Hey all,
Some folks on another forum are arguing about this question; I think it is badly written but fairly easy to figure out, but I am no logic master; can anyone give this a look?
Decide if these events are less than equally likely, equally likely, or more than equally likely that tomorrow there will be less than 5, exactly 5, or more than 5 accidents in your town?
The was I read it is basically that it is asking if it is likely all three events have the same chance of being true, correct? If it doesn't make sense at all, please say so!
Thanks.
I don't think it is being that specific; I think it is just asking if it is less likely, equally likely, or more likely that all these events have the same chance of happening? Or am I being crazy and it is just not written properly?
Obviously they are not equally likely. You can't have "more than equally likely" because that would be some type of "more" equality than exactly 1/3 each. Thus, although "less than equally likely" doesn't really make much sense, if this means that the events all have different probability, then it is the answer.
(I'm studying for a stat test right now btw)
I don't think that you are supposed to assign percentages; I think that you are supposed to describe the relative probability of each outcome. Without putting too much thought into it, I would guess that it is not possible for each of the outcomes to be "equally likely" -- that would require a 33.33% chance of each outcome happening. As to what the heck "less than equally likely" and "more than equally likely" mean when you have 3 possible outcomes, well, you got me there. Suppose the likelihood of each event happening is, respectively, 30%, 10% and 60%. Is that less than equally likely, or more than equally likely?
Dealing with the 2nd part first, I assume the probability of accidents is dependent upon a known rate of accidents per # of cars. I will also assume that "your town" has enough cars that the probability of one car accident a day is very high. Thirdly, I will assume that the # of cars involved in an accident is <2.
Let's call:
<5 accidents = A
exactly 5 accidents = B
>5 accidents = C
Probability of A = very high.
Probability of C = obviously less than A, and again, totally dependent upon the number of cars on the street.
Probability of B = Slim (I'm not a logics/stats person, but iirc, this means that neither A nor C happened.)
So, now let's look at the first part -- it makes no sense whatsoever. So, this whole exercise is therefore moot. (or Moo, if you're a JT fan.)
I live in Broward County, FL. My answer is Greater Than 30.
Those two cities at various times were the murder capital of the US. So it is more likely that DC will have > 5 accidents. Wait, murder isn't an accident. Still, there are a lot of accidents in DC. It's an accident waiting to happen. Gary Hart, he was an accident.
The question is seriously flawed. The frequency of accidents depends on the town and the definition of accident. Whoever wrote the question is a bonehead.
But generally, there are usually more than 5 accidents in a good sized town on any given day, if you count burnt toast as an accident.
What about the accidental tourist?
Am I the only one who took this to read that it's comparing some unknown "these events" to the second half of the sentence. And the second half of the sentence, due to OR statements which cover the whole set of options, is 100%. So whatever "these events" are - they obviously cannot be "more than equally likely", since that would be a probability of > 1.0.
OK, so I'm almost certain that's not what was meant, but that's how I first read it. Sounds like a ":" is missing or something.
The classic definition of a Poisson distribution is waiting time for an event where lambda equals the probability that an event will occur in a given time frame and t is the time frame of interest. Since you are asking about more than one event per day, t cannot be one day, it's probably best to think of t as something like a minute, perhaps an hour if it's a really small town. Lambda would then be the probability that in any given minute, an accident occurs. Without knowing lambda you can't really answer the question but you can set up the equation. My memory of the exact formula for a Poisson distribution is a bit rusty right now, but you could then solve for lambda with n=5 and go from there. Anybody else remember the Poission distribution formula at the moment? Poisson distributions are the only ones where the mean equals the variance.
...but still it is obvious that EXACTLY 5 is the least likely outcome, because it requires an exact number of accidents, while the other two choices are all SETS of numbers. If you live in NYC, perhaps fewer than 5 is less likely, but if you equalize the question by considering a smaller amount of time (such as an hour, or 4 hours, or whatever), you set up essentially the same situation.
If the number of accidents, in fact, makes a Gaussian distribution (which it probably does), the likelihood of A or C would exceed B even if 5 represented the peak of the curve. Duh.
Again, the question is fatally flawed.
Reading the statement, I wonder whether you're supposed to say that each scenario (<5,=5,>5) is more/less/equally likely than the others. Obviously, all three scenarios are not equally likely to each other (sets vs. individual outcomes, where the individual outcome is very unlikely to have a 1/3 probability), which would make some more likely and some less likely.
We should make this a poll. I vote for "Equally," but I have no clue why or even what's being asked. If that's the wrong answer, then I agree with everything Fish80 said.
Rich
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My stat/probability isn't too fresh, but I don't think you can have a truly Gaussian distro since you cannot have negative number of accidents. Plus in this case since we don't have a continuous distribution, I think it would be possible to have B be the most likely outcome. Consider this:
chance you're going to have # of accident in the town:
0 - 1%
1 - 2%
2 - 3%
3 - 4%
4 - 10%
5 - 40%
6 - 20%
7+ - 10%
Resembles a Bell Curve, but A=30%, B=40%, C=30%. If this was a continuous distribution, you're right we would have B=0%.
This is nit-picking but you can't really plot a line on a discrete distribution and think of it as a continuous distribution, mostly because you can't the area under the curve to be equal to 1.0. Hopefully, you're just using the line to illustrate the characteristics of the distribution.
I didn't understand this ^
So I read this: http://en.wikipedia.org/wiki/Poisson_process
Now I understand this much: 3%
I was in a committee meeting on teaching evals last year and one of the Biologists said that teaching evals follow a Poisson distribution. From the context, I thought he was saying that most of the data is crowded into one area of the scale (4s and 5s on a 1-5 scale) so that the difference between 4.4 and 4.1 ends up not being very significant. But now I'm thinking I had no idea what he was talking about.
Last edited by throatybeard; 10-18-2007 at 04:18 PM.
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