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RichardKennaway comments on binomial variance problem - Less Wrong
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Comments (14)
Um, B, but only by a hair. 55 is equidistant between 45 and 65, but the variance is smaller for A because 65 is farther from 50 than 45 is, so measured by the relevant standard deviations, 55 is closer to 45 than 65. (Making the obviously obvious assumption that children are assigned to classes independent of gender.)
I had to google up the source to find out why the "obvious" answer is supposed to be A.
What's the name of the principle that variance increases further from 50%?
Not having memorized the formula for variance in binomial distributions, but intuiting that said principle was true, was my weaker reason for concluding B.
More saliently, the problem statement contains the gratuitous information that boys are a majority in program A. It's Kahneman and Tversky, for FSM's sake; therefore this information is used to mislead. Therefore, B.
Decreases! Note that there's zero variance when p = 0 versus non-zero variance when p = 0.5.
No principle, just the fact that the variance of the binomial distribution is p(1-p), which peaks at p=0.5.
It looks like I approached the problem in exactly the same way you did. I'm very curious as to how common it is for people to think A is more likely; it really doesn't seem obvious to me either.
75% choose program A