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Probability Problem (Uniform Distribution)

  1. Sep 12, 2014 #1
    1. A harried passenger will miss by several minutes the scheduled 10 A.M. departure time of his fight to New York. Nevertheless, he might still make the flight, since boarding is always allowed until 10:10 A.M., and extended boarding is sometimes permitted as long as 20 minutes after that time. Assuming that extended boarding time is uniformly distributed over the above limits, find the probability that the passenger will make his flight, assuming he arrives at the boarding gate
    (a) at 10:05; (b) at 10:15; (c) at 10:25; and (d) at 10:35




    2. The answer to part (a) and (d) are 1 and 0, right?
    The answer to part (b) and (c) should be equal because we are talking about uniform distribution here, right?
     
  2. jcsd
  3. Sep 12, 2014 #2
    The uniform distribution means that the gate has an equal chance of closing at 10:15 as it does at 10:25. However, you are still more likely to make it on the plane if you arrive at 10:15 than if you arrive at 10:25, because a closure at 10:25 still allows you to make it at 10:15, while a closure at 10:15 doesn't allow you to make it at 10:25.
     
  4. Sep 12, 2014 #3

    Ray Vickson

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    Why do you think the probabilities are the same in (b) and (c)? Don't guess---think it through, carefully. In particular, if I arrive at 10:00 + t (t in minutes, t > 10) and the extended departure is X (X in minutes, 0 ≤ X ≤ 20), for what values of X can I still get on the plane? How can you translate that criterion into a probability statement?
     
  5. Sep 12, 2014 #4
    I think the distribution function will look like this,

    f(x)=1/20 , 10<x<30
    =0 , otherwise.

    The problem is Uniform Distribution is a continuous random variable. The probability at a certain value of x (e.g. 10:15 or 10:25) is meaningless. It will however give probability over a range of values.

    Please correct me.
     
  6. Sep 12, 2014 #5

    Ray Vickson

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    The issue is NOT whether x assumes a particular value, such as x = 15; the issue is whether a person arriving at t = 15 can get on the plane! You seem to be mixing up x and t.
     
  7. Sep 12, 2014 #6
    So how am I gonna approach the parts (b) and (c).
    Many thanks for the above posts.
     
  8. Sep 12, 2014 #7

    vela

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    Your distribution f(x) is correct if you're defining x to be the number of minutes at 10:00 that the plane departs.

    For (a), you know the probability of making the plane is equal to 1 because the plane always leaves after the traveler arrives at the boarding gate. Similarly, for (d), you know the probability is 0 because the plane always leaves before the traveler arrives. Think about how these two statements might be represented on a plot of f(x). If you understand that, you can figure out (b) and (c).
     
  9. Sep 12, 2014 #8

    Ray Vickson

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    Start by drawing a picture of a number line. Pick a value of t and draw it on the line; pick a feasible value of x and draw it. Now, what does that value of x actually mean? Draw it. Now you ought to be able to see what to do; if not, I give up.
     
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