range by six. However, if it does not state the nature of the data's distribution, then you divide the range by four. Both are only estimations at best, but it explains why there are two formulae floating around. :)