the three median line

[vce01]

im having a problem with this,

no problems with dividing up the number of points, and then finding the median. but what exactly do you do after that?
my text book says 'place your ruler so that it forms a line that passes through the two medians in the outer groups. keeping the ruler at the same slope, slide it one third of the way towards the middle median and draw in the line'.

how are you sposed to figure out you've slid it exactly 1/3rd of the way towards it? i'd have to be pretty much spot on because there have been questions in exams that ask you to find the slope of the line after you've figured it out.


and if anyone has checkpoints 08, can someone tell me how we're sposed to figure out the medians for Q47??

[ausyid]

If you need to find the actual three median line, just use Med-Med in the calculator.

You can find the slope using the gradient formula.... (y2-y1) / (x2-x1) . The two points you use are the median points of the two outer groups. Remember the slope of the graph doesn't change when you move it 1/3 of the way, which I don't think you can be asked to do in an exam as it is too inaccurate.

[vce01]

oh yeah, ofcourse *facepalm.

i haven't used the med-med function before though :/ where can i find it? (got a ti-83)

[jsimmo]

STAT -> CALC -> MED-MED


- Choose 2 points on the line (x1, y1) and (x2, y2)

- If the y intercept (c) can be read directly from the line then use the y=mx + c formula for the equation. Otherwise use the gradient and one point (x1, y1) in y-y1 = m(x - x1) for the equation.

But if you just have the data then it's easier just to use the calculator.

[vce01]

^thanks

[jsimmo]

Does anyone know how to use an accurate method to answer this question? The marks aren't positioned directly on one absolute point hence it's hard to determine the upper and lower medians?? I took a rough guess: (30, 15) and (145, 40) and got the answer 0.2 but the actual answer is B.. is there a more safer way to find the gradient without 'guessing' what the medians are because I don't want to take that risk in the exam.

[vce01]

yeah ive encountered that problem a couple of times, both times i got it wrong because ive had to guess the points. i dont have a way to work around it still

[RD]

Yeah I've screwed that type of question heaps of times.

[jsimmo]

Just found itutes (http://www.itute.com/mathline/exams_solutions.html) solutions for this question.

What they seemed to have done is drawn a line from the upper middle POINT to the lower middle point and then just used the 'slide 1/3rd of the way to the median point' technique.. and then used the intercept points of when the line hits the edge of the scatterplot.. I did it on my own paper and I got the points (0,8) (170,60)

(170 - 0) / (60 -

= 3.26

hmm.. so this works but it is still a bit risky because your using your ability to move the line 1/3rd of the way to the median point........ i think this is the only way to do it. hopefully nothing like this on the exam.. (unless they give us exact points)

[Mikey123]

My issue with further is bearing. I can't get the hang of them like any hints, links

[clinton_09]

My biggest killers are, three median line, bearings and especially volume ratios, i'll be sure to have many examples in my book

[plato]

The slope of the line before or after the "sliding" is the same and can be found from the slope between the two outer medians.
The "one-third slide" is a graphical approximation for the required line. The distance you must slide is often very small. Unless very accurate values are provided, any reasonable attempt to slide "about" one third should likely be sufficient in an exam.

[jsimmo]

The slope of the line before or after the "sliding" is the same and can be found from the slope between the two outer medians.
The "one-third slide" is a graphical approximation for the required line. The distance you must slide is often very small. Unless very accurate values are provided, any reasonable attempt to slide "about" one third should likely be sufficient in an exam.

yeah but the problem is finding what these two outer median values are.. by using the sliding technique you can determine a more accurate figure as to what points you should use to determine the gradient.

[Mikey123]

I need some practise exams if any of you guys have some.

[Noblesse]

I need some practise exams if any of you guys have some.

Here is a bunch provided by jessie0

http://vcenotes.com/forum/index.php/topic,6215.0.html

(just scroll down)