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Author Topic: 66-95-88.7% rule - 2010 VCAA Exam 1  (Read 1395 times)  Share 

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66-95-88.7% rule - 2010 VCAA Exam 1
« on: October 27, 2014, 07:12:26 pm »
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Hi guys, I am having a bit of trouble with Question 3 of section A on the 2010 further mathematics exam. I understand the 68-95-99.7% rule and have been going really well on my practice exams but this question is really confusing me. I have looked at the iTute worked solutions and still don't really understand what they have done. When I was doing the paper using my general knowledge I guessed the correct answer, but I would rather actually know how to do it. Any help would be greatly appreciated!

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010furmath1-w.pdf

Question 3

AbominableMowman

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #1 on: October 27, 2014, 07:45:40 pm »
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From the box plot we know that the mean is 180
Therefore looking at the histogram, you have to find the 68% interval, so 1 standard deviation above and below the mean which is 178. We can do this by looking at the percentages on the graph, so we need 32% on either side of 180. So therefore we have 178-182.5 which is approximately 68% of the data set and lie within one standard deviation from the mean. Now if we subtract 182.5-178 we get 4.5, and we know that 4.5 has to equal to two standard deviations because its one standard deviation above and one standard deviation below the mean. Then you divide 4.5/2 to get one standard deviation which is 2.25, round down to 2.
I had a bit of trouble with this question too, took me a while to understand it. Hope this helps!
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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #2 on: October 27, 2014, 08:10:48 pm »
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Thank you, that made perfect sense. I was trying to do that but with the damn boxplot. Your working out made heaps of sense though!

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #3 on: October 27, 2014, 10:03:44 pm »
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From the box plot we know that the mean is 180
Therefore looking at the histogram, you have to find the 68% interval, so 1 standard deviation above and below the mean which is 178. We can do this by looking at the percentages on the graph, so we need 32% on either side of 180. So therefore we have 178-182.5 which is approximately 68% of the data set and lie within one standard deviation from the mean. Now if we subtract 182.5-178 we get 4.5, and we know that 4.5 has to equal to two standard deviations because its one standard deviation above and one standard deviation below the mean. Then you divide 4.5/2 to get one standard deviation which is 2.25, round down to 2.
I had a bit of trouble with this question too, took me a while to understand it. Hope this helps!

how did you know the mean is 180? is it cause symmetrical?
and how did you know 1 standard deviation above and below the mean is 178?

thanks

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #4 on: October 27, 2014, 10:11:50 pm »
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how did you know the mean is 180? is it cause symmetrical?
and how did you know 1 standard deviation above and below the mean is 178?

thanks

We know the mean is 180 because of the box and whisker plot.

And he didn't know that - that was figured out using the 68-95-99.7% rule.

AbominableMowman

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #5 on: October 27, 2014, 11:03:16 pm »
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how did you know the mean is 180? is it cause symmetrical?
and how did you know 1 standard deviation above and below the mean is 178?

thanks
The mean is from the box plot and we can figure out 1 standard deviation below the mean is 178 by using the 68% rule from the histogram
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LiquidPaperz

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #6 on: October 28, 2014, 12:08:43 am »
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is the 180mean cause its symmetrical distribution?

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Re: 66-95-88.7% rule - 2010 VCAA Exam 1
« Reply #7 on: October 28, 2014, 12:26:58 am »
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The distribution is not completely symmetrical, but it's not far off it either.

The 180 degree mean comes from the "box and whisker plot". There is a shaded box which starts from the 25th percentile and ends on the 75th percentile. The vertical line in the middle of the box represents the 50th percentile (ie. the mean).

For question 3, you want to imagine another box that starts from the 16th percentile and goes until the 84th percentile (ie. it covers 68% of the data). This box is going to be slightly bigger than the "box and whisker plot" we're given, and looking at the distribution of data, we can roughly predict that it will start around the 178 degree mark and end around the 182 degree mark (without doing any real calculations). This means that the standard deviation is 2 degrees celsius, since 68% of our data falls between the 178 and 182 degree marks, with our mean being 180 degrees.
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