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# Statistics help, anyone?

I want to calculate a statistic involving several different factors, but don't know the math.

This person has different characteristics I wish to consider. One occurs in about 10% of the general population, one in about 15% of the general population and one in about 7% of the general population, with no relationship between the three. In other words, one is not more likely to be found in one of the other two groups by virtue of being in the third.

In a population of 34,000,000, how many such individuals are likely to occur?

Thanks much

I presume that you learned in your class about conditional probabilities: If A, B, and C are uncorrelated then P(A&B&C) = P(A)P(B)P(C).

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Yea, that'll make sense to him/her... L

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If they were paying attention in class, or have their textbook lying around to look at, it will. If not, well....

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What's the probability of that?

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Well, it's harder to calculate because the conditions are no longer uncorrelated. I'd need to know the full 2D (or n-D?) PDF to give you a quantitative result. However, since we're dealing with classical mechanics, I can tell you it's somewhere between zero and 100%.

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I'll defer to your obviously superior knowledge of stat's (I did mechanics instead) L

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Ah, the laws of mechanics. It either works or it doesn't. Probability is expressed based on the existence of snowballs in hell.

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...or how long it takes Satan to skate to work.

L

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Seriously, folks, I'm 50 years old and have studied some physics, chemistry, economics, algebra, but never statistics. I'm not some student trying to fudge on an assignment lol

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By the way, your profile here on I'bles is essentially empty. Since none of us are telepathic, there's no way to even guess who or what you are. Obviously you could put fake data into your profile, but putting that aside, with no data at all, and a question that looks very suspiciously like freshman homework, what do we have from which to guess?

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Okay, thanks. I'm sure you've seen the vast number of Questions posted which are obviously somebody working on homework :-)

My answer below was technically correct, if obscure. What it means is that, if your three probabilities are independent of one another (as you've assumed), just multiply them together (0.10 * 0.15 * 0.07), and multiply that by your initial population.

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Once I saw all the comments, I figured that was the case. Thanks for the explanation, rather than just an answer, BTW. that's not so bad. Why did everyone always gripe about stats courses? I seem to remember a lot of sighs and eye-rolling associated with their mention...

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Because statistics is FAR more complex than this. You've set up the most trivial possible example, asking for the composition of three independent PDFs. When the conditional probabilities don't factor, it gets more complicated. Then, if you're trying to deal with a real-world situation, with sampling and all the potential biases from sampling, it gets REALLY complicated.

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It's a homework question?

Start by applying the %'s to the population figure, then give us your numbers.

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