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Math Help!

MunacraMunacra Registered User regular
edited November 2007 in Help / Advice Forum
So I've been searching my notes and the internet for help with this math problem, but I can't find anything to save my life. It's probably rather simple, but I keep getting stumped on a step.
Suppose an earth-destroying-asteroid is known to be heading towards the earth, and will hit here any time between 30 hours from now and 40 hours from now. Any time is as likely in this interval as any other time.

a) What type of a variable is this?

b) What is the expected value for the time we expect this asteroid to destroy the earth?

c) What is the standard deviation for the time until the earth's destruction?

d) Draw a picture of this probability density curve. Label the expected value and +/- one standard deviation from the expected value on the x axis.

e) What is the probability that the earth is destroyed 41 hours or more from now?

f? between 30 and 40 hours from now?

g) after 38 hours from now?

h) between 20 and 35 hours from now?

i) exactly 35 hours from now?

This is what I have so far.

a) It is a uniform distribution variable. (I think this is where I go wrong)
b) my notes say that to find the expected value (mean) I should put (a+b)/2 so it comes out to 35 (30+40)/2 = 35
c)Standard deviation formula for uniform variable is (b-a squared divided by the square root of 12). so 100/square root of 12 = 28.87

And that's all I got because the rest I have no idea where to start. Am I thinking about this the wrong way? This problem doesn't seem very hard, but man I have no idea what to do now.:|

Munacra on

Posts

  • TopiaTopia Registered User regular
    edited November 2007
    I'm using more logic than math here but if "Any time is as likely in this interval as any other time." wouldn't then that mean that every single time frame they give you the exact same probability?

    Topia on
  • MunacraMunacra Registered User regular
    edited November 2007
    Yes. That's why I get all confused. is the answer to f,g,h,i the same thing? If so, why would I need to know the standard deviation and graph that?

    Munacra on
  • TopiaTopia Registered User regular
    edited November 2007
    In my mind I see those other questions as both a completely different set, and a way to trick you. I got shit like this a lot in my math classes (mind you, I've never taken university courses yet) in high school, so I just assumed lots of teachers do shit like this. It tests you in more than one way? It's late, we got some genius guys on this forum, hopefully you don't need this by morn, because if you can wait till wednesday night I'm sure you'll have a much better, more concrete answer.

    Topia on
  • SpoitSpoit *twitch twitch* Registered User regular
    edited November 2007
    I'm pretty sure that f-i are asking for the chance that it'd happen during those time periods, as opposed to some other part of the probablility distribution curve, so it'd in fact not be the same.

    EDIT: that is, any individual time during that period would have the same probablility of any other individual time, but the probability over a period would be greater.

    Spoit on
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  • TopiaTopia Registered User regular
    edited November 2007
    "What is the probability that the earth is destroyed 41 hours or more from now?"

    How are chance and probability different? I could see if like, it was limiting the impact to between 30-40 hours only, but the way it is worded is that it can be at any time, at all, which is wierd.

    If it did set that specific time frame, the probability for 41 hours would be 0/9, the probability for between 30 and 40 would be 9/9 (1/1) after 38 hours would 1/9, and so on and so forth.

    The question seems to contradict itself.

    "will hit here any time between 30 hours from now and 40 hours from now. " implies a specific time-frame, however "Any time is as likely in this interval as any other time." implies that any time at all, from now till infinity, is just as likely a time when the meteor will hit..?

    As the wording "between" would limit the possible hours of impact to 31, 32, 33, 34, 35, 36, 37, 38, and 39 only.

    Edit; it's late, so I just did a quick re-read. "Any time is as likely in this interval as any other time." is just worded badly, it actually does say that any time in this interval, is just as likely as any other time, in this interval. I think. Maybe it being late is messing with my brain here and this edit is actually more wrong than my post.

    Topia on
  • RocketScienceRocketScience Registered User regular
    edited November 2007
    Munacra wrote: »
    So I've been searching my notes and the internet for help with this math problem, but I can't find anything to save my life. It's probably rather simple, but I keep getting stumped on a step.

    This is what I have so far.

    a) It is a uniform distribution variable. (I think this is where I go wrong)
    b) my notes say that to find the expected value (mean) I should put (a+b)/2 so it comes out to 35 (30+40)/2 = 35
    c)Standard deviation formula for uniform variable is (b-a squared divided by the square root of 12). so 100/square root of 12 = 28.87

    And that's all I got because the rest I have no idea where to start. Am I thinking about this the wrong way? This problem doesn't seem very hard, but man I have no idea what to do now.:|

    Maybe start taking better notes. :P

    Your answer to a) is part right, but note that it is a continuous variable, rather than discrete like Topia assumed.

    b) is OK

    c) You got the formulas for std. dev and variance of a uniformly distributed continuous random variable mixed up.
    You can have a look through here for the right formula.
    http://people.hofstra.edu/Stefan_Waner/cprob/cprob3.html

    d) The probability density curve for uniform distribution looks like this
    350px-Uniform_distribution_PDF.png

    f-i) The probability of the variable falling in a given interval is equal to the area under the probability density curve for that interval.
    Remember for i) that the area under a single point is 0.

    RocketScience on
  • MunacraMunacra Registered User regular
    edited November 2007
    Thank a lot Rocket Science, I got it all figured out now.

    Topia, sorry. I should have specified it was a continous uniform variable, not a discrete one.:p

    Munacra on
  • Marty81Marty81 Registered User regular
    edited November 2007
    Your c is wrong. The formula for the SD of a uniform dist on [a,b] (I re-derived this because I thought something was wrong - it was harder than I expected!) is (b-a)/sqrt(12), not what you had.

    Marty81 on
  • MunacraMunacra Registered User regular
    edited November 2007
    thanks man.

    Munacra on
  • Marty81Marty81 Registered User regular
    edited November 2007
    Sure. And damn, I should have read RocketScience's post before I posted :P.

    Marty81 on
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