Timeline for Help me calculate how many people will come to my wedding! Can I attribute a percentage to each person and add them?
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| Apr 15, 2014 at 20:07 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 15, 2014 at 9:29 | comment | added | Glen_b | Actually, correction -- case D I worked out from first principles. Case C has a shortcut: it's 4 times a Bernoulli(0.5), so you just use the rule that $\text{Var}(kX) = k^2 \text{Var}(X)$. | |
| Apr 15, 2014 at 9:25 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 15, 2014 at 9:13 | comment | added | Glen_b | $\text{Var} = p(1-p)$ only applies to the Bernoulli. If you don't have a 0-1 variable, it can't be Bernoulli. So that one I simply worked out from first principles. | |
| Apr 15, 2014 at 8:29 | comment | added | Behacad | How do you arrive to variance when the mean is not a percentage? For example 2*(1-2)=-2 (for group C), and you mark the variance as 4 (and indeed variance can't be negative) | |
| Apr 15, 2014 at 8:03 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 15, 2014 at 5:54 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 14, 2014 at 11:46 | comment | added | Glen_b | Yes, it should be, as long as every individual is independent of every other individual. If there are any dependencies, you have to deal with those as discussed earlier. Once you're done with that, you can work out the chance (assuming all the previous calculations are okay) that the actual number of guests who turn up exceeds any given number -- but you must keep in mind that these numbers will be heavily dependent on the assumptions. | |
| Apr 14, 2014 at 11:41 | comment | added | Behacad | I'm not sure if I am on the right page, but here is what I did. I assigned everyone a probability in percentage and the sum is 165. I calculated variance based on p(1-p) for each person and the sum is 40, so the standard deviation is 6.3? | |
| Apr 14, 2014 at 9:31 | comment | added | Glen_b | Once you have that, you can use basic properties of variance | |
| Apr 14, 2014 at 9:28 | comment | added | Glen_b | A single person with probability of attendance $p$ has either 1 person attend (probability $p$) or 0 people attend (probability $1-p$). That doesn't have variance 0. | |
| Apr 14, 2014 at 8:01 | comment | added | Behacad | I'm confused. If its a group of 1 with a single number, wouldn't the mean be p and the variance 0? I'm not following you, sorry! | |
| Apr 14, 2014 at 5:30 | comment | added | Glen_b | If you have a single person with probability of attendance, $p$, do you know how to compute the mean and variance of the number of people attending (from that group of one)? | |
| Apr 14, 2014 at 3:48 | comment | added | Behacad | Thank you. I have a small table with percentages and amount of people with that percentage, but I don't know exactly what to do now. What means should I be adding? What variances? (100%-52, 90%-21, 80%-34, 70%-16,60%-32,50%-35,40%-25,30%-11,20%-22,10%-15,0%-9) | |
| Apr 14, 2014 at 1:14 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 14, 2014 at 0:18 | comment | added | Nick Stauner | @Behacad: If you know it's a question of all-or-none with a given group, you could just estimate the probability of the group coming as a single unit and weight the group by the number of individuals in it. I agree that margins of error would be good to include in your estimates too. | |
| Apr 13, 2014 at 23:56 | comment | added | Behacad | How could I easily take into account dependency? For example, if I know of a couple with two children, and I expect that the parents have about a 50% chance of coming. I know they will bring their children if they come. Is it save to attribute 50% to each person, and basically assume that 2 people are coming? | |
| Apr 13, 2014 at 15:36 | comment | added | Glen_b | @whuber Good point about other sources of dependence, such as weather. In some circumstances, such things can easily swamp the effects I mention. | |
| Apr 13, 2014 at 15:29 | comment | added | whuber♦ | +1 for mentioning dependencies. These arise for reasons other than interpersonal relationships, such as weather and travel conditions. Many of them induce positive correlations--which widen the range of uncertainty. If the estimates will be used to provide logistics (meals, seats, and so on), assessing the variation accurately is valuable. Although in a wedding application one can't do much more than make an educated guess, having a qualitative understanding of these statistical phenomena can lead to better guesses. | |
| Apr 13, 2014 at 15:21 | history | edited | Glen_b | CC BY-SA 3.0 |
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| Apr 13, 2014 at 8:17 | history | answered | Glen_b | CC BY-SA 3.0 |