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May 30[edit]

Apéry's constant[edit]

Is anything known about whether the following is rational or irrational??

Take Apery's constant, and find the value x in "Apery's constant equals pi cubed over x." Georgia guy (talk) 14:49, 30 May 2018 (UTC)[reply]

If x were known to be algebraic, then Apéry's constant would be transcendental, which is not known. Therefore x is not known to be algebraic, so it is not known to be rational.John Z (talk) 05:38, 31 May 2018 (UTC)[reply]

If the top 20% of a population has an average IQ of 130, how exactly do you calculate the average IQ of the entire population?[edit]

If the top 20% of a population has an average IQ of 130, how exactly do you calculate the average IQ of the entire population?

Also, for the record, I am assuming that one standard deviation is 15 IQ points. Futurist110 (talk) 22:58, 30 May 2018 (UTC)[reply]

You cannot, without some assumption about the distribution of IQ. The image at Intelligence_quotient#Current_tests seems to suggest that it follows a normal distribution, but it is unreferenced, and no quick search gave me satisfying evidence of that (either empirical evidence that the distribution of large populations fall on the Gaussian curve, or theoretical evidence that IQ scoring is made to produce normal distributions on large populations).

[EDIT: wrong answer, see comment below] If we do assume that it follows a normal distribution with mean μ (unknown, to determine) and standard deviation σ (known=15), then from the cumulative distribution function I copied from Normal distribution and elementary math, we want to find μ to solve

{\frac {1}{2}}\left[1+\operatorname {erf} \left({\frac {130-\mu }{\sigma {\sqrt {2}}}}\right)\right]=80\%\rightarrow \operatorname {erf} \left({\frac {130-\mu }{\sigma {\sqrt {2}}}}\right)=0.6\rightarrow \mu =130-\sigma {\sqrt {2}}\operatorname {erf} ^{-1}\left(0.6\right)

My mental approximations (15*sqrt(2)=20, erf^-1(0.6)=0.5) indicate a result close to 120. Tigraan^{Click here to contact me} 12:50, 31 May 2018 (UTC)[reply]

Tigraan, unless I misinterpreted what you were doing, you seem to be trying to say that 130 marks the 80th percentile, not that the top 20% has an average of 130. That will throw things off. –Deacon Vorbis (carbon • videos) 14:02, 31 May 2018 (UTC)[reply]

Indeed, I misread the question. Well, that's probably still solvable, but there already are correct answers. Tigraan^{Click here to contact me} 11:18, 1 June 2018 (UTC)[reply]

I ran a simulation of the problem and tried to find the mean through trial and error, and I got an answer around 109 (also assuming a normal distribution). How you would prove it mathematically is a tougher question, but a bit of searching around led me to truncated normal distribution, which may get you to a solution somehow. Iffy★Chat -- 13:43, 31 May 2018 (UTC)[reply]

This is some back-of-a-napkin work, so it might not be rigorous, but if we assume that the population is normal, then the right tail would follow a truncated normal distribution. Since we know the expected value of the right 20% tail, let $\alpha ={\dfrac {a-\mu }{\sigma }}=0.8$ and $\beta =\infty$ (since we are only truncating on the left). Then, $E[X]=\mu +{\dfrac {\phi (0.8)}{1-\Phi (0.8)}}(15)=130$ . Solving this (e.g., in R) yields $\mu \approx 109.49$ , which agrees with the result of ~~Deacon Vorbis's~~ Iffy's simulation above. --Kinu ^t/_c 19:17, 31 May 2018 (UTC)[reply]
I didn't actually do the simulation; that was Iffy. –Deacon Vorbis (carbon • videos) 19:35, 31 May 2018 (UTC)[reply]
I misread the signatures; thanks for the correction. --Kinu ^t/_c 01:18, 1 June 2018 (UTC)[reply]
IQ is currently defined to follow the Normal distribution with mean 100 and standard deviation 15. If we are talking about a subset then we should also probably be talking about having a smaller standard distribution which means the mean would be higher. Dmcq (talk) 22:34, 31 May 2018 (UTC)[reply]

From a normal distribution having mean value 0 and standard deviation 1, generate 10000 random numbers and sort them in descending order. The mean value of the first 2000 of these numbers is 1.4.

   (+/ % #)2000{.\:~+/(?-?)6 10000$0
1.40263

Bo Jacoby (talk) 22:45, 3 June 2018 (UTC).[reply]

If the proportion you take is p, then you're taking the average the normal distribution restricted to [a,∞) where a=Φ^-1(1-p) and Φ is the cumulative normal distribution function. According to my calculations, the mean works out to

{\frac {e^{-{\tfrac {a^{2}}{2}}}}{{\sqrt {2\pi }}p}}

, which is about 1.4 when p=.2, matching what you have. Not sure how that relates to the original question though. --RDBury (talk) 05:02, 5 June 2018 (UTC)[reply]