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Math Help - Portfolio Piece Question

  1. #1
    Newbie Danny51's Avatar
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    Question Portfolio Piece Question

    Ok, I was assigned the following question as a portfolio piece a few days back. Unfortunatly the due date was moved up an entire week due to some things my teacher has going on. Any help is greatly appreciated!

    The question :

    Based on the following model graph, who would you say is a safer driver : a 16 year old driver or a 70 year old driver? Explain.

    Driver's Age , Accidents per million miles driven
    25 , 430
    30 , 340
    40 , 220
    45 , 190
    55 , 190
    60 , 220

    (end question) EDIT : The chart came out strange, the two columns are supposed to be Age and Accidents with the respective data under each.

    I'm fairly certain that I need to find the equation used for this data, enter it into my graphing calc (TI-83), and compare the two ages. Does anyone have any ideas how I would go about doing this? Thanks!
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  2. #2
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    Hello there buddy

    I have attached a copy of the graph of the data to explain a little more what I am talking about. Basically, I think that it is unlikely that you would be asked to perform numerical curve fitting until you get to college as the process usually requires the computer number crunching or fairly technical statistical processes. However, that's not to say that you couldn't answer the question this way.
    You could, for example, do your best to actively interpolate your data to the y-axis and thus find the y-intercept and then guesstimate the a and b coefficients in the equation f(x)=ax^2+bx+c where c is the value of the y-intercept that you have found and a is positive. This is because your data looks a lot like a quadratic curve. It shouldn't take you too long to do this as your data looks fairly symmetrical around age 50.
    If you assume that the graph is symmetrical about age 50 (as it appears to be) you could then use the equation 2ax+b=0 where x=50 to find b in terms of a (in this case b=-100a. This would then leave you with the equation f(x)=ax(x-100)+c=0 and, since you have already estimated c, you would only need to vary a to obtain different values for your graph. I found that setting c at about 1200 and a at about 0.41 gave a pretty good result, though of course you will need to vary this. If you follow this method then you should then compare the fitted graph with the graph from your data and interperate from there
    However, you will need to be careful to state that the fitted model doesn't neccersarily reflect the data from real life and that you are only extrapolating from information that you have

    I hope that helped,

    Paul

    PS: If you didn't want to use the method above then you could simply do something similar to what I have done on the graph, although the method of fitting numerically will likely get you far more marks
    Attached Thumbnails Attached Thumbnails Portfolio Piece Question-untitled.jpg  
    Last edited by notyeteuler; March 6th 2008 at 05:33 AM.
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  3. #3
    Newbie Danny51's Avatar
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    Wow! Thanks very much! Do you think you could come and teach my math class sometime? (kidding, I wouldn't ask anyone to have to go through that...)

    f(x)=ax^2+bx+c is exactly what I needed (once again, thanks!!!). So many people were having problems with this question that today, my teacher went ahead and showed us that this is indeed the equation that should be used.

    I'm working on incorporating the numerical fitting method you described into a portfolio-ish fashion at this very moment. Those instruction really helped me move forward, thanks a ton for all your help!
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  4. #4
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    That's no problem at all
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