This isn't really surprising.I usually find myself wishing that Randall Munroe (author of XKCD) would teach most of my classes. Not only is he a polymath, but he's completely insane. Those are the best teachers.
I'd go so far as to say he's interested in a lot of stuff, but his actual level of knowledge is pretty much "dude with a Bachelor's Degree and access to wikipedia".
Nothing wrong with that, but let's not get too excited.
We show Exploits of a Mom (http://xkcd.com/327/) with the moms last speech bubble blanked out to candidates during interviews. You might be surprised how many have trouble explaining what is going on.
I didn't get it until the last bubble. I thought the son did the exploit and the mom was covering for him. I'm still unclear if that was actually the childs name or this was a one-off exploit by the mom.
I don't know how to optimize a function with an integral in R yet (unsurprising, just installed it 20 minutes ago), therefore I can't give a better answer. Now I want to learn R.
I find it odd that the test's adaptation removes the first two words from "standard creepiness rule: don't date under (age/2+7)". Is it objectionable to refer to dating with large gaps of age as creepy?
Edit: Just noticed that the Wolfram links break because of their syntax. HN then worsens the problem, by shortening the displayed links with '...', meaning copy/paste breaks! So I'm just removing the http prefix and you'll have to copy/paste to view any of the graphs.
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This seems to disagree with the gist of the comic, but I think the answer to the final part (largest dating pool) is 23 years old.
I arrived at that by doing the following, please correct me if you spot an error along the way!
The dating range for any age, t, would be defined as:
lower limit = .5t + 7
upper limit = 2t - 14
This agrees with the example in the XKCD strip. The dating range for an 18 year old is from .5(18)+7 = 16 to 2(18)-14 = 22. Because this is linear, the range will always be increasing with age.
Despite the range growing with age, we know that the proportion of singles is decreasing with age. Sigh. That's where the other model comes in. The author of this problem gives:
S(t) = e^(-0.05*t)
That looks like this: www.wolframalpha.com/input/?i=Plot[E^(-0.05+x),+{x,+85,18}]
I believe the largest dating pool would correspond with the greatest area yielded by taking the definite integral of this function from .5t+7 (the lower age limit) to 2t+14 (the upper age limit). To see this in pretty print, you can visit this link: www.wolframalpha.com/input/?i=integrate+(E^(-.05x))+dx+from+(.5x%2B7)+to+(2*x-14)
Evaluating that for any age would give the area under the curve corresponding to that age.
The next step, then, would be to find the maximum area for any age. To do this, we should be able to take the derivative of that previous equation, and set it equal to 0 in order to maximize it. Again, correct me if I'm wrong, but for the result of that, I get:
-2000(e^(-0.1t+0.7) - e^(-0.025t+0.35)) = 0
I plotted that to find that the max was located at age t = 23.
This graph illustrates the size of the dating pool corresponding to age along the x axis: www.wolframalpha.com/input/?i=Plot[-2000(e^(-0.1x%2B0.7)-e^(-0.025x%2B0.35)),+{x,+0,+100}]
Comments? Did I approach this totally wrong? Did I miss something along the way? Does that seem reasonable?
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[ 3.3 ms ] story [ 71.0 ms ] threadNothing wrong with that, but let's not get too excited.
Yeah, but that's actually pretty smart. There are a lot of people who haven't yet figured out that you can look up almost anything online....
The parent poster is saying that interviewees don't get what's going on with the exploit itself.
http://andromedayelton.com/dckx.php
Some topics clearly show up more often than others...
http://www.gocomics.com/calvinandhobbes/1995/08/23 I had to identify and analyze the survey bias that his response creates for a test in an introductory statistics class.
I don't know how to optimize a function with an integral in R yet (unsurprising, just installed it 20 minutes ago), therefore I can't give a better answer. Now I want to learn R.
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This seems to disagree with the gist of the comic, but I think the answer to the final part (largest dating pool) is 23 years old.
I arrived at that by doing the following, please correct me if you spot an error along the way!
The dating range for any age, t, would be defined as:
This agrees with the example in the XKCD strip. The dating range for an 18 year old is from .5(18)+7 = 16 to 2(18)-14 = 22. Because this is linear, the range will always be increasing with age.Despite the range growing with age, we know that the proportion of singles is decreasing with age. Sigh. That's where the other model comes in. The author of this problem gives:
That looks like this: www.wolframalpha.com/input/?i=Plot[E^(-0.05+x),+{x,+85,18}]I believe the largest dating pool would correspond with the greatest area yielded by taking the definite integral of this function from .5t+7 (the lower age limit) to 2t+14 (the upper age limit). To see this in pretty print, you can visit this link: www.wolframalpha.com/input/?i=integrate+(E^(-.05x))+dx+from+(.5x%2B7)+to+(2*x-14)
Evaluating that for any age would give the area under the curve corresponding to that age.
The next step, then, would be to find the maximum area for any age. To do this, we should be able to take the derivative of that previous equation, and set it equal to 0 in order to maximize it. Again, correct me if I'm wrong, but for the result of that, I get:
I plotted that to find that the max was located at age t = 23.This graph illustrates the size of the dating pool corresponding to age along the x axis: www.wolframalpha.com/input/?i=Plot[-2000(e^(-0.1x%2B0.7)-e^(-0.025x%2B0.35)),+{x,+0,+100}]
Comments? Did I approach this totally wrong? Did I miss something along the way? Does that seem reasonable?
Integrating, you get -20(e^(-0.05(2x-14)) - e^(-0.05(x/2+7)))
The derivative (eliminating the constant) is: (e^0.7)(-0.1)(e^(-0.1x))+(e^-0.35)(0.025)(e^(-0.025x))
Wolfram gives about 32.5 as the root. You can plot the integral (try www.coolmath.com/graphit) and mouse over to confirm.