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We have a lamp and it is in 2m distance. Its light intensity is 5. We want to know what amount of light is reaching to us. We can simply use a formula like: $ \mathrm{light} = \frac{\mathrm{intensity}}{\mathrm{distance}} $.

Now suppose that the distance is a random variable (i.e. it is in $ 2m \pm 1m $ distance, and the intensity is $ 5 \pm 4.7 $).

How we can change the formula in order to be able to use it in case we have random variables too?

P.S. Actually the question is how we can make use of probability distributions, variance, etc. to change the formula?

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    I would guess that the denominator should be squared unless something coherent is going on...2012-07-06
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    can't we work and make use of probability distributions here?2012-07-06
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    And there is no need to use **three** exclamation points! There was a time when typing an exclamation point required 3 keystrokes: you typed a vertical bar, a backspace, and then a period. That made people be somewhat parsimonious with their use. Maybe we should ask that anybody under a certain age be forced to use five or six keystrokes before typing one?2012-07-06
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    @ArturoMagidin: believe me, people of all ages are prone to this particular phenomenon!!2012-07-06
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    @Ben: And yet, it seems to infect the young more; multiple exclamation points, multiple question marks... Sigh.2012-07-06
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    Hmmm, trying to read other's handwriting was often a challenge...2012-07-06
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    This discussion is inappropriate here. The OP asks a serious question and we should try to answer it.2012-07-07
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    I would normally interpret $2 \pm 1$ as meaning that the value lies within the range $[1,3]$, but going by the rest of your question, I guess you mean that it has a certain probability distribution with mean $2$ and standard deviation $1$, is that right?2012-07-07
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    @Rahul: Yeah, considering a normal distribution with mean 2 and SD of 1 is what I meant.2012-07-09
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    Then you will have a hard time writing the result in the form $\text{mean} \pm \text{SD}$ because, as Michael Chernick's answer says, the result has infinite mean and standard deviation and does not follow a normal distribution.2012-07-10

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