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November 01, 2025, 10:35:43 am
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Author Topic: MAV 2008 Question...Help.  (Read 1826 times)  Share 

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StumbleBum

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MAV 2008 Question...Help.
« on: October 21, 2012, 04:59:09 pm »
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What would you guys put for this answer?
I estimated with my calculator and it gave me a different answer to theirs...
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brenden

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Re: MAV 2008 Question...Help.
« Reply #1 on: October 21, 2012, 05:06:45 pm »
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I would put that there is no outlier. I get mixed up with the skews so I check my summary book but I think that one is positive, so answer C?
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pi

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Re: MAV 2008 Question...Help.
« Reply #2 on: October 21, 2012, 05:10:27 pm »
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I haven't done further, but isn't that negatively skewed?
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StumbleBum

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Re: MAV 2008 Question...Help.
« Reply #3 on: October 21, 2012, 05:11:43 pm »
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Quote from: brenden on October 21, 2012, 05:06:45 pm
I would put that there is no outlier. I get mixed up with the skews so I check my summary book but I think that one is positive, so answer C?
It's negatively skewed, and yes no outlier. So the answer is A. However, when I put the data into my calculator and graphed a box-plot of it there was two outliers...
Why did you say that there isn't an outlier? Because it is not separated from the main group of data?
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Stick

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Re: MAV 2008 Question...Help.
« Reply #4 on: October 21, 2012, 05:20:17 pm »
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If you received two outliers, then option B is incorrect anyway. :P MAV just saved its butt in a very clever way. ^_^
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StumbleBum

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Re: MAV 2008 Question...Help.
« Reply #5 on: October 21, 2012, 05:25:46 pm »
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Quote from: Stick on October 21, 2012, 05:20:17 pm
If you received two outliers, then option B is incorrect anyway. :P MAV just saved its butt in a very clever way. ^_^
Yes, but that's only because I just estimated values from the histogram so I assumed that there would have been at least one outlier accounting for my estimations being slightly off.
Ehh, the answers say that it can't be tested mathematically because no values are given so you should assume no outliers. God I hate how further can be inaccurate at times. It isn't hard to make a sort of scale to use to estimate the values with a ruler...
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brenden

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Re: MAV 2008 Question...Help.
« Reply #6 on: October 21, 2012, 05:41:14 pm »
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Yeah my bad. Pesky skews. Sorry about that. I said no outlier because there doesn't look to be an extreme value at either end. Which one would you say was the outlier(s)?
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Daenerys Targaryen

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Re: MAV 2008 Question...Help.
« Reply #7 on: October 21, 2012, 08:36:03 pm »
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How did you put that data into your calculator?
I would choose A
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StumbleBum

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Re: MAV 2008 Question...Help.
« Reply #8 on: October 21, 2012, 09:06:36 pm »
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Quote from: HatersGonnaHate on October 21, 2012, 08:36:03 pm
How did you put that data into your calculator?
I would choose A
Made a scale as a reference and then ruled out all the heights to determine its frequency. In hindsight i shouldn't have bothered, there wasn't any values for a reason. Oh well, another 39/40. Just got to hope i don't do something stupid like that in the exam...
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Yendall

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Re: MAV 2008 Question...Help.
« Reply #9 on: October 21, 2012, 09:20:00 pm »
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On Histograms, outliers are usually quite obvious. I don't think VCAA would make you transform a historgram with no x and y axis values into a box plot. It doesn't seem like a logical approach because you're essentially guessing every single value.

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Re: MAV 2008 Question...Help.
« Reply #10 on: October 22, 2012, 08:30:31 pm »
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If you are asked to estimate, then the answer is negatively skewed, because you can't determine if there are outliers without calculating. If one segment is away from the group then you estimate that to be an outlier. Here they're all together so the answers is negatively skewed
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